Calculus Study Notes – Limits, Differentiation, Integration, Differential Equations | IIT JEE

📐 Mathematics · Calculus 🔥 Day 1

0/4 topics mastered

∂

Limits, Continuity & Differentiation

Standard limits · Indeterminate forms · Continuity · Derivatives · MVT

Limits — Standard Results

Limit	Value	Form
$\lim_{x\to 0} (\sin x)/x$	1	0/0 (trig)
$\lim_{x\to 0} (\tan x)/x$	1	0/0 (trig)
$\lim_{x\to 0} (1-\cos x)/x^2$	1/2	0/0
$\lim_{x\to 0} (\sin^{-1} x)/x$	1	0/0
$\lim_{x\to 0} (\tan^{-1} x)/x$	1	0/0
$\lim_{x\to 0} (e^x-1)/x$	1	0/0 (exponential)
$\lim_{x\to 0} (a^x-1)/x$	$\ln a$	0/0
$\lim_{x\to 0} \ln(1+x)/x$	1	0/0 (logarithm)
$\lim_{x\to\infty} (1+1/x)^x$	$e$	$1^\infty$
$\lim_{x\to 0} (1+x)^{1/x}$	$e$	$1^\infty$
$\lim_{x\to \infty} x^n/e^x$	0	exponential dominates
$\lim_{x\to \infty} (\ln x)/x^n$	0 ($n>0$)	power dominates log

Indeterminate Forms & Techniques

Seven Indeterminate Forms

$0/0,\ \infty/\infty,\ 0\cdot\infty,\ \infty-\infty,\ 0^0,\ \infty^0,\ 1^\infty$

All must be reduced before evaluating. The form $1^\infty$ is always indeterminate even though "1 to any power is 1" — the base and exponent both change.

L'Hôpital's Rule

If $\lim f(x)/g(x)$ gives $0/0$ or $\infty/\infty$, then:

$\lim f(x)/g(x) = \lim f'(x)/g'(x)$

Conditions: both $f$ and $g$ must be differentiable near $a$; $g'(x) \ne 0$ near $a$. May be applied repeatedly. Don't apply to non-indeterminate forms.

Sandwich (Squeeze) Theorem

If $g(x) \le f(x) \le h(x)$ near $x=a$ and $\lim g(x) = \lim h(x) = L$, then $\lim f(x) = L$.

Classic use: $\lim_{x\to 0} x^2\sin(1/x) = 0$ (squeeze between $-x^2$ and $x^2$).

$1^\infty$ Form Strategy

For $\lim [f(x)]^{g(x)}$ where $f\to 1, g\to\infty$:

$$ = e^{\lim g(x)\cdot[f(x)-1]} $$

Key identity: $\lim (1+\alpha)^{1/\alpha} = e$ as $\alpha\to 0$. So write $f(x)=1+\alpha$ where $\alpha\to 0$, then exponent $= g(x)\cdot\alpha \to$ take as the power of $e$.

Continuity

$f(x)$ is continuous at $x=a$ if: LHL = RHL = $f(a)$, i.e. $\lim_{x\to a^-} f(x) = \lim_{x\to a^+} f(x) = f(a)$

Type of discontinuity	Description	Example
Removable	LHL = RHL $\ne f(a)$ (or $f(a)$ undefined)	$f(x)=\sin x/x$ at $x=0$; fill with $f(0)=1$
Jump (non-removable)	LHL ≠ RHL (both finite)	Signum function at $x=0$; floor function at integers
Infinite (essential)	At least one limit is $\pm\infty$	$f(x)=1/x$ at $x=0$; $\tan x$ at $x=\pi/2$
Oscillatory	Limit does not exist (oscillates)	$\sin(1/x)$ at $x=0$

💡IVT (Intermediate Value Theorem): If $f$ is continuous on $[a,b]$ and $f(a)\cdot f(b)<0$, then $\exists\, c\in (a,b)$ such that $f(c)=0$. Used to prove existence of roots without finding them.

Differentiation from First Principles

$f'(x) = \lim_{h\to 0} [f(x+h) - f(x)]/h$

This definition is occasionally tested directly in JEE. Differentiability at a point requires the limit to exist and be equal from both sides. Every differentiable function is continuous, but not vice versa (e.g. $f(x)=|x|$ is continuous but not differentiable at $x=0$).

Standard Derivatives

f(x)	f'(x)	f(x)	f'(x)
$x^n$	$nx^{n-1}$	$e^x$	$e^x$
$\ln x$	$1/x$	$a^x$	$a^x \ln a$
$\sin x$	$\cos x$	$\cos x$	$-\sin x$
$\tan x$	$\sec^2 x$	$\cot x$	$-\operatorname{cosec}^2 x$
$\sec x$	$\sec x \tan x$	$\operatorname{cosec} x$	$-\operatorname{cosec} x \cot x$
$\sin^{-1} x$	$1/\sqrt{1-x^2}$	$\cos^{-1} x$	$-1/\sqrt{1-x^2}$
$\tan^{-1} x$	$1/(1+x^2)$	$\cot^{-1} x$	$-1/(1+x^2)$
$\|x\|$	$x/\|x\|\ (x\ne 0)$	$\sqrt x$	$1/(2\sqrt x)$

Differentiation Rules

Product & Quotient Rules

Product: $(uv)' = u'v + uv'$

Quotient: $\left(\frac{u}{v}\right)' = (u'v - uv')/v^{2}$

Mnemonic for quotient: "low d-high minus high d-low, over low squared."

Chain Rule

$\dfrac{d}{dx}[f(g(x))] = f'(g(x))\cdot g'(x)$

For three-layer: $\dfrac{d}{dx}[f(g(h(x)))] = f'(g(h(x)))\cdot g'(h(x))\cdot h'(x)$

Key: differentiate "outside-in," multiplying each inner derivative.

Implicit Differentiation

For $F(x,y)=0$, differentiate both sides w.r.t. $x$, treating $y$ as a function of $x$:

$$ \frac{d}{dx}[y^{2}] = 2y\cdot\frac{dy}{dx} $$

Then solve for $dy/dx$ algebraically. Used for circles, ellipses, folium of Descartes, etc.

Logarithmic Differentiation

For $y = [f(x)]^{g(x)}$ (variable base AND exponent):

$\ln y = g(x)\cdot\ln f(x) \to$ differentiate both sides $\to (1/y)\cdot y' = g'\ln f + g\cdot f'/f$

Also use for products/quotients of many functions to simplify computation.

Parametric Differentiation

If $x=f(t), y=g(t)$, then:

$dy/dx = (dy/dt)/(dx/dt) = g'(t)/f'(t)$

$$ \frac{d^{2}y}{dx^{2}} = \frac{d}{dx}\!\left(\frac{dy}{dx}\right) = \frac{d/dt\,(dy/dx)}{dx/dt} $$

Higher Derivatives

$y'' = d^2 y/dx^2$: second derivative (rate of change of slope → concavity)

Key results: if $y=e^x, y^{(n)}=e^x$; if $y=\sin x, y^{(n)}=\sin(x+n\pi/2)$; if $y=x^n, y^{(n)}=n!$ and $y^{(n+1)}=0$.

Leibniz rule: $(uv)^{(n)} = \sum C(n,r)\,u^{(n-r)}\,v^{(r)}$

Mean Value Theorems

Rolle's Theorem

Conditions: $f$ continuous on $[a,b]$, differentiable on $(a,b)$, $f(a) = f(b)$

Conclusion: $\exists\, c\in (a,b)$ such that $f'(c) = 0$

Geometric meaning: there exists a horizontal tangent between two points at the same height.

Lagrange's MVT

Conditions: $f$ continuous on $[a,b]$, differentiable on $(a,b)$

Conclusion: $\exists\, c\in (a,b)$ such that $f'(c) = [f(b)-f(a)]/(b-a)$

Geometric meaning: the instantaneous slope at $c$ equals the average slope of the chord. Rolle's theorem is the special case $f(a)=f(b)$.

🔢

Limits, Continuity & Differentiation — Key Formulas

Core Standard Limits

$\lim_{x\to 0} \sin x/x = 1$ | $\lim_{x\to 0} \tan x/x = 1$ | $\lim_{x\to 0} (1-\cos x)/x^2 = 1/2$
$\lim_{x\to 0} (e^x-1)/x = 1$ | $\lim_{x\to 0} (a^x-1)/x = \ln a$ | $\lim_{x\to 0} \ln(1+x)/x = 1$
$\lim_{x\to \infty} (1+1/x)^x = e$ | $\lim_{x\to 0} (1+x)^{1/x} = e$

L'Hôpital's Rule

If $\lim f(x)/g(x) = 0/0$ or $\infty/\infty$, then $\lim f(x)/g(x) = \lim f'(x)/g'(x)$

Apply only to $0/0$ or $\infty/\infty$ forms. Convert $0\cdot\infty \to 0/(1/\infty)$ or $\infty/(1/0)$ first. For $1^\infty$: take log first.

$1^\infty$ Form

$\lim [f(x)]^{g(x)} = e^{\lim g(x)\cdot[f(x)-1]}$ when $f\to 1, g\to\infty$

Equivalent to: take log, apply $\lim g\cdot\ln f$, use $\lim g\cdot(f-1) \approx \lim g\cdot\ln f$ when $f\to 1$.

First Principles (Definition of Derivative)

$f'(x) = \lim_{h\to 0} [f(x+h) - f(x)] / h$

Product, Quotient & Chain Rules

$(uv)' = u'v + uv'$
$(u/v)' = (u'v - uv')/v^2$
$\dfrac{d}{dx}[f(g(x))] = f'(g(x))\cdot g'(x)$

Parametric & Implicit Derivatives

Parametric: $dy/dx = (dy/dt)/(dx/dt)$
Implicit $F(x,y)=0$: $dy/dx = -(\partial F/\partial x)/(\partial F/\partial y)$

Logarithmic Differentiation (variable exponent)

$y = [f(x)]^{g(x)} \to \ln y = g(x)\cdot \ln f(x)$
$\to y'/y = g'\cdot\ln f + g\cdot f'/f \to y' = y\cdot(g'\cdot\ln f + g\cdot f'/f)$

Lagrange's Mean Value Theorem

$\exists\, c\in (a,b)$ such that $f'(c) = [f(b)-f(a)]/(b-a)$

Rolle's theorem: special case where $f(a)=f(b)$, giving $f'(c)=0$.

nth Derivative of Standard Functions

$D^n(e^x) = e^x$ | $D^n(\sin x) = \sin(x+n\pi/2)$ | $D^n(\cos x) = \cos(x+n\pi/2)$
$D^n(x^m) = n!$ if $m=n$; $= 0$ if $m<n$ | $D^n(\ln x) = (-1)^{n-1}(n-1)!/x^n$

🧮

Limits, Continuity & Differentiation — Solved Examples

Evaluate $\lim_{x\to 0} (\sin 3x)/(\sin 5x)$

Multiply and divide to create the standard form $\sin\theta/\theta = 1$:
$= \lim_{x\to 0} [(\sin 3x)/(3x)]\cdot (3x) / \{[(\sin 5x)/(5x)]\cdot (5x)\}$
$= [1\cdot 3x] / [1\cdot 5x]$ (as $x\to 0$ both bracketed parts $\to 1$)
= 3/5

Check continuity of $f(x) = \begin{cases} x^2+1 & x\le 1 \\ 3x-1 & x>1 \end{cases}$ at $x=1$

LHL: $\lim_{x\to 1^-} (x^2+1) = 1+1 = 2$
RHL: $\lim_{x\to 1^+} (3x-1) = 3-1 = 2$
$f(1)$: $= 1^2+1 = 2$
LHL = RHL = $f(1) = 2$ → f is continuous at $x=1$. ✓

Differentiate $y = \sin(x^2+1)$ using the chain rule

Outer function: $\sin(u)$, inner: $u = x^2+1$
$dy/dx = \cos(u) \cdot du/dx = \cos(x^{2}+1) \cdot 2x =$ $2x\cos(x^2+1)$

Evaluate $\lim_{x\to0} (e^x - e^{\sin x}) / (x - \sin x)$

Form: Put $x=0$: numerator $= e^0-e^0=0$; denominator $=0-0=0 \to 0/0$. Apply L'Hôpital.
Numerator differentiated: $e^x - e^{\sin x}\cdot\cos x$
Denominator differentiated: $1 - \cos x$
At $x=0$: num $= 1 - 1\cdot 1 = 0$; den $= 1 - 1 = 0$. Still $0/0$. Apply again.
Num'': $e^x - [e^{\sin x}\cdot \cos^{2}x + e^{\sin x}\cdot(-\sin x)] = e^x - e^{\sin x}(\cos^{2}x - \sin x)$
Den'': $\sin x$
At $x=0$: num $= 1 - 1\cdot(1-0) = 0$; den $= 0$. Still $0/0$ — apply once more.
Alternative: Write $e^x - e^{\sin x} = e^{\sin x}(e^{x-\sin x}-1) \approx e^{\sin x}\cdot(x-\sin x)$ for small $(x-\sin x)$. So limit $\approx e^{\sin x} \to$ $e^0 = 1$ as $x\to 0$.

Find $dy/dx$ if $x^3 + y^3 = 3axy$ (folium of Descartes)

Differentiate both sides with respect to $x$:
$$ 3x^{2} + 3y^{2}\frac{dy}{dx} = 3a\!\left[y + x\frac{dy}{dx}\right] $$
$$ 3y^{2}\frac{dy}{dx} - 3ax\frac{dy}{dx} = 3ay - 3x^{2} $$
$$ \frac{dy}{dx}(3y^{2} - 3ax) = 3ay - 3x^{2} $$
$dy/dx = (ay - x^{2}) / (y^{2} - ax) =$ $(ay-x^2)/(y^2-ax)$

Differentiate $y = x^{\sin x}$ and find the value at $x = \pi/2$

Variable base AND exponent → logarithmic differentiation:
$\ln y = \sin x \cdot \ln x$
Differentiate both sides w.r.t. $x$:
$(1/y) \cdot dy/dx = \cos x \cdot \ln x + \sin x \cdot (1/x)$
$$ dy/dx = y \cdot [\cos x \cdot \ln x + (\sin x)/x] = x^{\sin x} \cdot [\cos x \cdot \ln x + (\sin x)/x] $$
At $x = \pi/2$:
$$ y = (\pi/2)^{\sin \pi/2} = (\pi/2)^1 = \pi/2 $$
$dy/dx = (\pi/2) \cdot [\cos(\pi/2) \cdot \ln(\pi/2) + \sin(\pi/2)/(\pi/2)]$
$= (\pi/2) \cdot [0 \cdot \ln(\pi/2) + 1\cdot(2/\pi)] = (\pi/2) \cdot (2/\pi) =$ 1

📊

Limits, Continuity & Differentiation — Key Diagrams

Click any diagram to zoom

$\lim_{x\to 0} (\sin x)/x = 1$ — Geometric Proof

Geometric proof: the area of triangle OAP ≤ area of sector OAP ≤ area of triangle OAT gives $\sin x \le x \le \tan x$ for $0<x<\pi/2$. Dividing by $\sin x$: $1 \le x/\sin x \le 1/\cos x$. As $x\to 0$, $1/\cos x\to 1$, so by Sandwich theorem, $\sin x/x\to 1$.

Types of Discontinuity

Four types of discontinuity. Removable: fillable by redefining $f(a)$. Jump: sudden step (signum, floor functions). Infinite: vertical asymptote ($1/x, \tan x$). Oscillatory: $\sin(1/x)$ at origin — oscillates infinitely fast, no limit exists.

Derivative as Slope of Tangent & Mean Value Theorem

Left: the derivative $f'(x_0)$ is the slope of the tangent to the curve at point P — the limit of the secant slope as $h\to 0$. Right: Lagrange's MVT guarantees at least one point $c$ where the tangent (red) is parallel to the chord AB (grey dashed), i.e. instantaneous rate = average rate.

⚡

Limits, Continuity & Differentiation — Quick Quiz

Question 1

$\lim_{x\to 0} (1 - \cos x)/x^2$ equals:

Using the identity $1-\cos x = 2\sin^2(x/2)$: $(1-\cos x)/x^2 = 2\sin^2(x/2)/x^2 = (1/2)\cdot[\sin(x/2)/(x/2)]^2$. As $x\to 0$, the bracket $\to 1$, so limit $= (1/2)\cdot 1 =$ 1/2. Alternatively L'Hôpital twice: $\sin x/(2x) \to \cos x/2 = 1/2$.

Question 2

The function $f(x) = |x-1|$ is:

$f(x)=|x-1|$ is continuous everywhere (no breaks). At $x=1$: LHD $= \lim_{h\to 0^-}(|h|/h) = -1$; RHD $= \lim_{h\to 0^+}(h/h) = +1$. LHD ≠ RHD → not differentiable at $x=1$. A sharp corner (kink) is always a point of non-differentiability.

Question 3

If $y = \tan^{-1}(x^2)$, then $dy/dx =$

Chain rule: $\dfrac{d}{dx}[\tan^{-1}(u)] = \dfrac{1}{1+u^2}\cdot du/dx$. Here $u=x^2, du/dx=2x$. So $dy/dx = 1/(1+(x^2)^2) \cdot 2x =$ $2x/(1+x^4)$. Common error: forgetting to apply the chain rule and leaving $2x$ out.

Question 4

Rolle's theorem applied to $f(x) = x(x-1)(x-2)$ on $[0,2]$ guarantees $f'(c)=0$ for some $c$ in:

$f(0)=0, f(1)=0, f(2)=0$ — so $f(0)=f(2)$. $f$ is a polynomial (continuous + differentiable everywhere). Rolle's theorem applies on $[0,2]$ and guarantees at least one $c\in$ $(0,2)$ with $f'(c)=0$. In fact $f'(x)=3x^2-6x+2=0$ gives $c=(6\pm\sqrt{12})/6=1\pm 1/\sqrt 3$, both in $(0,2)$.

📉

Applications of Derivatives

Tangents & normals · Monotonicity · Maxima/minima · Concavity · Approximations

Tangent and Normal to a Curve

Tangent at $(x_1, y_1)$

Slope of tangent $= f'(x_1) = m_T$

Equation: $y - y_{1} = f'(x_{1})(x - x_{1})$

If $f'(x_1) = 0$: tangent is horizontal ($y = y_1$). If $f'(x_1) \to \infty$: tangent is vertical ($x = x_1$).

Normal at $(x_1, y_1)$

Slope of normal $= -1/f'(x_1) = m_N$ ($m_T \cdot m_N = -1$)

Equation: $y - y_{1} = -[1/f'(x_{1})](x - x_{1})$

Length of tangent, normal, subtangent, subnormal: standard results from coordinate geometry of curves.

Angle Between Two Curves

Curves intersect at a point where slopes are $m_1$ and $m_2$.

$$ \tan\phi = \left|\frac{m_{1} - m_{2}}{1 + m_{1}m_{2}}\right| $$

Orthogonal curves: $m_1 m_2 = -1$ (tangents perpendicular). Common JEE question: find $k$ so that $y=x^2$ and $y=k e^x$ are orthogonal.

Monotonicity

On an interval I:

Condition on f'	Behaviour	Example
$f'(x) > 0\ \forall x \in I$	Strictly increasing	$e^x$ on $\mathbb{R}$; $x^3$ on $\mathbb{R}$
$f'(x) < 0\ \forall x \in I$	Strictly decreasing	$e^{-x}$; $\cos x$ on $(0, \pi)$
$f'(x) = 0\ \forall x \in I$	Constant	$f(x) = c$
$f'(x) \ge 0$ (= 0 at isolated pts)	Monotonically increasing (non-strict)	$x^3$ at $x=0$: $f'=0$ but still increasing

💡For JEE: "monotonically increasing" usually means $f'(x) \ge 0$ with equality only at isolated points (not on any sub-interval). A function can be increasing even if $f'=0$ at a single point — $x^3$ is the standard example.

Critical Points and Extrema

A critical point is where $f'(c) = 0$ or $f'(c)$ is undefined. Every local extremum is a critical point, but not every critical point is an extremum.

First Derivative Test

Examine the sign of $f'$ on both sides of critical point $c$:

+ to − at $c$ → local maximum
− to + at $c$ → local minimum
No sign change → neither (inflection point)

Works for all cases including where $f''$ doesn't exist.

Second Derivative Test

If $f'(c) = 0$:

$f''(c) < 0$ → local maximum (concave down)
$f''(c) > 0$ → local minimum (concave up)
$f''(c) = 0$ → inconclusive (use first derivative test)

Faster but fails when $f''(c) = 0$. Classic fail: $f(x) = x^4$ at $c=0$ — minimum but $f''(0)=0$.

Absolute Extrema on $[a, b]$

Closed interval method:

Find all critical points in $(a, b)$
Evaluate $f$ at critical points AND at endpoints $a, b$
Largest value = absolute max; smallest = absolute min

Unlike local extrema, absolute extrema may occur at endpoints.

Concavity and Inflection Points

Concavity Rules

$f''(x) > 0$ on $I$ → concave up (curve bends upward, like a cup ∪)

$f''(x) < 0$ on $I$ → concave down (curve bends downward, like a cap ∩)

A local minimum lies on a concave-up region; a local maximum on concave-down.

Inflection Points

Point where concavity changes ($f''$ changes sign). Necessary: $f''(c) = 0$, but not sufficient (e.g. $f(x)=x^4$ at 0 has $f''=0$ but no inflection).

Procedure: solve $f''(x) = 0$; check sign change of $f''$ across solution; if sign changes, it's an inflection point.

Approximations (Linear / Differential)

For small $\Delta x$: $f(x + \Delta x) \approx f(x) + f'(x)\cdot\Delta x$

Written as: $\Delta y \approx dy = f'(x)\cdot dx$. Used to estimate $\sqrt{26} \approx \sqrt{25} + (1/(2\sqrt{25}))\cdot 1 = 5.1$, etc.

Related Rates

When two quantities $x(t)$ and $y(t)$ are related by an equation, differentiate both sides w.r.t. $t$:

Example: sphere volume $V=(4/3)\pi r^3 \to dV/dt = 4\pi r^2\cdot(dr/dt)$. If $dV/dt$ is given, find $dr/dt$ at a given $r$.

💡Optimisation shortcut: For symmetric expressions — if $x+y=k$ (constant), $xy$ is maximised when $x=y=k/2$. If $xy=k$, $x+y$ is minimised when $x=y=\sqrt k$. These follow from AM-GM and are often faster than calculus for JEE problems.

🔢

Applications of Derivatives — Key Formulas

Tangent and Normal Equations

Tangent: $y - y_{1} = f'(x_{1})(x - x_{1})$
Normal: $y - y_{1} = -[1/f'(x_{1})](x - x_{1})$
$m_T \cdot m_N = -1$

Monotonicity Conditions

$f'(x) > 0$ on $I$ → strictly increasing
$f'(x) < 0$ on $I$ → strictly decreasing
$f'(x) \ge 0$ (= 0 at isolated points only) → monotonically increasing

First and Second Derivative Tests

1st: $f'(c)=0$; sign + to − at $c$ → local max; − to + → local min
2nd: $f'(c)=0$ and $f''(c)<0$ → max; $f''(c)>0$ → min; $f''(c)=0$ → inconclusive

Concavity and Inflection

$f''(x) > 0$ → concave up ∪; $f''(x) < 0$ → concave down ∩
Inflection point: $f''(c) = 0$ AND sign of $f''$ changes at $c$

Linear Approximation

$f(x + \Delta x) \approx f(x) + f'(x)\cdot\Delta x$
$dy = f'(x)\cdot dx$

Angle Between Curves

$\tan\phi = \left|\dfrac{m_1 - m_2}{1 + m_1 m_2}\right|$
Orthogonal intersection: $m_1\cdot m_2 = -1$

🧮

Applications of Derivatives — Solved Examples

Find the equation of tangent to $y = x^3 - 3x$ at $x = 2$

At $x=2$: $y = 8-6 = 2$. Point: $(2, 2)$.
$f'(x) = 3x^2-3$; $f'(2) = 12-3 = 9$.
Tangent: $y-2 = 9(x-2) \to$ $y = 9x - 16$
Normal: slope $= -1/9$; $y-2 = -(1/9)(x-2) \to$ $9y + x = 20$

Find intervals where $f(x) = x^3 - 6x^2 + 9x + 1$ is increasing/decreasing

$$ f'(x) = 3x^{2}-12x+9 = 3(x^{2}-4x+3) = 3(x-1)(x-3) $$
Sign chart: $f'>0$ for $x<1$ (both factors $-\cdot-=+$), $f'<0$ for $1<x<3$ ($-\cdot+=-$), $f'>0$ for $x>3$ ($+\cdot+=+$).
Increasing on $(-\infty, 1) \cup (3, \infty)$; Decreasing on $(1, 3)$.
Local max at $x=1$: $f(1)=1-6+9+1=5$. Local min at $x=3$: $f(3)=27-54+27+1=1$.

A ladder 10 m long leans against a wall. Its base slides away at 2 m/s. Find the rate at which the top slides down when the base is 6 m from the wall.

$x^2 + y^2 = 100$. Differentiate w.r.t. $t$: $2x\cdot(dx/dt) + 2y\cdot(dy/dt) = 0$
When $x=6$: $y = \sqrt{100-36} = 8$. $dx/dt = 2$ m/s.
$2(6)(2) + 2(8)(dy/dt) = 0 \to dy/dt = -24/16 =$ $-3/2$ m/s
The top slides down at 1.5 m/s (negative = downward).

Find the maximum area of a rectangle inscribed in a semicircle of radius r

Let the rectangle have half-width $x$. Height $= \sqrt{r^2-x^2}$.
Area $A = 2x\cdot\sqrt{r^2-x^2}$. To maximise, set $dA/dx = 0$.
$$ dA/dx = 2\sqrt{r^{2}-x^{2}} + 2x\cdot\left(\frac{-x}{\sqrt{r^{2}-x^{2}}}\right) = \frac{2(r^{2}-2x^{2})}{\sqrt{r^{2}-x^{2}}} $$
Setting $= 0$: $r^2 - 2x^2 = 0 \to x = r/\sqrt 2$.
$A_{max} = 2\cdot(r/\sqrt2)\cdot\sqrt{r^2-r^2/2} = 2\cdot(r/\sqrt2)\cdot(r/\sqrt2) =$ $r^2$

Find all inflection points of $f(x) = x^4 - 4x^3 + 6x^2 - 4x + 1$

Note: $f(x) = (x-1)^4$. So $f'(x) = 4(x-1)^3, f''(x) = 12(x-1)^2$.
$f''(x) = 0$ at $x=1$. But $12(x-1)^2 \ge 0$ for all $x$ — it is never negative.
So $f''$ does NOT change sign at $x=1$. Therefore $x=1$ is NOT an inflection point.
This is a critical example: $f''=0$ is necessary but not sufficient for inflection. $x=1$ is actually the global minimum ($f(1)=0$).

📊

Applications of Derivatives — Key Diagrams

Click any diagram to zoom

Function Behaviour Summary: f, f', f'' Relationship

A cubic $f(x)$ showing: local maximum (red dot) where $f'=0$ and $f''<0$ (concave down); local minimum (green dot) where $f'=0$ and $f''>0$ (concave up); inflection point (orange dot) where $f''=0$ changes sign. Green arrows ↑ = increasing ($f'>0$); red ↓ = decreasing ($f'<0$).

Maxima/Minima Decision Flowchart

Decision flowchart for finding local extrema. Start: find $f'(c)=0$. Try second derivative test — it gives the answer immediately if $f''(c)\ne 0$. If $f''(c)=0$, fall back to first derivative test (sign analysis). All three outcomes (max/min/neither) are possible at an inconclusive second derivative.

⚡

Applications of Derivatives — Quick Quiz

Question 1

The slope of the normal to $y = x^2$ at the point $(1, 1)$ is:

$f'(x)=2x$; $f'(1)=2$. Slope of tangent $= 2$. Slope of normal $= -1/m_T =$ $-1/2$. The product of tangent and normal slopes always equals $-1$.

Question 2

$f(x) = 2x^3 - 9x^2 + 12x + 5$ has a local minimum at $x =$

$f'(x) = 6x^2-18x+12 = 6(x-1)(x-2) = 0$ at $x=1$ and $x=2$. $f''(x)=12x-18$. $f''(1)=-6<0$ → local max at $x=1$. $f''(2)=6>0$ → local minimum at $x=2$.

Question 3

The function $f(x) = x\cdot e^{-x}$ is increasing on:

$f'(x) = e^{-x} + x\cdot(-e^{-x}) = e^{-x}(1-x)$. Since $e^{-x}>0$ always, sign of $f'$ = sign of $(1-x)$. $f'>0$ when $1-x>0$, i.e. $x<1$. So $f$ is increasing on $(-\infty, 1)$.

∫

Integration & Applications

Indefinite & definite integrals · Techniques · Area · Volume of revolution

Standard Integrals

∫ f(x) dx	Result (+C)	∫ f(x) dx	Result (+C)
$\int x^n\, dx$	$x^{n+1}/(n+1)$	$\int 1/x\, dx$	$\ln\|x\|$
$\int e^x\, dx$	$e^x$	$\int a^x\, dx$	$a^x/\ln a$
$\int \sin x\, dx$	$-\cos x$	$\int \cos x\, dx$	$\sin x$
$\int \tan x\, dx$	$\ln\|\sec x\|$	$\int \cot x\, dx$	$\ln\|\sin x\|$
$\int \sec^2 x\, dx$	$\tan x$	$\int \operatorname{cosec}^2 x\, dx$	$-\cot x$
$\int \sec x\tan x\, dx$	$\sec x$	$\int \operatorname{cosec} x\cot x\, dx$	$-\operatorname{cosec} x$
$\int 1/\sqrt{1-x^2}\, dx$	$\sin^{-1} x$	$\int 1/(1+x^2)\, dx$	$\tan^{-1} x$
$\int 1/\sqrt{a^2-x^2}\, dx$	$\sin^{-1}(x/a)$	$\int 1/(a^2+x^2)\, dx$	$(1/a)\tan^{-1}(x/a)$
$\int 1/\sqrt{x^2-a^2}\, dx$	$\ln\|x+\sqrt{x^2-a^2}\|$	$\int 1/\sqrt{x^2+a^2}\, dx$	$\ln\|x+\sqrt{x^2+a^2}\|$

Integration Techniques

Substitution (u-sub)

If integrand has $f(g(x))\cdot g'(x)$, put $u = g(x), du = g'(x)\,dx$.

Example: $\int 2x\cdot \cos(x^2)\, dx$ — put $u=x^2, du=2x\, dx \to \int \cos u\, du = \sin u + C =$ $\sin(x^2) + C$

For definite integrals: also change limits to $u$-values.

Integration by Parts (ILATE)

$$ \int u\cdot v\, dx = u\cdot\int v\, dx - \int\left[u'\cdot\int v\, dx\right] dx $$

Choose $u$ first by ILATE: Inverse trig > Log > Algebraic > Trig > Exponential.

Special: $\int e^x[f(x)+f'(x)]\,dx = e^x f(x)+C$ — very common in JEE.

Partial Fractions

For rational $P(x)/Q(x)$ with $\deg P < \deg Q$, factor $Q(x)$ then split:

Linear $(x-a)$: $A/(x-a)$
Repeated $(x-a)^2$: $A/(x-a) + B/(x-a)^2$
Irreducible quadratic $(x^2+bx+c)$: $(Ax+B)/(x^2+bx+c)$

Trig Substitutions

Expression	Substitute
$\sqrt{a^2-x^2}$	$x = a\sin\theta$
$\sqrt{a^2+x^2}$	$x = a\tan\theta$
$\sqrt{x^2-a^2}$	$x = a\sec\theta$

Definite Integral Properties

Property	Formula	Use case
Reversal	$\int_a^b f = -\int_b^a f$	Swap limits changes sign
Additivity	$\int_a^c f = \int_a^b f + \int_b^c f$	Split at any point $b$
King's property	$\int_a^b f(x)\, dx = \int_a^b f(a+b-x)\, dx$	Most useful; eliminates x from numerator
Even function	$\int_{-a}^a f(x)\, dx = 2\int_0^a f(x)\, dx$ if $f(-x)=f(x)$	Symmetric limits + even → double half
Odd function	$\int_{-a}^a f(x)\, dx = 0$ if $f(-x)=-f(x)$	Symmetric limits + odd → zero
Period	$\int_0^{nT} f(x)\, dx = n\cdot\int_0^T f(x)\, dx$	Periodic $f$ with period $T$
Half-interval	$\int_0^{2a} f(x)\, dx = 2\int_0^a f(x)\, dx$ if $f(2a-x)=f(x)$; $= 0$ if $f(2a-x)=-f(x)$	Common for sin/cos integrals on $[0, \pi]$

💡King's property example: $I = \int_0^\pi x\cdot\sin x/(1+\cos^2 x)\, dx$. Let $x\to\pi-x$: $I = \int_0^\pi (\pi-x)\sin x/(1+\cos^2 x)\, dx$. Add: $2I = \pi\int_0^\pi \sin x/(1+\cos^2 x)\, dx = \pi\cdot[-\tan^{-1}(\cos x)]_0^\pi = \pi\cdot(\pi/4+\pi/4) = \pi^2/2$. So $I = \pi^2/4$.

Area Between Curves

Area bounded by $y=f(x), y=g(x), x=a, x=b$ ($f(x) \ge g(x)$ on $[a, b]$):

$A = \int_a^b [f(x) - g(x)]\, dx$

If curves intersect, split the interval at intersection points and take $|f-g|$ in each sub-interval.

Volume of Revolution

Disk method (about x-axis): $V = \pi\int_a^b [f(x)]^2\, dx$

Washer method (region between two curves): $V = \pi\int_a^b \{[f(x)]^2 - [g(x)]^2\}\, dx$

About y-axis (shell method): $V = 2\pi\int_a^b x\cdot f(x)\, dx$

Wallis Formula (useful for JEE)

$$ \int_{0}^{\pi/2} \sin^{n}x\, dx = \int_{0}^{\pi/2} \cos^{n}x\, dx $$

$n$	$\int_0^{\pi/2} \sin^n x\, dx$
odd ($n=2m+1$)	$[2\cdot 4\cdot 6\cdots(2m)] / [1\cdot 3\cdot 5\cdots(2m+1)]$
even ($n=2m$)	$[1\cdot 3\cdot 5\cdots(2m-1)] / [2\cdot 4\cdot 6\cdots(2m)] \cdot \pi/2$

Key values: $\int_0^{\pi/2} \sin^2 x\, dx = \pi/4$; $\int_0^{\pi/2} \sin^3 x\, dx = 2/3$; $\int_0^{\pi/2} \sin^4 x\, dx = 3\pi/16$.

🔢

Integration & Applications — Key Formulas

Integration by Parts

$\int u\cdot v\, dx = u\cdot\int v\, dx - \int[u'\cdot(\int v\, dx)]\, dx$
ILATE: Inverse trig > Log > Algebraic > Trig > Exponential
$\int e^x[f(x)+f'(x)]\, dx = e^x\cdot f(x) + C$

King's Property

$\int_a^b f(x)\, dx = \int_a^b f(a+b-x)\, dx$
Special ($a=0$): $\int_0^a f(x)\, dx = \int_0^a f(a-x)\, dx$

Replace $x$ with $(a+b-x)$. Used to simplify $\int_0^\pi x\cdot f(\sin x)\, dx \to (\pi/2)\int_0^\pi f(\sin x)\, dx$.

Even/Odd and Half-interval Properties

Even: $\int_{-a}^a f\, dx = 2\int_0^a f\, dx$ | Odd: $\int_{-a}^a f\, dx = 0$
$\int_0^{2a} f\, dx = 2\int_0^a f\, dx$ (if $f(2a-x)=f(x)$) or $0$ (if $f(2a-x)=-f(x)$)

Area and Volume Formulas

Area $= \int_a^b |f(x)-g(x)|\, dx$
Volume (disk, x-axis): $\pi\int_a^b [f(x)]^2\, dx$
Volume (shell, y-axis): $2\pi\int_a^b x\cdot f(x)\, dx$

Key Definite Integral Results

$\int_0^{\pi/2} \ln(\sin x)\, dx = -(\pi/2)\ln 2$
$\int_0^{\pi/2} \sin^2 x\, dx = \pi/4$ | $\int_0^{\pi/2} \sin^3 x\, dx = 2/3$
$\int_0^\infty e^{-x^2}\, dx = \sqrt\pi/2$ (Gaussian)

Important Indefinite Integrals

$\int \ln x\, dx = x\cdot\ln x - x + C$
$\int 1/(x^2-a^2)\, dx = (1/(2a))\cdot\ln|(x-a)/(x+a)| + C$
$\int \sqrt{a^2-x^2}\, dx = (x/2)\sqrt{a^2-x^2} + (a^2/2)\sin^{-1}(x/a) + C$

🧮

Integration & Applications — Solved Examples

Evaluate $\int x\cdot e^x\, dx$

Integration by parts. ILATE: $u = x$ (Algebraic), $dv = e^x\, dx$ (Exponential).
$du = dx, v = e^x$
$\int x\cdot e^x\, dx = x\cdot e^x - \int e^x\, dx = x\cdot e^x - e^x + C =$ $e^x(x - 1) + C$

Evaluate $\int_0^{\pi/2} \sin x/(\sin x + \cos x)\, dx$

Let $I = \int_0^{\pi/2} \sin x/(\sin x+\cos x)\, dx$. Apply King's property ($a=\pi/2$, replace $x$ with $\pi/2-x$):
$$ I = \int_0^{\pi/2} \cos x/(\cos x+\sin x)\, dx $$
Add: $2I = \int_0^{\pi/2} (\sin x+\cos x)/(\sin x+\cos x)\, dx = \int_0^{\pi/2} 1\, dx = \pi/2$
∴ $I =$ $\pi/4$

Find the area enclosed by $y = x^2$ and $y = x + 2$

Intersection: $x^2 = x+2 \to x^2-x-2=0 \to (x-2)(x+1)=0 \to x=-1, x=2$.
On $[-1, 2]$: $y=x+2$ is above $y=x^2$ (check at $x=0$: $2>0$).
$A = \int_{-1}^2 [(x+2)-x^2]\, dx = [x^2/2 + 2x - x^3/3]_{-1}^2$
$= (2+4-8/3) - (1/2-2+1/3) = (18/3-8/3) - (3/6-12/6+2/6)$
$= 10/3 - (-7/6) = 10/3 + 7/6 = 20/6 + 7/6 =$ $27/6 = 9/2$

Evaluate $\int e^x\cdot(\sin x + \cos x)\, dx$

Recognise the special form $\int e^x[f(x)+f'(x)]\, dx = e^x f(x)+C$.
Here $f(x) = \sin x$ (since $f'(x) = \cos x$).
$\therefore \int e^x(\sin x+\cos x)\, dx =$ $e^x\cdot\sin x + C$
Verify: $\dfrac{d}{dx}[e^x\sin x] = e^x\sin x + e^x\cos x = e^x(\sin x+\cos x)$ ✓

Evaluate $\int_0^\pi x\cdot\sin x/(1 + \cos^2 x)\, dx$

Let $I = \int_0^\pi x\cdot\sin x/(1+\cos^2 x)\, dx$. King's property with $[0, \pi]$: replace $x\to\pi-x$.
$\sin(\pi-x) = \sin x, \cos(\pi-x) = -\cos x$, so $\cos^2(\pi-x) = \cos^2 x$.
$$ I = \int_0^\pi (\pi-x)\cdot\sin x/(1+\cos^2 x)\, dx $$
Add: $2I = \pi\cdot\int_0^\pi \sin x/(1+\cos^2 x)\, dx$
Let $t = \cos x, dt = -\sin x\, dx$. Limits: $x=0\to t=1$; $x=\pi\to t=-1$.
$$ 2I = \pi\cdot\int_1^{-1} (-dt)/(1+t^2) = \pi\cdot\int_{-1}^1 dt/(1+t^2) = \pi\cdot[\tan^{-1} t]_{-1}^1 = \pi\cdot(\pi/4-(-\pi/4)) = \pi\cdot\pi/2 = \pi^2/2 $$
∴ $I =$ $\pi^2/4$

📊

Integration & Applications — Key Diagrams

Click any diagram to zoom

Area Between Two Curves and Fundamental Theorem

Left: area between curves $f(x)$ (blue, upper) and $g(x)$ (green, lower) equals $\int_a^b [f(x)-g(x)]\, dx$ (shaded region). Right: the Riemann sum (rectangles under the curve) in the limit becomes the definite integral — this is the Fundamental Theorem of Calculus: the integral equals the area under the curve, and $\dfrac{d}{dx}[\int_a^x f\, dt] = f(x)$.

Volume of Revolution — Disk vs Shell Method

Left (disk method): rotate $f(x)$ about x-axis; each slice is a disk of radius $f(x)$ and thickness $dx$; volume $= \pi\int [f(x)]^2\, dx$. Right (shell method): rotate about y-axis; each shell has radius $x$, height $f(x)$, thickness $dx$; volume $= 2\pi\int x\cdot f(x)\, dx$. Choose the method based on the axis of rotation.

⚡

Integration & Applications — Quick Quiz

Question 1

$\int_{-\pi}^\pi \sin^3 x\, dx$ equals:

$f(x)=\sin^3 x$ is an odd function: $f(-x)=\sin^3(-x)=-\sin^3 x=-f(x)$. For odd $f$, $\int_{-a}^a f\, dx =$ 0. No computation needed — just recognise the symmetry.

Question 2

$\int_0^{\pi/2} \sin x/(\sin x+\cos x)\, dx = $

King's property: $I = \int_0^{\pi/2}\sin x/(\sin x+\cos x)\, dx$. Replace $x\to\pi/2-x$: $I = \int_0^{\pi/2}\cos x/(\cos x+\sin x)\, dx$. Adding: $2I = \int_0^{\pi/2}1\, dx = \pi/2$. So $I =$ $\pi/4$. This is a classic JEE pattern.

Question 3

$\int e^x(1 + x\cdot\ln x)/x\, dx = $

Rewrite: $\int e^x[(1/x) + \ln x]\, dx$. Recognise the form $\int e^x[f'(x)+f(x)]\, dx = e^x f(x)+C$ where $f(x)=\ln x, f'(x)=1/x$. Result $=$ $e^x\cdot\ln x + C$. Verify by differentiating: $\dfrac{d}{dx}[e^x \ln x] = e^x \ln x + e^x/x = e^x(\ln x + 1/x)$ ✓

🌀

Differential Equations

Order & degree · Variable separable · Homogeneous · Linear first-order · Applications

Order, Degree, and Types of Solutions

Term	Definition	Example
Order	Highest derivative present	$d^2 y/dx^2 + y = 0 \to$ order 2
Degree	Power of highest-order derivative (after clearing radicals/fractions)	$(dy/dx)^3 + y = 0 \to$ degree 3
General solution	Contains arbitrary constant(s) equal to order	$y = Ce^x$ (for $dy/dx = y$)
Particular solution	Constants determined by initial/boundary conditions	$y = 2e^x$ if $y(0)=2$
Singular solution	Not obtainable from general solution for any $C$	Clairaut DE: $y = cx + f(c)$; singular: envelope of family

⚠️Degree is undefined if the DE cannot be made polynomial in derivatives (e.g. $dy/dx = \sin(d^2 y/dx^2)$ has undefined degree).

Method 1: Variable Separable

Form: $dy/dx = f(x)\cdot g(y)$. Separate: $dy/g(y) = f(x)\, dx$, then integrate both sides.

Example: $dy/dx = xy \to dy/y = x\, dx \to \ln|y| = x^2/2 + C \to y = Ae^{x^2/2}$

Method 2: Homogeneous DE

Form: $dy/dx = F(y/x)$ (coefficients homogeneous of same degree). Substitution: $y = vx$, so $dy/dx = v + x\cdot(dv/dx)$. The equation reduces to variable separable in $v$ and $x$.

Recognition test: replace $x$ by $\lambda x$ and $y$ by $\lambda y$ — if all $\lambda$ cancel, it is homogeneous.

💡After substituting $y=vx$, always bring the $v$ terms to one side before separating. The substitution $x=vy$ works if $dy/dx$ appears in a form where $x/y$ is natural.

Method 3: Linear First-Order DE

Standard form: $dy/dx + P(x)\cdot y = Q(x)$

Step 1: Compute integrating factor $\text{IF} = e^{\int P(x)\, dx}$

Step 2: Multiply both sides by IF. LHS becomes $\dfrac{d}{dx}[y\cdot\text{IF}]$.

Step 3: General solution: $y\cdot\text{IF} = \int Q(x)\cdot\text{IF}\, dx + C$

Similarly for $dx/dy + P(y)\cdot x = Q(y)$: $\text{IF} = e^{\int P(y)\, dy}$.

Method 4: Exact Differential Equations

Form: $M(x, y)\, dx + N(x, y)\, dy = 0$. Exact if $\partial M/\partial y = \partial N/\partial x$.

Solution: $F(x, y) = \int M\, dx$ (treating $y$ constant) + terms involving $y$ only (found by matching $\partial F/\partial y = N$).

Method 5: Bernoulli's Equation

Form: $dy/dx + P(x)y = Q(x)\cdot y^n$ ($n \ne 0, 1$). Divide by $y^n$, substitute $z = y^{1-n}$:

$dz/dx + (1-n)P(x)z = (1-n)Q(x)$ → reduces to linear in $z$.

Applications

Exponential Growth/Decay

$$ dN/dt = kN \to N = N_0 e^{kt} $$

$k > 0$: growth (population, compound interest). $k < 0$: decay (radioactive, drug elimination).

Half-life: $T_{1/2} = \ln 2/|k|$ (time for $N$ to halve).

Newton's Law of Cooling

$dT/dt = -k(T - T_a)$ where $T_a$ = ambient temperature, $k > 0$.

Solution: $T - T_a = (T_0 - T_a)\cdot e^{-kt}$

The object approaches ambient temperature exponentially. Used in JEE problems involving temperature at a specific time.

Orthogonal Trajectories

Given a family of curves $F(x, y, c)=0$ with slope $dy/dx=f(x, y)$, the orthogonal trajectories satisfy: $dy/dx = -1/f(x, y)$.

Classic: orthogonal trajectories of $x^2+y^2=c^2$ are straight lines through origin ($y=mx$).

Method Selection Summary

Pattern in DE	Method
$dy/dx = f(x)\cdot g(y)$	Variable separable
$dy/dx = F(y/x)$ or homogeneous degree	Substitution $y = vx$
$dy/dx + P(x)y = Q(x)$	Linear — integrating factor $e^{\int P\, dx}$
$M\, dx + N\, dy = 0,\ \partial M/\partial y = \partial N/\partial x$	Exact DE
$dy/dx + P(x)y = Q(x)y^n$	Bernoulli — substitute $z = y^{1-n}$

🔢

Differential Equations — Key Formulas

Variable Separable

$dy/dx = f(x)\cdot g(y)$ → $dy/g(y) = f(x)\, dx$ → $\int dy/g(y) = \int f(x)\, dx + C$

Homogeneous DE (y = vx substitution)

$y = vx$ → $dy/dx = v + x\cdot(dv/dx)$
Equation becomes separable in $v$ and $x$
Substitute back: $v = y/x$

Linear First-Order DE

$dy/dx + P(x)\cdot y = Q(x)$
$\text{IF} = e^{\int P\, dx}$
$y \cdot \text{IF} = \int Q\cdot\text{IF}\, dx + C$

Exact DE Condition and Solution

$M\, dx + N\, dy = 0$ exact iff $\partial M/\partial y = \partial N/\partial x$
Solution: $F(x, y)=C$ where $\partial F/\partial x=M, \partial F/\partial y=N$

Growth, Decay and Newton's Cooling

$dN/dt = kN \to N = N_0\cdot e^{kt}$ | $T_{1/2} = \ln 2/|k|$
Newton's cooling: $T - T_a = (T_0 - T_a)\cdot e^{-kt}$

Bernoulli Equation Reduction

$dy/dx + P\cdot y = Q\cdot y^n$ → divide by $y^n$
$z = y^{1-n}$ → $dz/dx + (1-n)P\cdot z = (1-n)Q$

🧮

Differential Equations — Solved Examples

Solve: $dy/dx = (1 + y^{2})/(1 + x^{2})$

Variable separable. Separate: $dy/(1+y^2) = dx/(1+x^2)$
Integrate both sides: $\tan^{-1} y = \tan^{-1} x + C$
Or: $\tan^{-1} y - \tan^{-1} x = C$ (general solution)
Using tan subtraction formula: $\tan^{-1}[(y-x)/(1+xy)] = C$, so $y-x = k(1+xy)$ for constant $k = \tan C$.

Solve: $dy/dx + y/x = x^{2}$ (linear DE)

Standard form: $dy/dx + (1/x)y = x^2$. Here $P(x) = 1/x, Q(x) = x^2$.
$\text{IF} = e^{\int(1/x)\,dx} = e^{\ln x} = x$
Multiply: $\dfrac{d}{dx}[xy] = x\cdot x^2 = x^3$
Integrate: $xy = x^4/4 + C$
∴ $y = x^3/4 + C/x$

Solve: $(x^{2} + y^{2})dx = 2xy dy$ (homogeneous)

Rewrite: $dy/dx = (x^2 + y^2)/(2xy)$. Check: both numerator and denominator are degree 2 — homogeneous. Put $y = vx$:
$$ dy/dx = v + x\cdot\frac{dv}{dx} = \frac{x^{2} + v^{2}x^{2}}{2x\cdot vx} = \frac{1 + v^{2}}{2v} $$
$$ x\cdot dv/dx = (1+v^{2})/(2v) - v = (1+v^{2}-2v^{2})/(2v) = (1-v^{2})/(2v) $$
Separate: $2v\cdot dv/(1-v^2) = dx/x$. Let $u = 1-v^2$: $du = -2v\, dv$
$$ -\int du/u = \int dx/x \to -\ln|1-v^{2}| = \ln|x| + \ln|k| $$
$|1-v^2|\cdot|x| = 1/k$, or $x(1-v^2) = A \to x(1 - y^2/x^2) = A \to$ $x^2 - y^2 = Ax$

A population doubles in 5 years. When will it be 8 times the original?

$N = N_0\cdot e^{kt}$. At $t=5$: $2N_0 = N_0\cdot e^{5k} \to e^{5k} = 2 \to 5k = \ln 2 \to k = (\ln 2)/5$
Find $t$ when $N = 8N_0$: $8N_0 = N_0\cdot e^{kt} \to e^{kt} = 8 \to kt = \ln 8 = 3\ln 2$
$t = 3\ln 2/k = 3\ln 2/[(\ln 2)/5] =$ 15 years

Find the curve passing through $(0, 1)$ such that the slope at any point equals the sum of the coordinates: $dy/dx = x + y$

Rearrange: $dy/dx - y = x$. Linear DE with $P = -1, Q = x$.
$\text{IF} = e^{\int(-1)dx} = e^{-x}$
$$ \frac{d}{dx}[y\cdot e^{-x}] = x\cdot e^{-x} $$
Integrate: $y\cdot e^{-x} = \int x\cdot e^{-x}\, dx$. Use IBP: $u=x, dv=e^{-x}dx \to v=-e^{-x}$
$$ \int x\cdot e^{-x}\, dx = -xe^{-x} - \int(-e^{-x})\, dx = -xe^{-x} - e^{-x} + C = -(x+1)e^{-x} + C $$
$$ y\cdot e^{-x} = -(x+1)e^{-x} + C \to y = -(x+1) + Ce^x $$
At $(0, 1)$: $1 = -1 + C \to C = 2$. Particular solution: $y = 2e^x - x - 1$

📊

Differential Equations — Key Diagrams

Click any diagram to zoom

DE Method Selection Flowchart

Decision flowchart for selecting the right method. Check top-to-bottom: separable → homogeneous → linear → exact → Bernoulli. Each "Yes" branch gives the solution technique. In JEE, most problems are variable separable or linear first-order.

Exponential Growth vs Decay and Newton's Cooling

Left: exponential growth ($k>0$) — curve increases without bound. Centre: exponential decay ($k<0$) — curve decreases toward zero; half-life $T_{1/2}=\ln 2/|k|$ is the time to halve. Right: Newton's cooling — temperature decays exponentially toward ambient temperature $T_a$ (dashed line); approaches $T_a$ asymptotically.

⚡

Differential Equations — Quick Quiz

Question 1

The order and degree of $(d^2 y/dx^2)^3 + (dy/dx)^2 = \sin x$ are:

Highest derivative is $d^2 y/dx^2$ → order = 2. The power of this highest derivative is 3 (it is cubed) → degree = 3. The $(dy/dx)^2$ term is lower order, so it doesn't affect order or degree.

Question 2

The integrating factor of $dy/dx - y/x = x^2$ is:

Standard form: $dy/dx + P\cdot y = Q$ where $P = -1/x$. $\text{IF} = e^{\int P\, dx} = e^{\int(-1/x)dx} = e^{-\ln x} = e^{\ln(1/x)} =$ $1/x$. Multiplying: $\dfrac{d}{dx}[y/x] = x$, so $y/x = x^2/2 + C$, giving $y = x^3/2 + Cx$.

Question 3

A radioactive substance decays to 1/8 of its original amount in 45 years. Its half-life is:

$1/8 = (1/2)^3$ — so the substance has undergone 3 half-lives in 45 years. Therefore each half-life $= 45/3 =$ 15 years. Using formula: $N=N_0\,(1/2)^{t/T_{1/2}} \to 1/8 = (1/2)^{45/T_{1/2}} \to 3 = 45/T_{1/2} \to T_{1/2}=15$.

Limit	Value	Form
\(\lim_{x\to 0} (\sin x)/x\)	1	0/0 (trig)
\(\lim_{x\to 0} (\tan x)/x\)	1	0/0 (trig)
\(\lim_{x\to 0} (1-\cos x)/x^2\)	1/2	0/0
\(\lim_{x\to 0} (\sin^{-1} x)/x\)	1	0/0
\(\lim_{x\to 0} (\tan^{-1} x)/x\)	1	0/0
\(\lim_{x\to 0} (e^x-1)/x\)	1	0/0 (exponential)
\(\lim_{x\to 0} (a^x-1)/x\)	\(\ln a\)	0/0
\(\lim_{x\to 0} \ln(1+x)/x\)	1	0/0 (logarithm)
\(\lim_{x\to\infty} (1+1/x)^x\)	\(e\)	\(1^\infty\)
\(\lim_{x\to 0} (1+x)^{1/x}\)	\(e\)	\(1^\infty\)
\(\lim_{x\to \infty} x^n/e^x\)	0	exponential dominates
\(\lim_{x\to \infty} (\ln x)/x^n\)	0 (\(n>0\))	power dominates log

f(x)	f'(x)	f(x)	f'(x)
\(x^n\)	\(nx^{n-1}\)	\(e^x\)	\(e^x\)
\(\ln x\)	\(1/x\)	\(a^x\)	\(a^x \ln a\)
\(\sin x\)	\(\cos x\)	\(\cos x\)	\(-\sin x\)
\(\tan x\)	\(\sec^2 x\)	\(\cot x\)	\(-\operatorname{cosec}^2 x\)
\(\sec x\)	\(\sec x \tan x\)	\(\operatorname{cosec} x\)	\(-\operatorname{cosec} x \cot x\)
\(\sin^{-1} x\)	\(1/\sqrt{1-x^2}\)	\(\cos^{-1} x\)	\(-1/\sqrt{1-x^2}\)
\(\tan^{-1} x\)	\(1/(1+x^2)\)	\(\cot^{-1} x\)	\(-1/(1+x^2)\)
\(\|x\|\)	\(x/\|x\|\ (x\ne 0)\)	\(\sqrt x\)	\(1/(2\sqrt x)\)

Condition on f'	Behaviour	Example
\(f'(x) > 0\ \forall x \in I\)	Strictly increasing	\(e^x\) on \(\mathbb{R}\); \(x^3\) on \(\mathbb{R}\)
\(f'(x) < 0\ \forall x \in I\)	Strictly decreasing	\(e^{-x}\); \(\cos x\) on \((0, \pi)\)
\(f'(x) = 0\ \forall x \in I\)	Constant	\(f(x) = c\)
\(f'(x) \ge 0\) (= 0 at isolated pts)	Monotonically increasing (non-strict)	\(x^3\) at \(x=0\): \(f'=0\) but still increasing

∫ f(x) dx	Result (+C)	∫ f(x) dx	Result (+C)
\(\int x^n\, dx\)	\(x^{n+1}/(n+1)\)	\(\int 1/x\, dx\)	\(\ln\|x\|\)
\(\int e^x\, dx\)	\(e^x\)	\(\int a^x\, dx\)	\(a^x/\ln a\)
\(\int \sin x\, dx\)	\(-\cos x\)	\(\int \cos x\, dx\)	\(\sin x\)
\(\int \tan x\, dx\)	\(\ln\|\sec x\|\)	\(\int \cot x\, dx\)	\(\ln\|\sin x\|\)
\(\int \sec^2 x\, dx\)	\(\tan x\)	\(\int \operatorname{cosec}^2 x\, dx\)	\(-\cot x\)
\(\int \sec x\tan x\, dx\)	\(\sec x\)	\(\int \operatorname{cosec} x\cot x\, dx\)	\(-\operatorname{cosec} x\)
\(\int 1/\sqrt{1-x^2}\, dx\)	\(\sin^{-1} x\)	\(\int 1/(1+x^2)\, dx\)	\(\tan^{-1} x\)
\(\int 1/\sqrt{a^2-x^2}\, dx\)	\(\sin^{-1}(x/a)\)	\(\int 1/(a^2+x^2)\, dx\)	\((1/a)\tan^{-1}(x/a)\)
\(\int 1/\sqrt{x^2-a^2}\, dx\)	\(\ln\|x+\sqrt{x^2-a^2}\|\)	\(\int 1/\sqrt{x^2+a^2}\, dx\)	\(\ln\|x+\sqrt{x^2+a^2}\|\)

Expression	Substitute
\(\sqrt{a^2-x^2}\)	\(x = a\sin\theta\)
\(\sqrt{a^2+x^2}\)	\(x = a\tan\theta\)
\(\sqrt{x^2-a^2}\)	\(x = a\sec\theta\)

Type of discontinuity	Description	Example
Removable	LHL = RHL \(\ne f(a)\) (or \(f(a)\) undefined)	\(f(x)=\sin x/x\) at \(x=0\); fill with \(f(0)=1\)
Jump (non-removable)	LHL ≠ RHL (both finite)	Signum function at \(x=0\); floor function at integers
Infinite (essential)	At least one limit is \(\pm\infty\)	\(f(x)=1/x\) at \(x=0\); \(\tan x\) at \(x=\pi/2\)
Oscillatory	Limit does not exist (oscillates)	\(\sin(1/x)\) at \(x=0\)

Property	Formula	Use case
Reversal	\(\int_a^b f = -\int_b^a f\)	Swap limits changes sign
Additivity	\(\int_a^c f = \int_a^b f + \int_b^c f\)	Split at any point \(b\)
King's property	\(\int_a^b f(x)\, dx = \int_a^b f(a+b-x)\, dx\)	Most useful; eliminates x from numerator
Even function	\(\int_{-a}^a f(x)\, dx = 2\int_0^a f(x)\, dx\) if \(f(-x)=f(x)\)	Symmetric limits + even → double half
Odd function	\(\int_{-a}^a f(x)\, dx = 0\) if \(f(-x)=-f(x)\)	Symmetric limits + odd → zero
Period	\(\int_0^{nT} f(x)\, dx = n\cdot\int_0^T f(x)\, dx\)	Periodic \(f\) with period \(T\)
Half-interval	\(\int_0^{2a} f(x)\, dx = 2\int_0^a f(x)\, dx\) if \(f(2a-x)=f(x)\); \(= 0\) if \(f(2a-x)=-f(x)\)	Common for sin/cos integrals on \([0, \pi]\)

\(n\)	\(\int_0^{\pi/2} \sin^n x\, dx\)
odd (\(n=2m+1\))	\([2\cdot 4\cdot 6\cdots(2m)] / [1\cdot 3\cdot 5\cdots(2m+1)]\)
even (\(n=2m\))	\([1\cdot 3\cdot 5\cdots(2m-1)] / [2\cdot 4\cdot 6\cdots(2m)] \cdot \pi/2\)

Term	Definition	Example
Order	Highest derivative present	\(d^2 y/dx^2 + y = 0 \to\) order 2
Degree	Power of highest-order derivative (after clearing radicals/fractions)	\((dy/dx)^3 + y = 0 \to\) degree 3
General solution	Contains arbitrary constant(s) equal to order	\(y = Ce^x\) (for \(dy/dx = y\))
Particular solution	Constants determined by initial/boundary conditions	\(y = 2e^x\) if \(y(0)=2\)
Singular solution	Not obtainable from general solution for any \(C\)	Clairaut DE: \(y = cx + f(c)\); singular: envelope of family

Pattern in DE	Method
\(dy/dx = f(x)\cdot g(y)\)	Variable separable
\(dy/dx = F(y/x)\) or homogeneous degree	Substitution \(y = vx\)
\(dy/dx + P(x)y = Q(x)\)	Linear — integrating factor \(e^{\int P\, dx}\)
\(M\, dx + N\, dy = 0,\ \partial M/\partial y = \partial N/\partial x\)	Exact DE
\(dy/dx + P(x)y = Q(x)y^n\)	Bernoulli — substitute \(z = y^{1-n}\)