Limits & Continuity · Differentiation · Applications of Derivatives · Integration & Applications · Differential Equations
| Limit | Value | Form |
|---|---|---|
| \(\lim_{x\to 0} \dfrac{\sin x}{x}\) | 1 | 0/0 (trig) |
| \(\lim_{x\to 0} \dfrac{\tan x}{x}\) | 1 | 0/0 (trig) |
| \(\lim_{x\to 0} \dfrac{1-\cos x}{x^2}\) | 1/2 | 0/0 |
| \(\lim_{x\to 0} \dfrac{\sin^{-1} x}{x}\) | 1 | 0/0 |
| \(\lim_{x\to 0} \dfrac{\tan^{-1} x}{x}\) | 1 | 0/0 |
| \(\lim_{x\to 0} \dfrac{e^x-1}{x}\) | 1 | 0/0 (exponential) |
| \(\lim_{x\to 0} \dfrac{a^x-1}{x}\) | \(\ln a\) | 0/0 |
| \(\lim_{x\to 0} \dfrac{\ln(1+x)}{x}\) | 1 | 0/0 (logarithm) |
| \(\lim_{x\to\infty} (1+\dfrac{1}{x})^x\) | \(e\) | \(1^\infty\) |
| \(\lim_{x\to 0} (1+x)^{1/x}\) | \(e\) | \(1^\infty\) |
| \(\lim_{x\to \infty} \dfrac{x^n}{e^x}\) | 0 | exponential dominates |
| \(\lim_{x\to \infty} \dfrac{\ln x}{x^n}\) | 0 (\(n>0\)) | power dominates log |
\(0/0,\ \infty/\infty,\ 0\cdot\infty,\ \infty-\infty,\ 0^0,\ \infty^0,\ 1^\infty\)
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.
If \(\lim \dfrac{f(x)}{g(x)}\) gives \(0/0\) or \(\infty/\infty\), then:
\(\lim \dfrac{f(x)}{g(x)} = \lim \dfrac{f'(x)}{g'(x)}\)
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.
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: \(\lim_{x\to 0} x^2\sin(\dfrac{1}{x}) = 0\) (squeeze between \(-x^2\) and \(x^2\)).
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 \(e\).
\(f(x)\) 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 \(\pm\infty\) | \(f(x)=\dfrac{1}{x}\) at \(x=0\); \(\tan x\) at \(x=\dfrac{\pi}{2}\) |
| Oscillatory | Limit does not exist (oscillates) | \(\sin(\dfrac{1}{x})\) at \(x=0\) |
\(f'(x) = \lim_{h\to 0} \dfrac{f(x+h) - f(x)}{h}\)
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 \(x=0\)).
| f(x) | f'(x) | f(x) | f'(x) |
|---|---|---|---|
| \(x^n\) | \(nx^{n-1}\) | \(e^x\) | \(e^x\) |
| \(\ln x\) | \(\dfrac{1}{x}\) | \(a^x\) | \(a^x \ln a\) |
| \(\sin x\) | \(\cos x\) | \(\cos x\) | \(-\sin x\) |
| \(\tan x\) | \(\sec^2 x\) | \(\cot x\) | \(-\operatorname{cosec}^2 x\) |
| \(\sec x\) | \(\sec x \tan x\) | \(\operatorname{cosec} x\) | \(-\operatorname{cosec} x \cot x\) |
| \(\sin^{-1} x\) | \(\dfrac{1}{\sqrt{1-x^2}}\) | \(\cos^{-1} x\) | \(-\dfrac{1}{\sqrt{1-x^2}}\) |
| \(\tan^{-1} x\) | \(\dfrac{1}{1+x^2}\) | \(\cot^{-1} x\) | \(-\dfrac{1}{1+x^2}\) |
| \(|x|\) | \(x/|x|\ (x\ne 0)\) | \(\sqrt x\) | \(\dfrac{1}{2\sqrt x}\) |
Product: \((uv)' = u'v + uv'\)
Quotient: \(\left(\frac{u}{v}\right)' = \dfrac{u'v - uv'}{v^{2}}\)
Mnemonic for quotient: "low d-high minus high d-low, over low squared."
\(\dfrac{d}{dx}[f(g(x))] = f'(g(x))\cdot g'(x)\)
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.
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 \(\dfrac{dy}{dx}\) algebraically. Used for circles, ellipses, folium of Descartes, etc.
For \(y = [f(x)]^{g(x)}\) (variable base AND exponent):
\(\ln y = g(x)\cdot\ln f(x) \to\) differentiate both sides \(\to (\dfrac{1}{y})\cdot y' = g'\ln f + g\cdot \dfrac{f'}{f}\)
Also use for products/quotients of many functions to simplify computation.
If \(x=f(t), y=g(t)\), then:
\(\dfrac{dy/dt}{dx/dt} = \dfrac{g'(t)}{f'(t)}\)
$$ \frac{d^{2}y}{dx^{2}} = \frac{d}{dx}\!\left(\frac{dy}{dx}\right) = \frac{d/dt\,(dy/dx)}{dx/dt} $$
\(y'' = d^2 \dfrac{y}{dx^2}\): second derivative (rate of change of slope → concavity)
Key results: if \(y=e^x, y^{(n)}=e^x\); if \(y=\sin x, y^{(n)}=\sin(x+n\dfrac{\pi}{2})\); if \(y=x^n, y^{(n)}=n!\) and \(y^{(n+1)}=0\).
Leibniz rule: \((uv)^{(n)} = \sum C(n,r)\,u^{(n-r)}\,v^{(r)}\)
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.
Conditions: \(f\) continuous on \([a,b]\), differentiable on \((a,b)\)
Conclusion: \(\exists\, c\in (a,b)\) such that \(f'(c) = \dfrac{f(b)-f(a)}{b-a}\)
Geometric meaning: the instantaneous slope at \(c\) equals the average slope of the chord. Rolle's theorem is the special case \(f(a)=f(b)\).
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 (\(x = x_1\)).
Slope of normal \(= -\dfrac{1}{f'(x_1)} = m_N\) (\(m_T \cdot m_N = -1\))
Equation: \(y - y_{1} = -[\dfrac{1}{f'(x_{1})}](x - x_{1})\)
Length of tangent, normal, subtangent, subnormal: standard results from coordinate geometry of 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.
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 |
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.
Examine the sign of \(f'\) on both sides of critical point \(c\):
Works for all cases including where \(f''\) doesn't exist.
If \(f'(c) = 0\):
Faster but fails when \(f''(c) = 0\). Classic fail: \(f(x) = x^4\) at \(c=0\) — minimum but \(f''(0)=0\).
Closed interval method:
Unlike local extrema, absolute extrema may occur at endpoints.
\(f''(x) > 0\) 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.
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.
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 \(\sqrt{26} \approx \sqrt{25} + (\dfrac{1}{2\sqrt{25}})\cdot 1 = 5.1\), etc.
When two quantities \(x(t)\) and \(y(t)\) are related by an equation, differentiate both sides w.r.t. \(t\):
Example: sphere volume \(V=(\dfrac{4}{3})\pi r^3 \to \dfrac{dV}{dt} = 4\pi r^2\cdot(\dfrac{dr}{dt})\). If \(\dfrac{dV}{dt}\) is given, find \(\dfrac{dr}{dt}\) at a given \(r\).
| ∫ f(x) dx | Result (+C) | ∫ f(x) dx | Result (+C) |
|---|---|---|---|
| \(\int x^n\, dx\) | \(\dfrac{x^{n+1}}{n+1}\) | \(\int \dfrac{1}{x}\, dx\) | \(\ln|x|\) |
| \(\int e^x\, dx\) | \(e^x\) | \(\int a^x\, dx\) | \(a^x/\ln a\) |
| \(\int \sin x\, dx\) | \(-\cos x\) | \(\int \cos x\, dx\) | \(\sin x\) |
| \(\int \tan x\, dx\) | \(\ln|\sec x|\) | \(\int \cot x\, dx\) | \(\ln|\sin x|\) |
| \(\int \sec^2 x\, dx\) | \(\tan x\) | \(\int \operatorname{cosec}^2 x\, dx\) | \(-\cot x\) |
| \(\int \sec x\tan x\, dx\) | \(\sec x\) | \(\int \operatorname{cosec} x\cot x\, dx\) | \(-\operatorname{cosec} x\) |
| \(\int \dfrac{1}{\sqrt{1-x^2}}\, dx\) | \(\sin^{-1} x\) | \(\int \dfrac{1}{1+x^2}\, dx\) | \(\tan^{-1} x\) |
| \(\int \dfrac{1}{\sqrt{a^2-x^2}}\, dx\) | \(\sin^{-1}(\dfrac{x}{a})\) | \(\int \dfrac{1}{a^2+x^2}\, dx\) | \((\dfrac{1}{a})\tan^{-1}(\dfrac{x}{a})\) |
| \(\int \dfrac{1}{\sqrt{x^2-a^2}}\, dx\) | \(\ln|x+\sqrt{x^2-a^2}|\) | \(\int \dfrac{1}{\sqrt{x^2+a^2}}\, dx\) | \(\ln|x+\sqrt{x^2+a^2}|\) |
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.
$$ \int u\cdot v\, dx = u\cdot\int v\, dx - \int\left[u'\cdot\int v\, dx\right] dx $$
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.
For rational \(\dfrac{P(x)}{Q(x)}\) with \(\deg P < \deg Q\), factor \(Q(x)\) then split:
| Expression | Substitute |
|---|---|
| \(\sqrt{a^2-x^2}\) | \(x = a\sin\theta\) |
| \(\sqrt{a^2+x^2}\) | \(x = a\tan\theta\) |
| \(\sqrt{x^2-a^2}\) | \(x = a\sec\theta\) |
| 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 \([0, \pi]\) |
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.
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): \(V = 2\pi\int_a^b x\cdot f(x)\, dx\)
$$ \int_{0}^{\pi/2} \sin^{n}x\, dx = \int_{0}^{\pi/2} \cos^{n}x\, dx $$
| \(n\) | \(\int_0^{\pi/2} \sin^n x\, dx\) |
|---|---|
| odd (\(n=2m+1\)) | \([2\cdot 4\cdot 6\cdots(2m)] / [1\cdot 3\cdot 5\cdots(2m+1)]\) |
| even (\(n=2m\)) | \([1\cdot 3\cdot 5\cdots(2m-1)] / [2\cdot 4\cdot 6\cdots(2m)] \cdot \dfrac{\pi}{2}\) |
Key values: \(\int_0^{\pi/2} \sin^2 x\, dx = \dfrac{\pi}{4}\); \(\int_0^{\pi/2} \sin^3 x\, dx = \dfrac{2}{3}\); \(\int_0^{\pi/2} \sin^4 x\, dx = 3\dfrac{\pi}{16}\).
| Term | Definition | Example |
|---|---|---|
| Order | Highest derivative present | \(d^2 \dfrac{y}{dx^2} + y = 0 \to\) order 2 |
| Degree | Power of highest-order derivative (after clearing radicals/fractions) | \((\dfrac{dy}{dx})^3 + y = 0 \to\) degree 3 |
| General solution | Contains arbitrary constant(s) equal to order | \(y = Ce^x\) (for \(\dfrac{dy}{dx} = y\)) |
| 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 |
Form: \(\dfrac{dy}{dx} = f(x)\cdot g(y)\). Separate: \(\dfrac{dy}{g(y)} = f(x)\, dx\), then integrate both sides.
Example: \(\dfrac{dy}{dx} = xy \to \dfrac{dy}{y} = x\, dx \to \ln|y| = \dfrac{x^2}{2} + C \to y = Ae^{x^2/2}\)
Form: \(\dfrac{dy}{dx} = F(\dfrac{y}{x})\) (coefficients homogeneous of same degree). Substitution: \(y = vx\), so \(\dfrac{dy}{dx} = v + x\cdot(\dfrac{dv}{dx})\). 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.
Standard form: \(\dfrac{dy}{dx} + P(x)\cdot y = Q(x)\)
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 \(\dfrac{dx}{dy} + P(y)\cdot x = Q(y)\): \(\text{IF} = e^{\int P(y)\, dy}\).
Form: \(M(x, y)\, dx + N(x, y)\, dy = 0\). Exact if \(\partial \dfrac{M}{\partial} y = \partial \dfrac{N}{\partial} x\).
Solution: \(F(x, y) = \int M\, dx\) (treating \(y\) constant) + terms involving \(y\) only (found by matching \(\partial \dfrac{F}{\partial} y = N\)).
Form: \(\dfrac{dy}{dx} + P(x)y = Q(x)\cdot y^n\) (\(n \ne 0, 1\)). Divide by \(y^n\), substitute \(z = y^{1-n}\):
\(\dfrac{dz}{dx} + (1-n)P(x)z = (1-n)Q(x)\) → reduces to linear in \(z\).
$$ \frac{dN}{dt} = kN \to N = N_0 e^{kt} $$
\(k > 0\): growth (population, compound interest). \(k < 0\): decay (radioactive, drug elimination).
Half-life: \(T_{1/2} = \ln 2/|k|\) (time for \(N\) to halve).
\(\dfrac{dT}{dt} = -k(T - T_a)\) 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.
Given a family of curves \(F(x, y, c)=0\) with slope \(\dfrac{dy}{dx}=f(x, y)\), the orthogonal trajectories satisfy: \(\dfrac{dy}{dx} = -\dfrac{1}{f(x, y)}\).
Classic: orthogonal trajectories of \(x^2+y^2=c^2\) are straight lines through origin (\(y=mx\)).
| Pattern in DE | Method |
|---|---|
| \(\dfrac{dy}{dx} = f(x)\cdot g(y)\) | Variable separable |
| \(\dfrac{dy}{dx} = F(\dfrac{y}{x})\) or homogeneous degree | Substitution \(y = vx\) |
| \(\dfrac{dy}{dx} + P(x)y = Q(x)\) | Linear — integrating factor \(e^{\int P\, dx}\) |
| \(M\, dx + N\, dy = 0,\ \partial \dfrac{M}{\partial} y = \partial \dfrac{N}{\partial} x\) | Exact DE |
| \(\dfrac{dy}{dx} + P(x)y = Q(x)y^n\) | Bernoulli — substitute \(z = y^{1-n}\) |