Quick answer: The chain rule gives d d x f ( g ( x ) ) = f ′ ( g ( x ) ) g ′ ( x ) \frac{d}{dx}f(g(x)) = f'(g(x))\,g'(x) d x d f ( g ( x )) = f ′ ( g ( x )) g ′ ( x ) implicit differentiation , differentiate both sides with respect to x, treating y as a function of x, then solve for d y d x \frac{dy}{dx} d x d y parametric curves x = f ( t ) x = f(t) x = f ( t ) y = g ( t ) y = g(t) y = g ( t ) d y d x = d y / d t d x / d t \frac{dy}{dx} = \frac{dy/dt}{dx/dt} d x d y = d x / d t d y / d t x x x^x x x logarithmic differentiation ).
This assumes the function is differentiable; see differentiability .
Standard derivatives
f(x) f′(x) x n x^n x n n x n − 1 nx^{n - 1} n x n − 1 e x e^x e x e x e^x e x a x a^x a x a x ln a a^x\ln a a x ln a ln x \ln x ln x 1 x \frac{1}{x} x 1 sin x \sin x sin x cos x \cos x cos x cos x \cos x cos x − sin x -\sin x − sin x tan x \tan x tan x sec 2 x \sec^2 x sec 2 x sec x \sec x sec x sec x tan x \sec x\tan x sec x tan x csc x \csc x csc x − csc x cot x -\csc x\cot x − csc x cot x cot x \cot x cot x − csc 2 x -\csc^2 x − csc 2 x sin − 1 x \sin^{-1}x sin − 1 x 1 1 − x 2 \frac{1}{\sqrt{1 - x^2}} 1 − x 2 1 cos − 1 x \cos^{-1}x cos − 1 x − 1 1 − x 2 -\frac{1}{\sqrt{1 - x^2}} − 1 − x 2 1 tan − 1 x \tan^{-1}x tan − 1 x 1 1 + x 2 \frac{1}{1 + x^2} 1 + x 2 1
Rules
Rule Formula Sum ( u + v ) ′ = u ′ + v ′ (u + v)' = u' + v' ( u + v ) ′ = u ′ + v ′ Product ( u v ) ′ = u ′ v + u v ′ (uv)' = u'v + uv' ( uv ) ′ = u ′ v + u v ′ Quotient ( u v ) ′ = u ′ v − u v ′ v 2 \left(\frac{u}{v}\right)' = \frac{u'v - uv'}{v^2} ( v u ) ′ = v 2 u ′ v − u v ′ Chain d y d x = d y d u ⋅ d u d x \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} d x d y = d u d y ⋅ d x d u
Implicit differentiation
Example: x 2 + y 2 = 25 x^2 + y^2 = 25 x 2 + y 2 = 25
Differentiate: 2 x + 2 y d y d x = 0 2x + 2y\frac{dy}{dx} = 0 2 x + 2 y d x d y = 0 d y d x = − x y \frac{dy}{dx} = -\frac{x}{y} d x d y = − y x
Parametric differentiation
d y d x = d y / d t d x / d t \frac{dy}{dx} = \frac{dy/dt}{dx/dt} d x d y = d x / d t d y / d t
Example: x = a cos t x = a\cos t x = a cos t y = a sin t y = a\sin t y = a sin t d y d x = a cos t − a sin t = − cot t \frac{dy}{dx} = \frac{a\cos t}{-a\sin t} = -\cot t d x d y = − a s i n t a c o s t = − cot t
Logarithmic differentiation
Use it for f ( x ) g ( x ) f(x)^{g(x)} f ( x ) g ( x )
Example: y = x x y = x^x y = x x
Step 1: ln y = x ln x \ln y = x\ln x ln y = x ln x
Step 2: 1 y d y d x = ln x + 1 \frac{1}{y}\frac{dy}{dx} = \ln x + 1 y 1 d x d y = ln x + 1 d y d x = x x ( 1 + ln x ) \frac{dy}{dx} = x^x(1 + \ln x) d x d y = x x ( 1 + ln x )
Inverse trig substitutions
Expression Substitute Simplifies to 1 − x 2 \sqrt{1 - x^2} 1 − x 2 x = sin θ x = \sin\theta x = sin θ cos θ \cos\theta cos θ 2 x 1 + x 2 \frac{2x}{1 + x^2} 1 + x 2 2 x 1 − x 2 1 + x 2 \frac{1 - x^2}{1 + x^2} 1 + x 2 1 − x 2 x = tan θ x = \tan\theta x = tan θ sin 2 θ \sin 2\theta sin 2 θ cos 2 θ \cos 2\theta cos 2 θ 2 x 1 − x 2 \frac{2x}{1 - x^2} 1 − x 2 2 x x = tan θ x = \tan\theta x = tan θ tan 2 θ \tan 2\theta tan 2 θ 3 x − 4 x 3 3x - 4x^3 3 x − 4 x 3 x = sin θ x = \sin\theta x = sin θ sin 3 θ \sin 3\theta sin 3 θ
These rely on the inverse trigonometric identities .
Example: y = tan − 1 2 x 1 − x 2 y = \tan^{-1}\frac{2x}{1 - x^2} y = tan − 1 1 − x 2 2 x ∣ x ∣ < 1 \lvert x\rvert < 1 ∣ x ∣ < 1 x = tan θ x = \tan\theta x = tan θ y = 2 θ = 2 tan − 1 x y = 2\theta = 2\tan^{-1}x y = 2 θ = 2 tan − 1 x d y d x = 2 1 + x 2 \frac{dy}{dx} = \frac{2}{1 + x^2} d x d y = 1 + x 2 2
Worked examples
Example 1: chain rule
Differentiate y = sin ( x 2 + 1 ) y = \sin(x^2 + 1) y = sin ( x 2 + 1 )
d y d x = cos ( x 2 + 1 ) ⋅ 2 x \frac{dy}{dx} = \cos(x^2 + 1) \cdot 2x d x d y = cos ( x 2 + 1 ) ⋅ 2 x
Example 2: quotient rule
Differentiate y = x 1 + x 2 y = \frac{x}{1 + x^2} y = 1 + x 2 x
d y d x = ( 1 + x 2 ) − x ⋅ 2 x ( 1 + x 2 ) 2 = 1 − x 2 ( 1 + x 2 ) 2 \frac{dy}{dx} = \frac{(1 + x^2) - x \cdot 2x}{(1 + x^2)^2} = \frac{1 - x^2}{(1 + x^2)^2} d x d y = ( 1 + x 2 ) 2 ( 1 + x 2 ) − x ⋅ 2 x = ( 1 + x 2 ) 2 1 − x 2
Example 3: implicit
Find d y d x \frac{dy}{dx} d x d y x 3 + y 3 = 3 x y x^3 + y^3 = 3xy x 3 + y 3 = 3 x y
3 x 2 + 3 y 2 y ′ = 3 y + 3 x y ′ 3x^2 + 3y^2y' = 3y + 3xy' 3 x 2 + 3 y 2 y ′ = 3 y + 3 x y ′
d y d x = y − x 2 y 2 − x \frac{dy}{dx} = \frac{y - x^2}{y^2 - x} d x d y = y 2 − x y − x 2
Example 4: derivative of one function with respect to another
Differentiate sin x \sin x sin x cos x \cos x cos x
d ( sin x ) d ( cos x ) = cos x − sin x = − cot x \frac{d(\sin x)}{d(\cos x)} = \frac{\cos x}{-\sin x} = -\cot x d ( c o s x ) d ( s i n x ) = − s i n x c o s x = − cot x
Summary table
Technique Use when Chain rule Function inside a function Implicit y not isolated Parametric x and y given in terms of t Logarithmic Variable in the power, or long products Trig substitution Inverse trig of algebraic expressions
Exam tips
Simplify before differentiating; inverse trig expressions often collapse to k tan − 1 x k\tan^{-1}x k tan − 1 x
In implicit problems, collect all d y d x \frac{dy}{dx} d x d y
For "derivative of f with respect to g", divide f ′ ( x ) f'(x) f ′ ( x ) g ′ ( x ) g'(x) g ′ ( x )
Next, study higher derivatives, Rolle's theorem and the mean value theorem , or see the NIMCET Calculus chapter .
Common mistakes
Forgetting the inner derivative in the chain rule.
Reversing the numerator terms in the quotient rule.
Treating y as a constant in implicit differentiation.
Using d d x x x = x ⋅ x x − 1 \frac{d}{dx}x^x = x \cdot x^{x - 1} d x d x x = x ⋅ x x − 1
Frequently Asked Questions
What is the chain rule?
d/dx f(g(x)) = f′(g(x)) · g′(x).
How do you differentiate implicitly?
Differentiate both sides with respect to x, treating y as a function of x, and solve for dy/dx.
What is dy/dx for parametric equations?
dy/dx = (dy/dt)/(dx/dt).
What is the derivative of xˣ?
xˣ(1 + ln x), found by logarithmic differentiation.
What is the derivative of tan⁻¹x?
1/(1 + x²).
