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JEE Mains Maths · Differentiation

Chain Rule and Inverse-Trig Simplification

Differentiating composite and inverse functions, and simplifying an inverse-trig or algebraic expression first so that its derivative is short.

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Why this matters

Thirteen PYQs, eleven of them multiple choice, and one from 2026. Seven simplify first — four collapse an inverse-trig expression and three rewrite a trigonometric or algebraic fraction — and then differentiate; six apply the chain rule to a composite, three of them to an inverse function through g′(f(x)) = 1/f′(x). Two ideas cover the page.

Concept 1 of 2: Simplify before you differentiate

Many of these functions look heavy but collapse. A substitution x=tan⁡θx=\tan\theta or x=sin⁡θx=\sin\theta turns an inverse-trig expression into a multiple of θ\theta. Dividing top and bottom by cos⁡x\cos x turns a ratio of sines and cosines into tan⁡(α−x)\tan(\alpha-x). An algebraic fraction often shares a factor. Differentiate only after the collapse, and read the interval: it decides which branch the angle is on.

Definition

  • tan⁡−12x1−x2=2tan⁡−1x\tan^{-1}\frac{2x}{1-x^2}=2\tan^{-1}x for ∣x∣<1|x|<1.
  • sin⁡−12x1+x2=2tan⁡−1x\sin^{-1}\frac{2x}{1+x^2}=2\tan^{-1}x for ∣x∣≤1|x|\le1.
  • tan⁡−1acos⁡x−bsin⁡xbcos⁡x+asin⁡x=tan⁡−1ab−x\tan^{-1}\frac{a\cos x-b\sin x}{b\cos x+a\sin x}=\tan^{-1}\frac ab-x when the right side lies in (−π2,π2)\left(-\frac\pi2,\frac\pi2\right).
  • tan⁡−1(tan⁡u)=u\tan^{-1}(\tan u)=u only for u∈(−π2,π2)u\in\left(-\frac\pi2,\frac\pi2\right).

A standard substitution

x=tan⁡θ:tan⁡−12x1−x2=2θ=2tan⁡−1x(∣x∣<1)x=\tan\theta:\quad\tan^{-1}\frac{2x}{1-x^2}=2\theta=2\tan^{-1}x\quad(|x|<1)

Worked example

Find dydx\frac{dy}{dx} if y=tan⁡−12x1−x2y=\tan^{-1}\frac{2x}{1-x^2}, ∣x∣<1|x|<1.
Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2026 · 4 Apr 2026 Shift 1 · Q66Moderate

Example 1 · Differentiation · Chain Rule and Inverse-Trig Simplification

If y=tan⁡−1(3cos⁡x−4sin⁡x4cos⁡x+3sin⁡x)+2tan⁡−1(x1+1−x2)y=\tan^{- 1}\left( \frac{3\cos x- 4\sin x}{4\cos x+ 3\sin x} \right)+ 2\tan^{- 1}\left( \frac{x}{1 +\sqrt{1 -x^{2}}} \right), then dydx\frac{dy}{dx} at x=32x =\frac{\sqrt{3}}{2} is equal to :

The branch decides the answer

cos⁡−1(cos⁡x)=x\cos^{-1}(\cos x)=x only on [0,π][0,\pi]. On [π,2π][\pi,2\pi] it is 2π−x2\pi-x, whose derivative is −1-1, not 11. Read the stated interval before you simplify.
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Concept 2 of 2: The chain rule and inverse functions

Differentiate from the outside in, multiplying the derivative of each layer at its own input. If gg undoes ff, then g(f(x))=xg(f(x))=x, and the chain rule gives g′(f(x)) f′(x)=1g'(f(x))\,f'(x)=1. To find g′g' at a number kk, first solve f(a)=kf(a)=k; then g′(k)=1f′(a)g'(k)=\frac{1}{f'(a)}.

Definition

  • (f(g(x)))′=f′(g(x)) g′(x)\big(f(g(x))\big)'=f'(g(x))\,g'(x).
  • If g(f(x))=xg(f(x))=x and f(a)=kf(a)=k, then g(k)=ag(k)=a and g′(k)=1f′(a)g'(k)=\frac{1}{f'(a)}.
  • log⁡ab=ln⁡bln⁡a\log_ab=\frac{\ln b}{\ln a}: change the base before differentiating.

Derivative of an inverse

g′(k)=1f′(a),f(a)=kg'(k)=\frac{1}{f'(a)},\qquad f(a)=k

Worked example

Let f(x)=x3+2x+1f(x)=x^3+2x+1 and let gg be its inverse. Find g′(4)g'(4).
Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2024 · 5 Apr 2024 · Q77Moderate

Example 2 · Differentiation · Chain Rule and Inverse-Trig Simplification

Let f(x)=x5+2x3+3x+1,x∈Rf(x) =x^{5}+ 2x^{3}+ 3x + 1,x \in R, and g(x)g(x) be a function such that g(f(x))=xg(f(x)) = x for all x∈Rx \in R. Then g(7)g′(7)\frac{g(7)}{g^{'}(7)} is equal to:

Invert at the right point

g′(k)g'(k) is 1f′(a)\frac{1}{f'(a)} where f(a)=kf(a)=k, not 1f′(k)\frac{1}{f'(k)}. Solve f(a)=kf(a)=k first; the root is usually a small integer.
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Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (2)

  • Simplify before you differentiate

    A standard substitution

    x=tan⁡θ:tan⁡−12x1−x2=2θ=2tan⁡−1x(∣x∣<1)x=\tan\theta:\quad\tan^{-1}\frac{2x}{1-x^2}=2\theta=2\tan^{-1}x\quad(|x|<1)
  • The chain rule and inverse functions

    Derivative of an inverse

    g′(k)=1f′(a),f(a)=kg'(k)=\frac{1}{f'(a)},\qquad f(a)=k

Watch out for (2)

Test yourself on Differentiation

20 past JEE Mains questions from this chapter, timed at 48 minutes and marked the way the exam marks it. You see your score and every answer the moment you finish. Free to start.

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