Let $f(x) = -x$ for $-2\le x<0$; $f(x)=2x$ for $0\le x\le2$; $f(x)=\dfrac{2x-4}{3}$ for $2<x\le7$. And $g(x)=6-3x$ for $0\le x\le2$; $g(x)=\dfrac{18-x}{4}$ for $2<x\le7$. Let $u(x)=f[g(x)]$, $v(x)=g[f(x)]$ and $w(x)=g[g(x)]$. Find each derivative at $x=1$, i.e. find $u'(1)$, $v'(1)$ and $w'(1)$, if it exists; if it doesn't exist then explain why.

**Setup.** On $0\le x\le2$: $f(x)=2x$ (slope $2$), $g(x)=6-3x$ (slope $-3$). On $2<x\le7$: $f(x)=\dfrac{2x-4}{3}$ (slope $\dfrac23$), $g(x)=\dfrac{18-x}{4}$ (slope $-\dfrac14$). At $x=2$ the two pieces of $f$ have slopes $2$ and $\dfrac23$ (unequal — $f$ has a corner at $x=2$), and the two pieces of $g$ have slopes $-3$ and $-\dfrac14$ (unequal — $g$ also has a corner at $x=2$). At $x=1$: $g(1)=6-3(1)=3$ (using $0\le x\le2$), and $g'(1)=-3$ (safely inside the piece, away from the corner). Also $f(1)=2(1)=2$ (using $0\le x\le2$), and $f'(1)=2$. **$u(x)=f[g(x)]$ at $x=1$:** the inner value is $g(1)=3$, which lies in $(2,7]$ — away from $f$'s corner at $2$, so $f$ is differentiable there with $f'(3)=\dfrac23$ (slope of the $2 0$, $f(1+h)=2(1+h)=2+2h>2$, so $g$ uses its right piece there, giving a right-hand derivative $f'(1)\cdot\left(-\dfrac14\right)=2\cdot\left(-\dfrac14\right)=-\dfrac12$. For small $h<0$, $f(1+h)=2+2h<2$, so $g$ uses its left piece there, giving a left-hand derivative $f'(1)\cdot(-3)=2(-3)=-6$. Since $-\dfrac12\ne-6$, the left- and right-hand derivatives disagree, so **$v'(1)$ does not exist**. **$w(x)=g[g(x)]$ at $x=1$:** the inner value is $g(1)=3\in(2,7]$ — away from $g$'s corner, so $g'(3)=-\dfrac14$ (slope of the $2 [!ANSWER] $u'(1)=-2$, $v'(1)$ does not exist (left derivative $-6\ne$ right derivative $-\dfrac12$), $w'(1)=\dfrac34$

Let f(x) = -x for -2≤ x<0; f(x)=2x for 0≤ x≤2; f(x)=(2x-4)/(3) for 2<x≤7. And g(x)=6-3x for 0≤ x≤2; g(x)=(18-x

MISCELLANEOUS EXERCISE 1 (II) · Q226

f(x) = -x

Maharashtra MsbshseTextbookSubjectiveImportance★★★★★

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✓ Free question

Concept understanding — Chain Rule

The Chain Rule: Why It Makes Sense

Imagine you're assembling a toy. First you put part A into part B, then you put that combined piece into part C. The final toy's position depends on how you moved A, which then affected B, which then affected C. That's exactly what the chain rule captures — how a change in the first variable ripples through a sequence of functions to affect the final output.

Let's make this concrete. Suppose you have a function $f$ that depends on $g$ , and $g$ itself depends on $x$ :

y = f(g(x))

You want to know: if $x$ changes by a tiny amount, how much does $y$ change? The answer isn't just $f'(g(x))$ — because $g(x)$ itself changes when $x$ changes. You have to multiply the two rates:

How fast does $g$ change with respect to $x$ ? That's $g'(x)$ .
How fast does $f$ change with respect to its input $g$ ? That's $f'(g(x))$ .

The total effect is the product:

\frac{dy}{dx} = f'(g(x)) \cdot g'(x)

Note

In Leibniz notation, this looks even more natural: $\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}$ , where $u = g(x)$ . The $du$ 's "cancel" like fractions — though this is just a helpful memory aid, not a rigorous proof.

The Precise Statement

Chain Rule (single variable): If $g$ is differentiable at $x$ and $f$ is differentiable at $g(x)$ , then the composite function $h(x) = f(g(x))$ is differentiable at $x$ , and

h'(x) = f'(g(x)) \cdot g'(x)

That's it. One multiplication. But the power is enormous — it lets you differentiate almost any nested function.

A Simple Example

Differentiate $h(x) = \sin(3x^2)$ .

Here $f(u) = \sin u$ and $g(x) = 3x^2$ . Then:

$f'(u) = \cos u$ , so $f'(g(x)) = \cos(3x^2)$
$g'(x) = 6x$

Multiply: $h'(x) = \cos(3x^2) \cdot 6x = 6x \cos(3x^2)$

Watch out

The most common mistake is forgetting to multiply by the inner derivative. Students often write $\frac{d}{dx} \sin(3x^2) = \cos(3x^2)$ and stop — that's wrong. The chain rule demands you also multiply by $6x$ .

Why It's Called a "Chain"

Think of a chain of links: $x \to g \to f$ . Each link has its own rate of change. To find the total rate from $x$ to $f$ , you multiply the rates of each link. If you had three functions — say $h(x) = f(g(k(x)))$ — you'd multiply three derivatives:

h'(x) = f'(g(k(x))) \cdot g'(k(x)) \cdot k'(x)

The chain can be as long as you like. Each new function adds one more factor.

The Intuition in One Sentence

The chain rule says: the rate of change of the whole is the product of the rates of change of the parts, evaluated at the right places.

Important

The chain rule is not optional — it's the backbone of calculus. Every derivative of a trigonometric, exponential, logarithmic, or power function that isn't just $x^n$ uses it. Master this, and you master differentiation.

The chain rule is one of the most heavily tested formulas in the NCERT Class 12 Continuity and Differentiability chapter, and it underlies nearly every differentiation problem in CBSE boards, JEE Main and JEE Advanced. Whether you're searching 'chain rule differentiation class 12 examples' or 'chain rule important questions for JEE', this f'(g(x))·g'(x) pattern is the formula every subsequent derivative rule in the syllabus builds on.

Setup. On $0\le x\le2$ : $f(x)=2x$ (slope $2$ ), $g(x)=6-3x$ (slope $-3$ ). On $2<x\le7$ : $f(x)=\dfrac{2x-4}{3}$ (slope $\dfrac23$ ), $g(x)=\dfrac{18-x}{4}$ (slope $-\dfrac14$ ). At $x=2$ the two pieces of $f$ have slopes $2$ and $\dfrac23$ (unequal — $f$ has a corner at $x=2$ ), and the two pieces of $g$ have slopes $-3$ and $-\dfrac14$ (unequal — $g$ also has a corner at $x=2$ ).

At $x=1$ : $g(1)=6-3(1)=3$ (using $0\le x\le2$ ), and $g'(1)=-3$ (safely inside the piece, away from the corner). Also $f(1)=2(1)=2$ (using $0\le x\le2$ ), and $f'(1)=2$ .

$u(x)=f[g(x)]$ at $x=1$ : the inner value is $g(1)=3$ , which lies in $(2,7]$ — away from $f$ 's corner at $2$ , so $f$ is differentiable there with $f'(3)=\dfrac23$ (slope of the $2<x\le7$ piece).

u'(1)=f'(g(1))\cdot g'(1)=f'(3)\cdot g'(1)=\frac23\cdot(-3)=-2.

$v(x)=g[f(x)]$ at $x=1$ : the inner value is $f(1)=2$ — this lands exactly on $g$ 's corner point $x=2$ , where $g'(2)$ does not exist (left slope $-3\ne$ right slope $-\dfrac14$ ). To see this directly for $v$ , check the one-sided derivatives of $v$ at $x=1$ : for small $h>0$ , $f(1+h)=2(1+h)=2+2h>2$ , so $g$ uses its right piece there, giving a right-hand derivative $f'(1)\cdot\left(-\dfrac14\right)=2\cdot\left(-\dfrac14\right)=-\dfrac12$ . For small $h<0$ , $f(1+h)=2+2h<2$ , so $g$ uses its left piece there, giving a left-hand derivative $f'(1)\cdot(-3)=2(-3)=-6$ . Since $-\dfrac12\ne-6$ , the left- and right-hand derivatives disagree, so $v'(1)$ does not exist.

$w(x)=g[g(x)]$ at $x=1$ : the inner value is $g(1)=3\in(2,7]$ — away from $g$ 's corner, so $g'(3)=-\dfrac14$ (slope of the $2<x\le7$ piece).

w'(1)=g'(g(1))\cdot g'(1)=g'(3)\cdot g'(1)=\left(-\frac14\right)(-3)=\frac34.

✓Final answer

u'(1)=-2

Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

Showing the 12 most recent of 115 on this concept.

CBSE 2026Set A1 markMCQ
$\frac{d}{dx}\sqrt{x^2+ax+1} =$
(a) $\frac{x+a}{2\sqrt{x^2+ax+1}}$
(b) $\frac{2x+a}{2\sqrt{x^2+ax+1}}$
(c) $\frac{2x+a}{\sqrt{x^2+ax+1}}$
(d) $\frac{1}{2\sqrt{x^2+ax+1}}$
›Reveal solutionSolution

$\frac{d}{dx}\sqrt{x^2+ax+1} = \frac{2x+a}{2\sqrt{x^2+ax+1}}$ .

Let $u = x^2+ax+1$ , so $\frac{du}{dx} = 2x+a$ .

Using $\frac{d}{dx}\sqrt{u} = \frac{1}{2\sqrt{u}}\cdot\frac{du}{dx}$ :

$\frac{d}{dx}\sqrt{x^2+ax+1} = \frac{2x+a}{2\sqrt{x^2+ax+1}}.$

✓Final answer
$\frac{2x+a}{2\sqrt{x^2+ax+1}}$
CBSE 2026Set A1 markMCQ
$\frac{d}{dx}(\sin x^2) =$
(a) $2x\cos x^2$
(b) $\cos x^2$
(c) $x^2\cos x^2$
(d) $x\cos x^2$
›Reveal solutionSolution

$\frac{d}{dx}\sin(x^2) = 2x\cos(x^2)$ .

Let $u = x^2$ , so $\frac{du}{dx} = 2x$ .

By the chain rule,

$\frac{d}{dx}\sin u = \cos u\cdot\frac{du}{dx} = \cos(x^2)\cdot 2x = 2x\cos(x^2).$

✓Final answer
$2x\cos x^2$
CBSE 2026Set A1 markMCQ
$\frac{d}{dx}\sqrt{\cot x} =$
(a) $\frac{1}{2\sqrt{\cot x}}$
(b) $\sqrt{\csc^2 x}$
(c) $\frac{-\csc^2 x}{2\sqrt{\cot x}}$
(d) $\frac{\csc^2 x}{2\sqrt{\cot x}}$
›Reveal solutionSolution

$\frac{d}{dx}\sqrt{\cot x} = \frac{-\csc^2 x}{2\sqrt{\cot x}}$ .

Let $u = \cot x$ , so $\frac{du}{dx} = -\csc^2 x$ .

Using $\frac{d}{dx}\sqrt{u} = \frac{1}{2\sqrt{u}}\cdot\frac{du}{dx}$ :

$\frac{d}{dx}\sqrt{\cot x} = \frac{1}{2\sqrt{\cot x}}\cdot(-\csc^2 x) = \frac{-\csc^2 x}{2\sqrt{\cot x}}.$

✓Final answer
$\frac{-\csc^2 x}{2\sqrt{\cot x}}$
CBSE 2026Set A1 markMCQ
$\frac{d}{dx}(\cos\sqrt{x}) =$
(a) $\sin\sqrt{x}$
(b) $\frac{-\sin\sqrt{x}}{2\sqrt{x}}$
(c) $\frac{\sin\sqrt{x}}{2\sqrt{x}}$
(d) $\frac{1}{2\sqrt{x}}$
›Reveal solutionSolution

$\frac{d}{dx}\cos\sqrt{x} = \frac{-\sin\sqrt{x}}{2\sqrt{x}}$ .

Let $u = \sqrt{x}$ , so $\frac{du}{dx} = \frac{1}{2\sqrt{x}}$ .

By the chain rule,

$\frac{d}{dx}\cos u = -\sin u\cdot\frac{du}{dx} = -\sin\sqrt{x}\cdot\frac{1}{2\sqrt{x}} = \frac{-\sin\sqrt{x}}{2\sqrt{x}}.$

✓Final answer
$\frac{-\sin\sqrt{x}}{2\sqrt{x}}$
CBSE 2026Set A1 markMCQ
$\frac{d}{dx}(\cos x^3) =$
(a) $-3x^2\sin x^3$
(b) $\sin x^3$
(c) $3x^2\sin x^3$
(d) $3x^2$
›Reveal solutionSolution

Chain rule on $\cos(x^3)$ gives $-3x^2\sin x^3$ .

Let the inner function be $u=x^3$ , so $y=\cos u$ .

$\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}=(-\sin u)(3x^2)=-3x^2\sin x^3$ .

✓Final answer
$-3x^2\sin x^3$
CBSE 2026Set ANNUAL1 markMCQ
$\log(\cos e^x)$
(a) $-\tan e^x$
(b) $e^x\tan e^x$
(c) $-e^x\tan e^x$
(d) $\tan e^x$
›Reveal solutionSolution

Using the chain rule twice, $\dfrac{d}{dx}\log(\cos e^x)=-e^x\tan e^x$ .

Let $y=\log(\cos e^x)$ . We differentiate using the chain rule, working from the outside in.

Step 1: Differentiate $\log(u)$ where $u=\cos e^x$ :

$\frac{dy}{dx}=\frac{1}{\cos e^x}\cdot\frac{d}{dx}(\cos e^x)$

Step 2: Differentiate $\cos(v)$ where $v=e^x$ :

$\frac{d}{dx}(\cos e^x)=-\sin(e^x)\cdot\frac{d}{dx}(e^x)=-\sin(e^x)\cdot e^x$

Step 3: Combine:

$\frac{dy}{dx}=\frac{-e^x\sin e^x}{\cos e^x}=-e^x\tan e^x$

✓Final answer
$-e^x\tan e^x$
CBSE 2026Set ANNUAL1 markMCQ
$y = \cos^{-1}\left(\frac{1-x^2}{1+x^2}\right)$
(a) $\frac{1}{1+x^2}$
(b) $\frac{2}{4+x}$
(c) $\frac{2}{1+x^2}$
(d) $\frac{2}{x+x^2}$
›Reveal solutionSolution

Put $x=\tan\phi$ so the expression simplifies to $\cos^{-1}(\cos2\phi)=2\phi=2\tan^{-1}x$ , whose derivative is standard.

Let $x=\tan\phi$ . Then $\dfrac{1-x^2}{1+x^2} = \dfrac{1-\tan^2\phi}{1+\tan^2\phi} = \cos2\phi$ .

So $y=\cos^{-1}(\cos2\phi)=2\phi=2\tan^{-1}x$ (valid for $0<x<1$ , i.e. $0<\phi<\pi/4$ , so $2\phi$ is in the principal range).

$\dfrac{dy}{dx} = \dfrac{2}{1+x^2}$ .

✓Final answer
$\dfrac{2}{1+x^2}$
CBSE 2026Set ANNUAL1 mark
$y = e^x + e^{x^2} + \ldots + e^{x^5}$
›Reveal solutionSolution

Differentiate each term $e^{x^n}$ using the chain rule: $\dfrac{d}{dx}e^{x^n}=nx^{n-1}e^{x^n}$ .

$y=e^x+e^{x^2}+e^{x^3}+e^{x^4}+e^{x^5}$

$\dfrac{dy}{dx}=e^x+2xe^{x^2}+3x^2e^{x^3}+4x^3e^{x^4}+5x^4e^{x^5}$ .

✓Final answer
$\dfrac{dy}{dx}=e^x+2xe^{x^2}+3x^2e^{x^3}+4x^3e^{x^4}+5x^4e^{x^5}$
CBSE 2026Set ANNUAL1 markMCQ
$\dfrac{d}{dx}(\cos 3x) =$
(a) $\sin 3x$
(b) $-3\sin 3x$
(c) $\cos 3x$
(d) $-3\cos 3x$
›Reveal solutionSolution

Differentiate $\cos(3x)$ using the chain rule: derivative of $\cos u$ is $-\sin u$ , times $\dfrac{du}{dx}$ .

$\dfrac{d}{dx}(\cos 3x) = -\sin(3x) \cdot \dfrac{d}{dx}(3x) = -\sin(3x)\cdot 3 = -3\sin 3x$ .

✓Final answer
$-3\sin 3x$
CBSE 2026Set ANNUAL1 markMCQ
$\dfrac{d}{dx} \tan^{-1}(x^2) =$
(a) $\dfrac{2x}{1+x^4}$
(b) $\dfrac{x}{1+x^2}$
(c) $\dfrac{x^3}{1+x^2}$
(d) None of these
›Reveal solutionSolution

Use $\dfrac{d}{dx}\tan^{-1}u = \dfrac{1}{1+u^2}\cdot\dfrac{du}{dx}$ with $u=x^2$ .

$\dfrac{d}{dx}\tan^{-1}(x^2) = \dfrac{1}{1+(x^2)^2}\cdot \dfrac{d}{dx}(x^2) = \dfrac{2x}{1+x^4}$ .

✓Final answer
$\dfrac{2x}{1+x^4}$
CBSE 2026Set ANNUAL1 markMCQ
$\dfrac{d}{dx} e^{\sin x} =$
(a) $e^{\sin x} \cdot \cos x$
(b) $e^{\sin x}$
(c) $\cos x$
(d) None of these
›Reveal solutionSolution

Differentiate $e^{u}$ with $u=\sin x$ : derivative is $e^u \cdot \dfrac{du}{dx}$ .

$\dfrac{d}{dx}e^{\sin x} = e^{\sin x} \cdot \dfrac{d}{dx}(\sin x) = e^{\sin x}\cos x$ .

✓Final answer
$e^{\sin x}\cdot\cos x$
CBSE 2026Set ANNUAL1 markMCQ
$\dfrac{d}{dx}(\log \sec x) =$
(a) $\tan x$
(b) $\cot x$
(c) $\csc x$
(d) None of these
›Reveal solutionSolution

Differentiate $\log(\sec x)$ using $\dfrac{d}{dx}\log u = \dfrac{u'}{u}$ with $u=\sec x$ .

$\dfrac{d}{dx}\log(\sec x) = \dfrac{1}{\sec x}\cdot \dfrac{d}{dx}(\sec x) = \dfrac{1}{\sec x}\cdot \sec x \tan x = \tan x$ .

✓Final answer
$\tan x$

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