Class 12 Maths NCERT Solutions Chapter 5 Ex 5.2 – The Chain Rule

Let y=\sin(x^2+5). This is a composite of u(x)=x^2+5 and v(t)=\sin t.

By the chain rule: \dfrac{dy}{dx}=\cos(x^2+5)\cdot\dfrac{d}{dx}(x^2+5)=\cos(x^2+5)\cdot 2x.

Let y=\cos(\sin x), a composite of u(x)=\sin x and v(t)=\cos t.

By the chain rule: \dfrac{dy}{dx}=-\sin(\sin x)\cdot\dfrac{d}{dx}(\sin x)=-\sin(\sin x)\cdot\cos x.

Let y=\sin(ax+b), a composite of u(x)=ax+b and v(t)=\sin t.

By the chain rule: \dfrac{dy}{dx}=\cos(ax+b)\cdot\dfrac{d}{dx}(ax+b)=\cos(ax+b)\cdot a.

This is a chain of three functions: u=\sqrt{x}, v=\tan u, y=\sec v.

Differentiate outward: \dfrac{dy}{dv}=\sec v\tan v, \dfrac{dv}{du}=\sec^2 u, \dfrac{du}{dx}=\dfrac{1}{2\sqrt{x}}.

Multiply by the extended chain rule: \dfrac{dy}{dx}=\sec(\tan\sqrt{x})\tan(\tan\sqrt{x})\cdot\sec^2(\sqrt{x})\cdot\dfrac{1}{2\sqrt{x}}.

Let u=\sin(ax+b) and v=\cos(cx+d). Then u'=a\cos(ax+b) and v'=-c\sin(cx+d) (chain rule on each).

Apply the quotient rule: \dfrac{dy}{dx}=\dfrac{u'v-uv'}{v^2}.

\dfrac{dy}{dx}=\dfrac{a\cos(ax+b)\cos(cx+d)-\sin(ax+b)\big(-c\sin(cx+d)\big)}{\cos^2(cx+d)}.

Let y=\cos(x^3)\cdot\sin^2(x^5). Use the product rule with each factor differentiated by the chain rule.

\dfrac{d}{dx}\cos(x^3)=-\sin(x^3)\cdot 3x^2=-3x^2\sin(x^3).

\dfrac{d}{dx}\sin^2(x^5)=2\sin(x^5)\cos(x^5)\cdot 5x^4=10x^4\sin(x^5)\cos(x^5).

Product rule: \dfrac{dy}{dx}=-3x^2\sin(x^3)\sin^2(x^5)+\cos(x^3)\cdot 10x^4\sin(x^5)\cos(x^5).

Let y=2\left[\cot(x^2)\right]^{1/2}.

\dfrac{dy}{dx}=2\cdot\dfrac{1}{2}\left[\cot(x^2)\right]^{-1/2}\cdot\dfrac{d}{dx}\cot(x^2).

\dfrac{d}{dx}\cot(x^2)=-\csc^2(x^2)\cdot 2x=-2x\csc^2(x^2).

Substitute: \dfrac{dy}{dx}=\left[\cot(x^2)\right]^{-1/2}\cdot\big(-2x\csc^2(x^2)\big).

Let y=\cos(\sqrt{x}), a composite of u(x)=\sqrt{x} and v(t)=\cos t.

By the chain rule: \dfrac{dy}{dx}=-\sin(\sqrt{x})\cdot\dfrac{1}{2\sqrt{x}}.

Left hand derivative: \text{LHD}=\lim_{h\to 0^-}\dfrac{f(1+h)-f(1)}{h}=\lim_{h\to 0^-}\dfrac{|h|-0}{h}.

For h \lt 0, |h|=-h, so \text{LHD}=\lim_{h\to 0^-}\dfrac{-h}{h}=-1.

Right hand derivative: \text{RHD}=\lim_{h\to 0^+}\dfrac{|h|}{h}=\lim_{h\to 0^+}\dfrac{h}{h}=1 (since h \gt 0 \Rightarrow |h|=h).

At x = 1: \text{RHD}=\lim_{h\to 0^+}\dfrac{[1+h]-[1]}{h}. For small h \gt 0, 1 \lt 1+h \lt 2, so [1+h]=1, giving \text{RHD}=\dfrac{1-1}{h}=0.

\text{LHD}=\lim_{h\to 0^-}\dfrac{[1+h]-[1]}{h}.

For small h \lt 0, 0 \lt 1+h \lt 1, so [1+h]=0, giving \text{LHD}=\dfrac{0-1}{h}=-\dfrac{1}{h}.

This grows without bound as h\to 0^- — the limit does not exist.

Since LHD does not exist (and certainly ≠ RHD), f is not differentiable at x = 1.

At x = 2: by an identical argument, \text{RHD}=\lim_{h\to 0^+}\dfrac{[2+h]-[2]}{h}=\dfrac{2-2}{h}=0, while \text{LHD}=\lim_{h\to 0^-}\dfrac{[2+h]-[2]}{h}=\dfrac{1-2}{h}=-\dfrac{1}{h}, which does not exist as a finite limit.