Ex 5.2 Class 12 Maths Ncert Solutions

Given: \( \sin \left(x^{2}+5\right) \)
Let \( y=\sin \left(x^{2}+5\right) \)
\(=\frac{d y}{d x}=\frac{d}{d x} \sin \left(x^{2}+5\right)\)
\(=\cos \left(x^{2}+5\right) \cdot \frac{d}{d x} \sin \left(x^{2}+5\right)\)
\(=\cos \left(x^{2}+5\right) \cdot\left[\frac{d}{d x}(x)^{2}+\frac{d}{d x}(5)\right]\)
\(=\cos \left(x^{2}+5\right) \cdot(2 x+0)\)
\(=\cos \left(x^{2}+5\right) \cdot(2 x)\)
\(=2 x \cdot \cos \left(x^{2}+5\right)\)

Given: \( \cos (\sin x) \)
Let \( y=\cos (\sin x) \)
\(=\frac{d y}{d x}=\frac{d}{d x}(\cos (\sin x))\)
\(=-\sin (\sin x) \cdot \frac{d}{d x}(\sin x)\)
\(=-\sin (\sin x) \cdot \cos x\)
\(=-\cos x \cdot \sin (\sin x)\)

Given: \( \sin (a x+b) \)
Let \( y=\sin (\mathrm{a}x+\mathrm{b}) \)
\(=\frac{d y}{d x}=\frac{d}{d x}(\sin (\mathrm{a}x+\mathrm{b}))\)
\(=\cos (\mathrm{a}x+\mathrm{b}) \cdot \frac{d}{d x}(\mathrm{a}x+\mathrm{b})\)
\(=\cos (\mathrm{a}x+\mathrm{b}) \cdot\left(\frac{d}{d x}\left(\mathrm{a}x+\frac{d}{d x}(b)\right)\right.\)
\(=\cos (\mathrm{a}x+\mathrm{b}) \cdot(\mathrm{a}+0)\)
\(=\cos (\mathrm{a}x+\mathrm{b}) \cdot(\mathrm{a})\)
\(=\mathrm{a} \cdot \cos (\mathrm{a}x+\mathrm{b})\)

Given: \( \sec (\tan (\sqrt{x})) \)
Let \( y=\sec (\tan (\sqrt{x})) \)
\( =\frac{d y}{d x}=\frac{d}{d x}(\sec (\tan (\sqrt{x}))) \)
\( =\sec (\tan (\sqrt{x})) \cdot \tan (\tan (\sqrt{x}))\left(\frac{d}{d x}(\tan \sqrt{x})\right) \)
\( =\sec (\tan (\sqrt{x})) \cdot \tan (\tan (\sqrt{x})) \cdot \operatorname{Sec}^{2}(\sqrt{x}) \cdot \frac{d}{d x}(\sqrt{x}) \)
\( \left.=\sec (\tan (\sqrt{x})) \cdot \tan (\tan (\sqrt{x})) \cdot \sec ^{2}(\sqrt{x})\right) \cdot \frac{1}{2(\sqrt{x})} \)
\( =\frac{1}{2(\sqrt{x})}\left(\sec (\tan (\sqrt{x})) \cdot \tan (\tan (\sqrt{x})) \cdot \sec ^{2}(\sqrt{x})\right) \)

Given: \( \frac{\sin (a x+b)}{\cos (c x+d)} \)
Let \( y=\frac{\sin (a x+b)}{\cos (c x+d)} \)
\(=\frac{d y}{d x}=\frac{d}{d x}\left(\frac{\sin (a x+b)}{\cos (c x+d 0}\right)\)
We know that \( \frac{d}{d x}\left(\frac{u}{v}\right)=\frac{v d(u)-u d(v)}{v^{2}} \)
\(=\frac{[\cos (c x+d) \cdot d(\sin (a x+b))-\sin (\mathrm{ax}+\mathrm{b}) \cdot \mathrm{d}(\cos (c x+d))]}{[\cos (c x+d)]^{2}}\)
\(=\frac{[\cos (c x+d) \cdot(\cos (a x+b)) \cdot d(a x+b)-\sin (a x+b) \cdot(-\sin (c x+d) d(c x+d)]}{\left[\cos (c x+d]^{2}\right.}\)
\(=\frac{[\cos (c x+d) \cdot(\cos (a x+b)) \cdot(a)-\sin (a x+b) \cdot(-\sin (c x+d)(c)]}{[\cos (c x+d)]^{2}}\)
\(=\frac{[a \cos (c x+d) \cos (a x+b)]}{[\cos (c x+d)]^{2}}+\frac{[c \sin (c x+d) \sin (a x+b)}{[\cos (c x+d)]^{2}}\)
\(=\frac{[a \cos (a x+b)]}{[\cos (c x+d)]}+\frac{[c \sin (c x+d) \sin (a x+b)]}{[\cos (c x+d)] \cos (c x+d)]}\)
\(=\mathrm{a} \cos (\mathrm{a}x+\mathrm{b}) \sec (c x+d)+\mathrm{c} \mathrm{sin}(\mathrm{a}x+\mathrm{b}) \tan (\mathrm{c} x+\mathrm{d}) \sec (\mathrm{c} x+\mathrm{d})\)

Given: \( \cos x^{3} \cdot \sin ^{2}\left(x^{5}\right) \)
Let \( y=\cos x^{3} \cdot \sin ^{2}\left(x^{5}\right) \)
\(=\frac{d y}{d x}=\frac{d}{d x}\left(\cos x^{3} \cdot \sin ^{2}\left(x^{5}\right)\right)\)
We know that, \( \frac{d y}{d x}(\mathrm{u} . \mathrm{v})=\mathrm{u} . \mathrm{d}(\mathrm{v})+\mathrm{v} . \mathrm{d}(\mathrm{u}) \)
\(=\cos x^{3} \cdot \frac{d}{d x} \sin ^{2}\left(x^{5}\right)+\sin ^{2}\left(x^{5}\right) \cdot \frac{d}{d x}\left(\cos x^{3}\right)\)
\(=\cos x^{3} \cdot 2 \sin \left(x^{5}\right) \cdot\left(\frac{d}{d x} \sin \left(x^{5}\right)\right)+\sin 2\left(x^{5}\right) \cdot\left(-\sin x^{3}\right) \cdot\left(\frac{d}{d x} x^{3}\right)\)
\(=\cos x^{3} \cdot 2 \sin \left(x^{5}\right) \cdot \cos \left(x^{5}\right)\left(\frac{d}{d x} x^{5}\right)+\sin 2\left(x^{5}\right) \cdot\left(-\sin x^{3}\right) \cdot\left(3 x^{2}\right)\)
\(=\cos x^{3} \cdot 2 \sin \left(x^{5}\right) \cdot \cos \left(x^{5}\right)\left(5 x^{4}\right)+\sin 2\left(x^{5}\right) \cdot\left(-\sin x^{3}\right) \cdot\left(3 x^{2}\right)\)
\(=10 x^{4} \cdot \cos x^{3} \cdot \sin \left(x^{5}\right) \cdot \cos \left(x^{5}\right)-\left(3 x^{2}\right) \cdot \sin 2\left(x^{5}\right) \cdot\left(\sin x^{3}\right)\)

Let \( y=2 \sqrt{\cos \left(x^{3}\right)} \)
Apply derivative both the sides with respect to \( x \).\(\frac{d y}{d x}-2 \cdot \frac{1}{2}\left\{\cot \left(x^{2}\right)\right\}^{\frac{-1}{2}} \cdot \frac{d}{d x} \cot \left(x^{2}\right)\)
\(=\frac{1}{\sqrt{\cot \left(x^{2}\right)}} \cdot\left(-\cos \theta c\left(x^{2}\right)\right) \frac{d}{d x} x^{2}\)
\(=\frac{1}{\sqrt{\cot \left(x^{2}\right)}}\left(-\cos \theta c\left(x^{2}\right)\right\}(2 x)\)
\(=\frac{-2 x \operatorname{cosec}\left(x^{2}\right)}{\sqrt{\cot \left(x^{2}\right)}} \)

Given: \( \cos \sqrt{x} \)
Let \( y=\cos \sqrt{x} \)
\(=\frac{d y}{d x}=\frac{d}{d x}(\cos \sqrt{x})\)
\(=-\sin (\sqrt{x}) \cdot\left(\frac{d}{d x} \sqrt{x}\right)\)
\(=-\sin (\sqrt{x}) \cdot \frac{1}{2} \cdot\left(x^{-\frac{1}{2}}\right)\)
\(=-\sin (\sqrt{x}) \cdot \frac{1}{2 \sqrt{x}}\)
\(=-\frac{\sin (\sqrt{x})}{2 \sqrt{x}}\)

Given: \( \mathrm{f}(x)=|x-1|, x \in \mathrm{R} \)
because a function f is differentiable at a point \( x=\mathrm{c} \) in its domain if both its limits as:
\( \lim _{h \rightarrow 0^{-}} \frac{[f(c+h)-f(c)]}{h} \) and \( \lim _{h \rightarrow 0^{+}} \frac{[f(c+h)-f(c)]}{h} \) are finite and equal.
Now, to check the differentiability of the given function at \( x=1 \),
Let we consider the left hand limit of function f at \( x=1 \)
\(=\lim _{h \rightarrow 0^{-}} \frac{[f(1+h)-f(1)]}{h}\)
\(=\lim _{h \rightarrow 0^{-}}[|1+h-1|-|1-1|]\)
\(=\lim _{h \rightarrow 0^{-}} \frac{[|h|-0]}{h}\)
\(=\lim _{h \rightarrow 0^{-}} \frac{[-h]}{h} \text { because, }\{\mathrm{h} < 0 \Rightarrow|\mathrm{h}|=-\mathrm{h}\}\)
\(=-1\)
Now, let we consider the right hand limit of function f at \( x=1 \)
\(=\lim _{h \rightarrow 0^{+}}[f(1+h)-f(1)]\)
\(=\lim _{h \rightarrow 0^{+}}[|1+h-1|-|1-1|]\)
\(=\lim _{h \rightarrow 0^{+}}[|h|-0]\)
\(=\lim _{h \rightarrow 0^{+}} \frac{[h]}{h} \text { because, }\{\mathrm{h} > 0 \Rightarrow \mid \mathrm{h}=\mathrm{h}\}\)
\(=1\)
Because, left hand limit is not equal to right hand limit of function f at \( x=1 \), so f is not differentiable at \( x=1 \).

Given: \( \mathrm{f}(x)=[x], 0 < x < 3 \)
because a function f is differentiable at a point \( x=\mathrm{c} \) in its domain if both its limits as:
\( \lim _{h \rightarrow 0^{-}} \frac{[f(c+h)-f(c)]}{h} \) and \( \lim _{h \rightarrow 0^{+}} \frac{[f(c+h)-f(c)]}{h} \) are finite and equal.
Now, to check the differentiability of the given function at \( x=1 \),
Let we consider the left-hand limit of function f at \( x=1 \)
\(=\lim _{h \rightarrow 0^{-}} \frac{[f(1+h)-f(1)]}{h}\)
\(=\lim _{h \rightarrow 0^{-}} \frac{[|1+h|-|1|]}{h}\)
\(=\lim _{h \rightarrow 0^{-}} \frac{[1+h-1-1]}{h}\)
\(=\lim _{h \rightarrow 0^{-}} \frac{[h-1]}{h} \text { because, }\{\mathrm{h} < 0= > |\mathrm{h}|=-\mathrm{h}\}\)
\(=-\frac{1}{0}=\infty\)
Let we consider the right hand limit of function f at \( x=1 \)
\(=\lim _{h \rightarrow 0^{+}} \frac{[f(1+h)-f(1)]}{h}\)
\(=\lim _{h \rightarrow 0^{+}} \frac{[1+h]-[1]}{h}\)
\(=\lim _{h \rightarrow 0^{+}} \frac{[1-1]}{h}\)
\(=\lim _{h \rightarrow 0^{+}} \frac{[0]}{h}\)
\(=0\)
Because, left hand limit is not equal to right hand limit of function f at \( x=1 \), so f is not differentiable at \( x=1 \).
Let we consider the left hand limit of function f at \( x=2 \)
\(=\lim _{h \rightarrow 0^{-}} \frac{[f(2+h)-f(2)]}{h}\)
\(=\lim _{h \rightarrow 0^{-}} \frac{[2+h]-[2]}{h}\)
\(=\lim _{h \rightarrow 0^{-}} \frac{[2+h-1-2]}{h}\)
\(=\lim _{h \rightarrow 0^{-}} \frac{[h+1-2]}{h}\)
\(=-\frac{1}{0}=\infty\)
Now, let we consider the right hand limit of function f at \( x=2 \)
\(=\lim _{h \rightarrow 0^{+}} \frac{[f(2+h)-f(2)]}{h}\)
\(=\lim _{h \rightarrow 0^{+}} \frac{[2+h]-[2]}{h}\)
\(=\lim _{h \rightarrow 0^{+}} \frac{[2-2]}{h}\)
\(=\lim _{h \rightarrow 0^{+}} \frac{[0]}{h}\)
\(=0\)
Because, left hand limit is not equal to right hand limit of function f at \( x=2 \), so f is not differentiable at \( x=2 \).