# Ncert Solutions For Class 12 Maths Ex 5.2

## Ncert Solutions For Class 12 Maths Chapter 5 Ex 5.2

Q1: Differentiate the function with respect to $$x$$

$$f(x)=\ sin (x^{2}+5)$$

Sol:The given function is $$f(x)=\ sin (x^{2}+5)=y$$

Let $$t=(x^{2}+5)$$

$$So f(t)=\sin t$$

$$\frac{\mathrm{d} y}{\mathrm{d} x}=\frac{\mathrm{d} y}{\mathrm{d} t}\times \frac{\mathrm{d} t}{\mathrm{d} x}$$ ……..(i)

$$\frac{\mathrm{d} y}{\mathrm{d} t}=\cos t$$ ……..(ii)

And $$\frac{\mathrm{d} t}{\mathrm{d} x}=2x$$ …….(iii)

Substituting equation (ii) and (iii) in (i) we have,

$$\frac{\mathrm{d} y}{\mathrm{d} x}=\cos t \times (2x)$$

And we know $$t=x^{2}+5$$

Thus $$\frac{\mathrm{d} y}{\mathrm{d} x}=(2x)\times \cos x$$

Q2: Differentiate the function with respect to $$x$$

$$f(x)=\cos (\sin x)$$

Sol:

Let $$f(x)=\cos (\sin x)$$

Here $$f$$ is a composite function which can be written in the form of two composite function $$u and v$$.

$$u(x)=\sin x\; \; and\; \; v(t)=\cos t$$ $$(vou)(x)=v(u(x))=v(\sin x)=\cos (\sin x)=f(x)$$

Put $$t=u(x)=\sin x$$

$$therefore,\; \frac{\mathrm{d} v}{\mathrm{d} t}=\frac{\mathrm{d} (\cos t)}{\mathrm{d} x}=-\sin t = -\sin(\sin x)$$ $$\frac{\mathrm{d} t}{\mathrm{d} x}=\frac{\mathrm{d} (\sin x)}{\mathrm{d} x}=\cos x$$

By chain rule ,

$$\frac{\mathrm{d} f}{\mathrm{d} x}=\frac{\mathrm{d} v}{\mathrm{d} t}\times \frac{\mathrm{d} t}{\mathrm{d} x}$$ $$\frac{\mathrm{d} f}{\mathrm{d} x}=-\sin (\sin x)\times \ cos x=-\cos x\sin (\sin x)$$

Q3: Differentiate the function with respect to $$x$$

$$f(x)=\sin (ax+b)$$

Sol:

Let $$f(x)=\sin (ax+b)$$

Here $$f$$ is a composite function which can be written in the form of two composite function $$u and v$$.

$$u(x)=ax+b\; \; and\; \; v(t)=\sin t$$ $$(vou)(x)=v(u(x))=v(ax+b)=\sin (ax+b)=f(x)$$

Put $$t=u(x)=ax+b$$

$$therefore,\; \frac{\mathrm{d} v}{\mathrm{d} t}=\frac{\mathrm{d} (\sin t)}{\mathrm{d} x}=\cos t = \cos(ax+b)$$ $$\frac{\mathrm{d} t}{\mathrm{d} x}=\frac{\mathrm{d} (ax+b)}{\mathrm{d} x}=a$$

By chain rule ,

$$\frac{\mathrm{d} f}{\mathrm{d} x}=\frac{\mathrm{d} v}{\mathrm{d} t}\times \frac{\mathrm{d} t}{\mathrm{d} x}$$ $$\frac{\mathrm{d} f}{\mathrm{d} x}=a\times \ cos (ax+b)$$

Q4: Differentiate the function with respect to $$x$$

$$\sec (\tan (\sqrt{x}))$$.

Sol:

Let $$\sec (\tan (\sqrt{x}))$$

Here $$f$$ is a composite function which can be written in the form of three composite function $$u,v and w$$.

$$u(x)=\sqrt{x}\; \; v(t)=\tan t\; \; and w(s)=\sec s$$ $$(wovou)(x)=w[v(u(x))]=w[v(\sqrt{x})]=w[\tan( \sqrt{x})]=\sec (\tan (\sqrt{x}))=f(x)$$

Put $$s=v(t)=\tan t and t=u(x)=\sqrt{x}$$

then ,$$\frac{\mathrm{d} w}{\mathrm{d} s}=\frac{\mathrm{d} (\sec s)}{\mathrm{d} s}=\sec s\tan s =\sec(\tan t). \tan (\tan t)$$ ( as $$s=\tan t$$

$$=\sec(\tan \sqrt{x}). \tan (\tan \sqrt{x})$$ $$\frac{\mathrm{d} s}{\mathrm{d} t}=\frac{\mathrm{d} (\tan t)}{\mathrm{d} t}=\sec ^{2}t=\sec ^{2}\sqrt{t}$$ $$\frac{\mathrm{d} t}{\mathrm{d} x}=\frac{\mathrm{d} \sqrt{x}}{\mathrm{d} x}=\frac{\mathrm{d} (x^{\frac{1}{2}})}{\mathrm{d} x}=\frac{1}{2}.x^{\frac{1}{2}-1}=\frac{1}{2\sqrt{x}}$$

By chain rule ,

$$\frac{\mathrm{d} t}{\mathrm{d} x}=\frac{\mathrm{d} w}{\mathrm{d} s}\times \frac{\mathrm{d} s}{\mathrm{d} t}\times \frac{\mathrm{d} t}{\mathrm{d} x}$$

=$$\sec (\tan\sqrt{x}).\tan (\tan\sqrt{x})\times \sec ^{2} \sqrt{x}\times \frac{1}{2\sqrt{x}}$$

=$$\frac{\sec ^{2}\sqrt{x}.\sec (\tan \sqrt{x}.\tan (\tan \sqrt{x}))}{2\sqrt{x}}$$

Q5: Differentiate the function with respect to $$x$$

$$f(x)=\frac{\sin (ax+b)}{\cos(cx+b)}$$

Sol:

The given function is $$f(x)=\frac{\sin (ax+b)}{\cos(cx+b)}= \frac{g(x)}{h(x)}$$,

where $$g(x)= \sin (ax+b)$$ and $$h(x)=\cos (cx+d)$$

Consider $$g(x)=\sin (ax+b)$$

Here $$g$$ is a composite function which can be written in the form of two composite function $$u and v$$.

$$u(x)=ax+b\; \; v(t)=\sin t$$ $$(vou)(x)=v(u(x))=v(ax+b)=\sin (ax+b)]=g(x)$$

Put $$t=u(x)=ax+b$$

$$\frac{\mathrm{d} v}{\mathrm{d} t}=\frac{\mathrm{d} (\sin t)}{\mathrm{d} t}=\cos t=\cos (ax+b)$$ $$\frac{\mathrm{d} t}{\mathrm{d} x}=\frac{\mathrm{d} (ax+b)}{\mathrm{d} x}=a$$

By chain rule ,

$$g^{‘}=\frac{\mathrm{d} g}{\mathrm{d} x}=\frac{\mathrm{d} v}{\mathrm{d} t}\times \frac{\mathrm{d} t}{\mathrm{d} x}=\cos (ax+b).a=a \cos (ax+b)$$

Consider $$h(x)=\cos (cx+d)$$

Here $$h$$ is a composite function which can be written in the form of two composite function $$p and q$$.

$$p(x)=cx+d\; \; q(y)=\cos y$$ $$(qop)(x)=q(p(x))=q(cx+d)=\cos (cx+d)]=h(x)$$

Put $$y=p(x)=cx+d$$

$$\frac{\mathrm{d} q}{\mathrm{d} y}=\frac{\mathrm{d} (\cos y)}{\mathrm{d} y}=-\sin y=-\sin (cx+d)$$ $$\frac{\mathrm{d} y}{\mathrm{d} x}=\frac{\mathrm{d} (cx+d)}{\mathrm{d} x}=c$$

By chain rule ,

$$h^{‘}=\frac{\mathrm{d} h}{\mathrm{d} x}=\frac{\mathrm{d} q}{\mathrm{d} y}\times \frac{\mathrm{d} y}{\mathrm{d} x}=-\sin (cx+d).c=-c \sin (cx+d)$$

Therefore by chain rule , we obtain

$$f^{‘}=\frac{a\cos (ax+b).\ cos (cx+d)-\sin (ax+b)(-\sin(cx+d))}{[\cos(cx+d)]^{2}}$$

=$$=\frac{a\cos (ax+b)}{\ cos (cx+d)}+c\sin (ax+b).\frac{\sin (cx+d)}{\cos (cx+d)}\times \frac{1}{\cos (cx+d)}$$

=$$=a\cos (ax+b)\ sec (cx+d)+c\sin (ax+b).\tan (cx+d).\sec (cx+d)$$

Q6: Differentiate the function with respect to $$x$$

$$f(x)=\cos x^{3}.\sin ^{2}(x^{5})$$

Sol:

The given function is $$f(x)=\cos x^{3}.\sin ^{2}(x^{5})$$

$$\frac{\mathrm{d} }{\mathrm{d} x}\left [ \cos x^{3}.\sin ^{2}(x^{5}) \right ]=\sin ^{2}(x^{5})\times \frac{\mathrm{d} }{\mathrm{d} x}\cos x^{3}+\cos x^{3}\times \frac{\mathrm{d} }{\mathrm{d} x}\sin ^{2}(x^{5})$$

=$$\sin ^{2}(x^{5})\times (-\sin x^{3})\times \frac{\mathrm{d} }{\mathrm{d} x}(x^{3})+\cos x^{3}\times 2\sin (x^{5})\times \frac{\mathrm{d} }{\mathrm{d} x}\left [ \sin (x^{5}) \right ]$$

$$=-\sin x^{3}.\sin ^{2}(x^{5}) \times 3x^{2}+2\sin (x^{5}).\cos x^{3}.\cos x^{5}\frac{\mathrm{d} }{\mathrm{d} x} (x^{5})$$ $$=-\sin x^{3}.\sin ^{2}(x^{5}) \times 3x^{2}+2\sin (x^{5}).\cos x^{3}.\cos x^{5}\times 5x^{4}$$ $$=10x^{4}\sin x^{5}.\cos x^{3}.\cos x^{5}\times -3x^{2}\sin x^{3}.\sin ^{2}(x^{5} )$$

Q7: Differentiate the functions with respect to x.

$$2\sqrt{\cot (x^{2})}$$

Sol:

The given function is $$2\sqrt{\cot (x^{2})}$$

$$\frac{\mathrm{d} }{\mathrm{d} x}2\sqrt{\cot (x^{2})}$$ $$=2.\frac{1}{2\sqrt{\cot (x^{2})}}\times \frac{\mathrm{d} }{\mathrm{d} x}\left [ \cot (x^{2}) \right ]$$ $$=\sqrt{\frac{\sin (x^{2})}{\cos (x^{2})}}\times -cosec^{2} (x^{2})\times \frac{\mathrm{d} }{\mathrm{d} x}(x^{2})$$ $$=-\sqrt{\frac{\sin (x^{2})}{\cos (x^{2})}}\times \frac{1}{\sin ^{2(x^{2})}}\times(2x)$$ $$=\frac{-2x}{\sqrt{\cos (x^{2})}\sqrt{\sin (x^{2})}\sin (x^{2})}$$ $$=\frac{-2\sqrt{2}x}{\sqrt{2\cos (x^{2})\sin (x^{2})}\sin (x^{2})}$$ $$=\frac{-2\sqrt{2}x}{\sin (x^{2})\sqrt{\sin 2(x^{2})}}$$

Q8: Differentiate the functions with respect to x.

$$\cos\sqrt{x}$$

Sol:

The given function $$f(x)$$ is $$\cos\sqrt{x}$$.

Let $$u(x)=\sqrt{x}$$

And $$v(t)=\cos t$$

$$(vou)(x)=v(u(x))$$ $$=v(\sqrt{x})$$ $$=\cos (\sqrt{x})$$

=$$f(x)$$

Clearly, $$f$$ is a composite function of two functions, $$u\; and \; v$$, such that

$$t=u(x)=\sqrt{x}$$

Then, $$\frac{\mathrm{d} t}{\mathrm{d} x}=\frac{\mathrm{d} }{\mathrm{d} x}(\sqrt{x})=\frac{\mathrm{d} }{\mathrm{d} x}(x^{\frac{1}{2}})=\frac{1}{2}x^{-\frac{1}{2}}$$

$$=\frac{1}{2\sqrt{x}}$$

And, $$\frac{\mathrm{d} v}{\mathrm{d} t}=\frac{\mathrm{d} }{\mathrm{d} t}(\cos t)=-\sin t$$

$$=-\sin (\sqrt{x})$$

By chain rule we have,

$$\frac{\mathrm{d} v}{\mathrm{d} x}=\frac{\mathrm{d} v}{\mathrm{d} t}.\frac{\mathrm{d} t}{\mathrm{d} x}$$.

$$=-\sin (\sqrt{x}).\frac{1}{2\sqrt{x}}$$ $$=-\frac{1}{2\sqrt{x}}\sin (\sqrt{x})$$ $$=-\frac{\sin (\sqrt{x})}{2\sqrt{x}}$$

Q9: Prove that the function f given by $$f(x)=\left | x-1 \right |,x\in \mathbb{R}$$, is not differentiable at $$x=1$$.

Sol:

The given function is $$f(x)=\left | x-1 \right |,x\in \mathbb{R}$$.

It is known that a function $$f$$ is differentiable at a point $$x=c$$ in its domain if the right hand limit and the left hand limit are finite and equal.

To check the differentiability of the given function at x=1,

The right hand and the left hand limits where x=c are

$$\lim_{h \to 0^{+}}\frac{f(c+h)-f(c)}{h}$$ and $$\lim_{h \to 0^{-}}\frac{f(c+h)-f(c)}{h}$$

Considering the right hand limit of the given function at x=1

$$\lim_{h \to 0^{+}}\frac{f(1+h)-f(1)}{h}$$ $$\lim_{h \to 0^{+}}\frac{f(c+h)-f(c)}{h}=\lim_{h \to 0^{+}}\frac{\left | 1+h-1 \right |-\left | 1-1 \right |}{h}$$ $$=\lim_{h \to 0^{+}}\frac{\left | h \right |-\left | 0 \right |}{h}$$ $$=\lim_{h \to 0^{+}}\frac{h}{h}$$ $$=1$$

Considering the left hand limit of the given function at x=1

$$\lim_{h \to 0^{-}}\frac{f(1+h)-f(1)}{h}$$ $$\lim_{h \to 0^{-}}\frac{f(c+h)-f(c)}{h}=\lim_{h \to 0^{-}}\frac{\left | 1+h-1 \right |-\left | 1-1 \right |}{h}$$ $$=\lim_{h \to 0^{-}}\frac{\left | h \right |-\left | 0 \right |}{h}$$ $$=\lim_{h \to 0^{-}}\frac{-h}{h}$$ $$=-1$$

Since the left and right hand limits of f at x = 1 are not equal, f is not differentiable at x = 1

Q10: Prove that the greatest integer function defined by $$f(x)=\left [ x \right ], 0<x<3$$ is not differentiable at x = 1 and x = 2.

Sol:

The function f is $$f(x)=\left [ x \right ], 0<x<3$$

It is known that a function f is differentiable at a point $$x=c$$ in its domain if both the left hand and the left hand limit are equal

$$\lim_{h \to 0^{+}}\frac{f(c+h)-f(c)}{h}$$ and $$\lim_{h \to 0^{-}}\frac{f(c+h)-f(c)}{h}$$ are finite and equal.

To check the differentiability of the given function at $$x=1$$, consider the right hand limit of f at $$x=1$$

$$\lim_{h \to 0^{+}}\frac{f(1+h)-f(1)}{h}$$ $$\lim_{h \to 0^{+}}\frac{\left [ 1+h \right ]-\left [ 1 \right ]}{h}$$ $$=\lim_{h \to 0^{+}}\frac{1-1}{h}$$ $$=\lim_{h \to 0^{+}}\frac{1-1}{h}=\lim_{h \to 0^{+}}(0)=0$$

Now consider the left hand limit of f at $$x=1$$

$$\lim_{h \to 0^{-}}\frac{f(1+h)-f(1)}{h}$$ $$\lim_{h \to 0^{-}}\frac{\left [ 1+h \right ]-\left [ 1 \right ]}{h}$$ $$=\lim_{h \to 0^{-}}\frac{0-1}{h}$$ $$=\lim_{h \to 0^{-}}\frac{-1}{h}=\infty$$

Since left hand and the right hand limit of f at x=1 are not equal, f is not differentiable at x=1.

Now to check the differentiability of the given function at x=2,

consider the left hand limit at x=2.

$$\lim_{h \to 0^{-}}\frac{f(2+h)-f(2)}{h}$$ $$\lim_{h \to 0^{-}}\frac{\left [ 2+h \right ]-\left [ 2 \right ]}{h}$$ $$=\lim_{h \to 0^{-}}\frac{1-2}{h}$$ $$=\lim_{h \to 0^{-}}\frac{-1}{h}=\infty$$

Now consider the right hand limit of f at $$x=2$$

$$\lim_{h \to 0^{+}}\frac{f(2+h)-f(2)}{h}$$ $$\lim_{h \to 0^{+}}\frac{\left [ 2+h \right ]-\left [ 2 \right ]}{h}$$ $$=\lim_{h \to 0^{+}}\frac{2-2}{h}$$ $$=\lim_{h \to 0^{+}}\frac{0}{h}=\lim_{h \to 0^{+}}(0)=0$$

Since the left and right hand limits of f at x = 2 are not equal, f is not differentiable at x = 2