Solution

Let $y=(3 x^{2}-9 x+5)^{9}$

Using chain rule, we obtain

$ \begin{aligned} \frac{d y}{d x} & =\frac{d}{d x}(3 x^{2}-9 x+5)^{9} \\ & =9(3 x^{2}-9 x+5)^{8} \cdot \frac{d}{d x}(3 x^{2}-9 x+5) \\ & =9(3 x^{2}-9 x+5)^{8} \cdot(6 x-9) \\ & =9(3 x^{2}-9 x+5)^{8} \cdot 3(2 x-3) \\ & =27(3 x^{2}-9 x+5)^{8}(2 x-3) \end{aligned} $

Solution

$ \begin{aligned} & \text{ Let } y=\sin ^{3} x+\cos ^{6} x \\ & \therefore \frac{d y}{d x}=\frac{d}{d x}(\sin ^{3} x)+\frac{d}{d x}(\cos ^{6} x) \\ & =3 \sin ^{2} x \cdot \frac{d}{d x}(\sin x)+6 \cos ^{5} x \cdot \frac{d}{d x}(\cos x) \\ & =3 \sin ^{2} x \cdot \cos x+6 \cos ^{5} x \cdot(-\sin x) \\ & =3 \sin x \cos x(\sin x-2 \cos ^{4} x) \end{aligned} $

Solution

Let $y=(5 x)^{3 \cos 2 x}$

Taking logarithm on both the sides, we obtain

$\log y=3 \cos 2 x \log 5 x$

Differentiating both sides with respect to $x$, we obtain

$\frac{1}{y} \frac{d y}{d x}=3[\log 5 x \cdot \frac{d}{d x}(\cos 2 x)+\cos 2 x \cdot \frac{d}{d x}(\log 5 x)]$

$\Rightarrow \frac{d y}{d x}=3 y[\log 5 x(-\sin 2 x) \cdot \frac{d}{d x}(2 x)+\cos 2 x \cdot \frac{1}{5 x} \cdot \frac{d}{d x}(5 x)]$

$\Rightarrow \frac{d y}{d x}=3 y[-2 \sin 2 x \log 5 x+\frac{\cos 2 x}{x}]$

$\Rightarrow \frac{d y}{d x}=3 y[\frac{3 \cos 2 x}{x}-6 \sin 2 x \log 5 x]$

$\therefore \frac{d y}{d x}=(5 x)^{3 \cos 2 x}[\frac{3 \cos 2 x}{x}-6 \sin 2 x \log 5 x]$

Solution

Let $y=\sin ^{-1}(x \sqrt{x})$

Using chain rule, we obtain

$ \begin{aligned} \frac{d y}{d x} & =\frac{d}{d x} \sin ^{-1}(x \sqrt{x}) \\ & =\frac{1}{\sqrt{1-(x \sqrt{x})^{2}}} \times \frac{d}{d x}(x \sqrt{x}) \\ & =\frac{1}{\sqrt{1-x^{3}}} \cdot \frac{d}{d x}(x^{\frac{3}{2}}) \\ & =\frac{1}{\sqrt{1-x^{3}}} \times \frac{3}{2} \cdot x^{\frac{1}{2}} \\ & =\frac{3 \sqrt{x}}{2 \sqrt{1-x^{3}}} \\ & =\frac{3}{2} \sqrt{\frac{x}{1-x^{3}}} \end{aligned} $

Solution

Let $y=\frac{\cos ^{-1} \frac{x}{2}}{\sqrt{2 x+7}}$

By quotient rule, we obtain

$ \begin{aligned} & \frac{d y}{d x}=\frac{\sqrt{2 x+7} \frac{d}{d x}(\cos ^{-1} \frac{x}{2})-(\cos ^{-1} \frac{x}{2}) \frac{d}{d x}(\sqrt{2 x+7})}{(\sqrt{2 x+7})^{2}} \\ & =\frac{\sqrt{2 x+7}[\frac{-1}{\sqrt{1-(\frac{x}{2})^{2}}} \cdot \frac{d}{d x}(\frac{x}{2})]-(\cos ^{-1} \frac{x}{2}) \frac{1}{2 \sqrt{2 x+7}} \cdot \frac{d}{d x}(2 x+7)}{2 x+7} \\ & =\frac{\sqrt{2 x+7} \frac{-1}{\sqrt{4-x^{2}}}-(\cos ^{-1} \frac{x}{2}) \frac{2}{2 \sqrt{2 x+7}}}{2 x+7} \\ & =\frac{-\sqrt{2 x+7}}{\sqrt{4-x^{2}} \times(2 x+7)}-\frac{\cos ^{-1} \frac{x}{2}}{(\sqrt{2 x+7})(2 x+7)} \\ & =-[\frac{1}{\sqrt{4-x^{2}} \sqrt{2 x+7}}+\frac{\cos ^{-1} \frac{x}{2}}{(2 x+7)^{\frac{3}{2}}}] \end{aligned} $

Solution

Let $y=\cot ^{-1}[\frac{\sqrt{1+\sin x}+\sqrt{1-\sin x}}{\sqrt{1+\sin x}-\sqrt{1-\sin x}}]$

Then, $\frac{\sqrt{1+\sin x}+\sqrt{1-\sin x}}{\sqrt{1+\sin x}-\sqrt{1-\sin x}}$

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

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

$=\frac{2+2 \sqrt{1-\sin ^{2} x}}{2 \sin x}$

$=\frac{1+\cos x}{\sin x}$

$=\frac{2 \cos ^{2} \frac{x}{2}}{2 \sin \frac{x}{2} \cos \frac{x}{2}}$

$=\cot \frac{x}{2}$

Therefore, equation (1) becomes

$y=\cot ^{-1}(\cot \frac{x}{2})$

$\Rightarrow y=\frac{x}{2}$

$\therefore \frac{d y}{d x}=\frac{1}{2} \frac{d}{d x}(x)$

$\Rightarrow \frac{d y}{d x}=\frac{1}{2}$

Solution

Let $y=(\log x)^{\log x}$

Taking logarithm on both the sides, we obtain

$\log y=\log x \cdot \log (\log x)$

Differentiating both sides with respect to $x$, we obtain

$\frac{1}{y} \frac{d y}{d x}=\frac{d}{d x}[\log x \cdot \log (\log x)]$

$\Rightarrow \frac{1}{y} \frac{d y}{d x}=\log (\log x) \cdot \frac{d}{d x}(\log x)+\log x \cdot \frac{d}{d x}[\log (\log x)]$

$\Rightarrow \frac{d y}{d x}=y[\log (\log x) \cdot \frac{1}{x}+\log x \cdot \frac{1}{\log x} \cdot \frac{d}{d x}(\log x)]$

$\Rightarrow \frac{d y}{d x}=y[\frac{1}{x} \log (\log x)+\frac{1}{x}]$

$\therefore \frac{d y}{d x}=(\log x)^{\log x}[\frac{1}{x}+\frac{\log (\log x)}{x}]$

Solution

Let $y=\cos (a \cos x+b \sin x)$

By using chain rule, we obtain

$ \begin{aligned} & \frac{d y}{d x}=\frac{d}{d x} \cos (a \cos x+b \sin x) \\ & \begin{aligned} \Rightarrow \frac{d y}{d x} & =-\sin (a \cos x+b \sin x) \cdot \frac{d}{d x}(a \cos x+b \sin x) \\ & =-\sin (a \cos x+b \sin x) \cdot[a(-\sin x)+b \cos x] \\ & =(a \sin x-b \cos x) \cdot \sin (a \cos x+b \sin x) \end{aligned} \end{aligned} $

Solution

Let $y=(\sin x-\cos x)^{(\sin x-\cos x)}$

Taking logarithm on both the sides, we obtain

$\log y=\log [(\sin x-\cos x)^{(\sin x-\cos x)}]$

$\Rightarrow \log y=(\sin x-\cos x) \cdot \log (\sin x-\cos x)$

Differentiating both sides with respect to $x$, we obtain

$\frac{1}{y} \frac{d y}{d x}=\frac{d}{d x}[(\sin x-\cos x) \log (\sin x-\cos x)]$

$\Rightarrow \frac{1}{y} \frac{d y}{d x}=\log (\sin x-\cos x) \cdot \frac{d}{d x}(\sin x-\cos x)+(\sin x-\cos x) \cdot \frac{d}{d x} \log (\sin x-\cos x)$

$\Rightarrow \frac{1}{y} \frac{d y}{d x}=\log (\sin x-\cos x) \cdot(\cos x+\sin x)+(\sin x-\cos x) \cdot \frac{1}{(\sin x-\cos x)} \cdot \frac{d}{d x}(\sin x-\cos x)$

$\Rightarrow \frac{d y}{d x}=(\sin x-\cos x)^{(\sin x-\cos x)}[(\cos x+\sin x) \cdot \log (\sin x-\cos x)+(\cos x+\sin x)]$

$\therefore \frac{d y}{d x}=(\sin x-\cos x)^{(\sin x-\cos x)}(\cos x+\sin x)[1+\log (\sin x-\cos x)]$

Solution

Let $y=x^{x}+x^{a}+a^{x}+a^{a}$

Also, let $x^{x}=u, x^{a}=v, a^{x}=w$, and $a^{a}=s$

$\therefore y=u+v+w+s$

$\Rightarrow \frac{d y}{d x}=\frac{d u}{d x}+\frac{d v}{d x}+\frac{d w}{d x}+\frac{d s}{d x}$

$u=x^{x}$

$\Rightarrow \log u=\log x^{x}$

$\Rightarrow \log u=x \log x$

Differentiating both sides with respect to $x$, we obtain $\frac{1}{u} \frac{d u}{d x}=\log x \cdot \frac{d}{d x}(x)+x \cdot \frac{d}{d x}(\log x)$

$\Rightarrow \frac{d u}{d x}=u[\log x \cdot 1+x \cdot \frac{1}{x}]$

$\Rightarrow \frac{d u}{d x}=x^{x}[\log x+1]=x^{x}(1+\log x)$

$v=x^{a}$

$\therefore \frac{d v}{d x}=\frac{d}{d x}(x^{a})$

$\Rightarrow \frac{d v}{d x}=a x^{a-1}$

$w=a^{x}$

$\Rightarrow \log w=\log a^{x}$

$\Rightarrow \log w=x \log a$

Differentiating both sides with respect to $x$, we obtain

$\frac{1}{w} \cdot \frac{d w}{d x}=\log a \cdot \frac{d}{d x}(x)$

$\Rightarrow \frac{d w}{d x}=w \log a$

$\Rightarrow \frac{d w}{d x}=a^{x} \log a$

$s=a^{a}$

Since $a$ is constant, $a^{a}$ is also a constant.

$\therefore \frac{d s}{d x}=0$

From (1), (2), (3), (4), and (5), we obtain

$ \begin{aligned} \frac{d y}{d x} & =x^{x}(1+\log x)+a x^{a-1}+a^{x} \log a+0 \\ & =x^{x}(1+\log x)+a x^{a-1}+a^{x} \log a \end{aligned} $

Solution

Let $y=x^{x^{2}-3}+(x-3)^{x^{2}}$

Also, let $u=x^{x^{2}-3}$ and $v=(x-3)^{x^{2}}$

$\therefore y=u+v$

Differentiating both sides with respect to $x$, we obtain

$\frac{d y}{d x}=\frac{d u}{d x}+\frac{d v}{d x}$

$u=x^{x^{2}-3}$

$\therefore \log u=\log (x^{x^{2}-3})$

$\log u=(x^{2}-3) \log x$

Differentiating with respect to $x$, we obtain

$\frac{1}{u} \cdot \frac{d u}{d x}=\log x \cdot \frac{d}{d x}(x^{2}-3)+(x^{2}-3) \cdot \frac{d}{d x}(\log x)$

$\Rightarrow \frac{1}{u} \frac{d u}{d x}=\log x \cdot 2 x+(x^{2}-3) \cdot \frac{1}{x}$

$\Rightarrow \frac{d u}{d x}=x^{x^{2}-3} \cdot[\frac{x^{2}-3}{x}+2 x \log x]$

Also,

$v=(x-3)^{x^{2}}$

$\therefore \log v=\log (x-3)^{x^{2}}$

$\Rightarrow \log v=x^{2} \log (x-3)$

Differentiating both sides with respect to $x$, we obtain $\frac{1}{v} \cdot \frac{d v}{d x}=\log (x-3) \cdot \frac{d}{d x}(x^{2})+x^{2} \cdot \frac{d}{d x}[\log (x-3)]$

$\Rightarrow \frac{1}{v} \frac{d v}{d x}=\log (x-3) \cdot 2 x+x^{2} \cdot \frac{1}{x-3} \cdot \frac{d}{d x}(x-3)$

$\Rightarrow \frac{d v}{d x}=v[2 x \log (x-3)+\frac{x^{2}}{x-3} \cdot 1]$

$\Rightarrow \frac{d v}{d x}=(x-3)^{x^{2}}[\frac{x^{2}}{x-3}+2 x \log (x-3)]$

Substituting the expressions of $\frac{d u}{d x}$ and $\frac{d v}{d x}$ in equation (1), we obtain

$\frac{d y}{d x}=x^{x^{2}-3}[\frac{x^{2}-3}{x}+2 x \log x]+(x-3)^{x^{2}}[\frac{x^{2}}{x-3}+2 x \log (x-3)]$

Solution

It is given that, $y=12(1-\cos t), x=10(t-\sin t)$

$\therefore \frac{d x}{d t}=\frac{d}{d t}[10(t-\sin t)]=10 \cdot \frac{d}{d t}(t-\sin t)=10(1-\cos t)$

$\frac{d y}{d t}=\frac{d}{d t}[12(1-\cos t)]=12 \cdot \frac{d}{d t}(1-\cos t)=12 \cdot[0-(-\sin t)]=12 \sin t$

$\therefore \frac{d y}{d x}=\frac{(\frac{d y}{d t})}{(\frac{d x}{d t})}=\frac{12 \sin t}{10(1-\cos t)}=\frac{12 \cdot 2 \sin \frac{t}{2} \cdot \cos \frac{t}{2}}{10 \cdot 2 \sin ^{2} \frac{t}{2}}=\frac{6}{5} \cot \frac{t}{2}$

Solution

It is given that, $y=\sin ^{-1} x+\sin ^{-1} \sqrt{1-x^{2}}$

$\therefore \frac{d y}{d x}=\frac{d}{d x}[\sin ^{-1} x+\sin ^{-1} \sqrt{1-x^{2}}]$

$\Rightarrow \frac{d y}{d x}=\frac{d}{d x}(\sin ^{-1} x)+\frac{d}{d x}(\sin ^{-1} \sqrt{1-x^{2}})$

$\Rightarrow \frac{d y}{d x}=\frac{1}{\sqrt{1-x^{2}}}+\frac{1}{\sqrt{1-(\sqrt{1-x^{2}})^{2}}} \cdot \frac{d}{d x}(\sqrt{1-x^{2}})$

$\Rightarrow \frac{d y}{d x}=\frac{1}{\sqrt{1-x^{2}}}+\frac{1}{x} \cdot \frac{1}{2 \sqrt{1-x^{2}}} \cdot \frac{d}{d x}(1-x^{2})$

$\Rightarrow \frac{d y}{d x}=\frac{1}{\sqrt{1-x^{2}}}+\frac{1}{2 x \sqrt{1-x^{2}}}(-2 x)$

$\Rightarrow \frac{d y}{d x}=\frac{1}{\sqrt{1-x^{2}}}-\frac{1}{\sqrt{1-x^{2}}}$

$\therefore \frac{d y}{d x}=0$

Solution

It is given that,

$x \sqrt{1+y}+y \sqrt{1+x}=0$ $\Rightarrow x \sqrt{1+y}=-y \sqrt{1+x}$

Squaring both sides, we obtain

$x^{2}(1+y)=y^{2}(1+x)$

$\Rightarrow x^{2}+x^{2} y=y^{2}+x y^{2}$

$\Rightarrow x^{2}-y^{2}=x y^{2}-x^{2} y$

$\Rightarrow x^{2}-y^{2}=x y(y-x)$

$\Rightarrow(x+y)(x-y)=x y(y-x)$

$\therefore x+y=-x y$

$\Rightarrow(1+x) y=-x$

$\Rightarrow y=\frac{-x}{(1+x)}$

Differentiating both sides with respect to $x$, we obtain

$y=\frac{-x}{(1+x)}$

$\frac{d y}{d x}=-\frac{(1+x) \frac{d}{d x}(x)-x \frac{d}{d x}(1+x)}{(1+x)^{2}}=-\frac{(1+x)-x}{(1+x)^{2}}=-\frac{1}{(1+x)^{2}}$

Hence, proved.

Solution

It is given that, $(x-a)^{2}+(y-b)^{2}=c^{2}$

Differentiating both sides with respect to $x$, we obtain

$$ \begin{align*} & \frac{d}{d x}[(x-a)^{2}]+\frac{d}{d x}[(y-b)^{2}]=\frac{d}{d x}(c^{2}) \\ & \Rightarrow 2(x-a) \cdot \frac{d}{d x}(x-a)+2(y-b) \cdot \frac{d}{d x}(y-b)=0 \\ & \Rightarrow 2(x-a) \cdot 1+2(y-b) \cdot \frac{d y}{d x}=0 \\ & \Rightarrow \frac{d y}{d x}=\frac{-(x-a)}{y-b} \tag{1}\\ & \therefore \frac{d^{2} y}{d x^{2}}=\frac{d}{d x}[\frac{-(x-a)}{y-b}] \end{align*} $$

$ \begin{aligned} & =-[\frac{(y-b) \cdot \frac{d}{d x}(x-a)-(x-a) \cdot \frac{d}{d x}(y-b)}{(y-b)^{2}}] \\ & =-[\frac{(y-b)-(x-a) \cdot \frac{d y}{d x}}{(y-b)^{2}}] \\ & =-[\frac{(y-b)-(x-a) \cdot{\frac{-(x-a)}{y-b}}}{(y-b)^{2}}] \\ & =-[\frac{(y-b)^{2}+(x-a)^{2}}{(y-b)^{3}}] \\ & \therefore[\frac{1+(\frac{d y}{d x})^{2}}{\frac{d^{2} y}{d x^{2}}}]^{\frac{3}{2}}=\frac{[1+\frac{(x-a)^{2}}{(y-b)^{2}}]^{\frac{3}{2}}}{-[\frac{(y-b)^{2}+(x-a)^{2}}{(y-b)^{3}}]}=\frac{[\frac{(y-b)^{2}+(x-a)^{2}}{(y-b)^{2}}]^{\frac{3}{2}}}{-[\frac{(y-b)^{2}+(x-a)^{2}}{(y-b)^{3}}]} \\ & =\frac{[\frac{c^{2}}{(y-b)^{2}}]^{\frac{3}{2}}}{-\frac{c^{2}}{(y-b)^{3}}}=\frac{\frac{c^{3}}{(y-b)^{3}}}{-\frac{c^{2}}{(y-b)^{3}}} \end{aligned} $

$ =-c \text{, which is constant and is independent of } a \text{ and } b $

Hence, proved.

Solution

It is given that, $\cos y=x \cos (a+y)$

$\therefore \frac{d}{d x}[\cos y]=\frac{d}{d x}[x \cos (a+y)]$

$\Rightarrow-\sin y \frac{d y}{d x}=\cos (a+y) \cdot \frac{d}{d x}(x)+x \cdot \frac{d}{d x}[\cos (a+y)]$

$\Rightarrow-\sin y \frac{d y}{d x}=\cos (a+y)+x \cdot[-\sin (a+y)] \frac{d y}{d x}$

$\Rightarrow[x \sin (a+y)-\sin y] \frac{d y}{d x}=\cos (a+y)$

Since $\cos y=x \cos (a+y), x=\frac{\cos y}{\cos (a+y)}$

Then, equation (1) reduces to

$[\frac{\cos y}{\cos (a+y)} \cdot \sin (a+y)-\sin y] \frac{d y}{d x}=\cos (a+y)$

$\Rightarrow[\cos y \cdot \sin (a+y)-\sin y \cdot \cos (a+y)] \cdot \frac{d y}{d x}=\cos ^{2}(a+y)$

$\Rightarrow \sin (a+y-y) \frac{d y}{d x}=\cos ^{2}(a+b)$

$\Rightarrow \frac{d y}{d x}=\frac{\cos ^{2}(a+b)}{\sin a}$

Hence, proved.

Solution

It is given that, $x=a(\cos t+t \sin t)$ and $y=a(\sin t-t \cos t)$

$\therefore \frac{d x}{d t}=a \cdot \frac{d}{d t}(\cos t+t \sin t)$

$=a[-\sin t+\sin t \cdot \frac{d}{d x}(t)+t \cdot \frac{d}{d t}(\sin t)]$

$=a[-\sin t+\sin t+t \cos t]=a t \cos t$

$\frac{d y}{d t}=a \cdot \frac{d}{d t}(\sin t-t \cos t)$

$=a[\cos t-{\cos t \cdot \frac{d}{d t}(t)+t \cdot \frac{d}{d t}(\cos t)}]$

$=a[\cos t-{\cos t-t \sin t}]=a t \sin t$

$\therefore \frac{d y}{d x}=\frac{(\frac{d y}{d t})}{(\frac{d x}{d t})}=\frac{a t \sin t}{a t \cos t}=\tan t$

Then, $\frac{d^{2} y}{d x^{2}}=\frac{d}{d x}(\frac{d y}{d x})=\frac{d}{d x}(\tan t)=\sec ^{2} t \cdot \frac{d t}{d x}$

$ \begin{aligned} & =\sec ^{2} t \cdot \frac{1}{a t \cos t} \quad[\frac{d x}{d t}=a t \cos t \Rightarrow \frac{d t}{d x}=\frac{1}{a t \cos t}] \\ & =\frac{\sec ^{3} t}{a t}, 0<t<\frac{\pi}{2} \end{aligned} $

Solution

It is known that, $|x|= \begin{cases}x, & \text{ if } x \geq 0 \\ -x, & \text{ if } x<0\end{cases}$

Therefore, when $x \geq 0, f(x)=|x|^{3}=x^{3}$

In this case, $f^{\prime}(x)=3 x^{2}$ and hence, $f^{\prime \prime}(x)=6 x$

When $x<0, f(x)=|x|^{3}=(-x)^{3}=-x^{3}$

In this case, $f^{\prime}(x)=-3 x^{2}$ and hence, $f^{\prime \prime}(x)=-6 x$

Thus, for $f(x)=|x|^{3}, f^{\prime \prime}(x)$ exists for all real $x$ and is given by,

$f^{\prime \prime}(x)= \begin{cases}6 x, & \text{ if } x \geq 0 \\ -6 x, & \text{ if } x<0\end{cases}$

Solution

$\sin (A+B)=\sin A \cos B+\cos A \sin B$

Differentiating both sides with respect to $x$, we obtain

$ \begin{aligned} & \frac{d}{d x}[\sin (A+B)]=\frac{d}{d x}(\sin A \cos B)+\frac{d}{d x}(\cos A \sin B) \\ & \Rightarrow \cos (A+B) \cdot \frac{d}{d x}(A+B)=\cos B \cdot \frac{d}{d x}(\sin A)+\sin A \cdot \frac{d}{d x}(\cos B) \\ & +\sin B \cdot \frac{d}{d x}(\cos A)+\cos A \cdot \frac{d}{d x}(\sin B) \\ & \Rightarrow \cos (A+B) \cdot \frac{d}{d x}(A+B)=\cos B \cdot \cos A \frac{d A}{d x}+\sin A(-\sin B) \frac{d B}{d x} \\ & +\sin B(-\sin A) \cdot \frac{d A}{d x}+\cos A \cos B \frac{d B}{d x} \\ & \Rightarrow \cos (A+B) \cdot[\frac{d A}{d x}+\frac{d B}{d x}]=(\cos A \cos B-\sin A \sin B) \cdot[\frac{d A}{d x}+\frac{d B}{d x}] \\ & \therefore \cos (A+B)=\cos A \cos B-\sin A \sin B \end{aligned} $

Solution

$ \begin{aligned} & y= \begin{vmatrix} f(x) & g(x) & h(x) \\ l & m & n \\ a & b & c \end{vmatrix} \\ & \Rightarrow y=(m c-n b) f(x)-(l c-n a) g(x)+(l b-m a) h(x) \end{aligned} $

Then, $\frac{d y}{d x}=\frac{d}{d x}[(m c-n b) f(x)]-\frac{d}{d x}[(l c-n a) g(x)]+\frac{d}{d x}[(l b-m a) h(x)]$

$ =(m c-n b) f^{\prime}(x)-(l c-n a) g^{\prime}(x)+(l b-m a) h^{\prime}(x) $

$ = \begin{vmatrix} f^{\prime}(x) & g^{\prime}(x) & h^{\prime}(x) \\ l & m & n \\ a & b & c \end{vmatrix} $

Thus, $\frac{d y}{d x}= \begin{vmatrix} f^{\prime}(x) & g^{\prime}(x) & h^{\prime}(x) \\ l & m & n \\ a & b & c\end{vmatrix} $

Solution

It is given that, $y=e^{a \cos ^{-1} x}$

Taking logarithm on both the sides, we obtain

$\log y=a \cos ^{-1} x \log e$

$\log y=a \cos ^{-1} x$

Differentiating both sides with respect to $x$, we obtain

$\frac{1}{y} \frac{d y}{d x}=a \times \frac{-1}{\sqrt{1-x^{2}}}$

$\Rightarrow \frac{d y}{d x}=\frac{-a y}{\sqrt{1-x^{2}}}$

By squaring both the sides, we obtain

$(\frac{d y}{d x})^{2}=\frac{a^{2} y^{2}}{1-x^{2}}$

$\Rightarrow(1-x^{2})(\frac{d y}{d x})^{2}=a^{2} y^{2}$

$(1-x^{2})(\frac{d y}{d x})^{2}=a^{2} y^{2}$

Again differentiating both sides with respect to $x$, we obtain

$(\frac{d y}{d x})^{2} \frac{d}{d x}(1-x^{2})+(1-x^{2}) \times \frac{d}{d x}[(\frac{d y}{d x})^{2}]=a^{2} \frac{d}{d x}(y^{2})$

$\Rightarrow(\frac{d y}{d x})^{2}(-2 x)+(1-x^{2}) \times 2 \frac{d y}{d x} \cdot \frac{d^{2} y}{d x^{2}}=a^{2} \cdot 2 y \cdot \frac{d y}{d x}$

$\Rightarrow(\frac{d y}{d x})^{2}(-2 x)+(1-x^{2}) \times 2 \frac{d y}{d x} \cdot \frac{d^{2} y}{d x^{2}}=a^{2} \cdot 2 y \cdot \frac{d y}{d x}$

$\Rightarrow-x \frac{d y}{d x}+(1-x^{2}) \frac{d^{2} y}{d x^{2}}=a^{2} \cdot y \quad[\frac{d y}{d x} \neq 0]$

$\Rightarrow(1-x^{2}) \frac{d^{2} y}{d x^{2}}-x \frac{d y}{d x}-a^{2} y=0$

Hence, proved.