Solution

Let $y=x^{2}+3 x+2$

Then,

$\frac{d y}{d x}=\frac{d}{d x}(x^{2})+\frac{d}{d x}(3 x)+\frac{d}{d x}(2)=2 x+3+0=2 x+3$

$\therefore \frac{d^{2} y}{d x^{2}}=\frac{d}{d x}(2 x+3)=\frac{d}{d x}(2 x)+\frac{d}{d x}(3)=2+0=2$

Solution

Let $y=x^{20}$

Then,

$\frac{d y}{d x}=\frac{d}{d x}(x^{20})=20 x^{19}$

$\therefore \frac{d^{2} y}{d x^{2}}=\frac{d}{d x}(20 x^{19})=20 \frac{d}{d x}(x^{19})=20 \cdot 19 \cdot x^{18}=380 x^{18}$

Solution

Let $y=x \cdot \cos x$

Then,

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

Solution

Let $y=\log x$

Then,

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

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

Solution

Let $y=x^{3} \log x$

Then,

$ \begin{aligned} & \frac{d y}{d x}=\frac{d}{d x}[x^{3} \log x]=\log x \cdot \frac{d}{d x}(x^{3})+x^{3} \cdot \frac{d}{d x}(\log x) \\ & =\log x \cdot 3 x^{2}+x^{3} \cdot \frac{1}{x}=\log x \cdot 3 x^{2}+x^{2} \\ & =x^{2}(1+3 \log x) \\ & \therefore \frac{d^{2} y}{d x^{2}}=\frac{d}{d x}[x^{2}(1+3 \log x)] \\ & =(1+3 \log x) \cdot \frac{d}{d x}(x^{2})+x^{2} \frac{d}{d x}(1+3 \log x) \\ & =(1+3 \log x) \cdot 2 x+x^{2} \cdot \frac{3}{x} \\ & =2 x+6 x \log x+3 x \\ & =5 x+6 x \log x \\ & =x(5+6 \log x) \end{aligned} $

Solution

Let $y=e^{x} \sin 5 x$

Then,

$ \begin{aligned} \frac{d y}{d x} & =\frac{d}{d x}(e^{x} \sin 5 x)=\sin 5 x \cdot \frac{d}{d x}(e^{x})+e^{x} \frac{d}{d x}(\sin 5 x) \\ & =\sin 5 x \cdot e^{x}+e^{x} \cdot \cos 5 x \cdot \frac{d}{d x}(5 x)=e^{x} \sin 5 x+e^{x} \cos 5 x \cdot 5 \\ & =e^{x}(\sin 5 x+5 \cos 5 x) \\ \therefore \frac{d^{2} y}{d x^{2}} & =\frac{d}{d x}[e^{x}(\sin 5 x+5 \cos 5 x)] \\ & =(\sin 5 x+5 \cos 5 x) \cdot \frac{d}{d x}(e^{x})+e^{x} \cdot \frac{d}{d x}(\sin 5 x+5 \cos 5 x) \\ & =(\sin 5 x+5 \cos 5 x) e^{x}+e^{x}[\cos 5 x \cdot \frac{d}{d x}(5 x)+5(-\sin 5 x) \cdot \frac{d}{d x}(5 x)] \\ & =e^{x}(\sin 5 x+5 \cos 5 x)+e^{x}(5 \cos 5 x-25 \sin 5 x) \\ & =e^{x}(10 \cos 5 x-24 \sin 5 x)=2 e^{x}(5 \cos 5 x-12 \sin 5 x) \end{aligned} $

Solution

Let $y=e^{6 x} \cos 3 x$

Then,

$ \begin{aligned} \frac{d y}{d x} & =\frac{d}{d x}(e^{6 x} \cdot \cos 3 x)=\cos 3 x \cdot \frac{d}{d x}(e^{6 x})+e^{6 x} \cdot \frac{d}{d x}(\cos 3 x) \\ & =\cos 3 x \cdot e^{6 x} \cdot \frac{d}{d x}(6 x)+e^{6 x} \cdot(-\sin 3 x) \cdot \frac{d}{d x}(3 x) \\ & =6 e^{6 x} \cos 3 x-3 e^{6 x} \sin 3 x \\ \therefore \frac{d^{2} y}{d x^{2}} & =\frac{d}{d x}(6 e^{6 x} \cos 3 x-3 e^{6 x} \sin 3 x)=6 \cdot \frac{d}{d x}(e^{6 x} \cos 3 x)-3 \cdot \frac{d}{d x}(e^{6 x} \sin 3 x) \\ & =6 \cdot[6 e^{6 x} \cos 3 x-3 e^{6 x} \sin 3 x]-3 \cdot[\sin 3 x \cdot \frac{d}{d x}(e^{6 x})+e^{6 x} \cdot \frac{d}{d x}(\sin 3 x)] \\ & =36 e^{6 x} \cos 3 x-18 e^{6 x} \sin 3 x-3[\sin 3 x \cdot e^{6 x} \cdot 6+e^{6 x} \cdot \cos 3 x \cdot 3] \\ & =36 e^{6 x} \cos 3 x-18 e^{6 x} \sin 3 x-18 e^{6 x} \sin 3 x-9 e^{6 x} \cos 3 x \\ & =27 e^{6 x} \cos 3 x-36 e^{6 x} \sin 3 x \\ & =9 e^{6 x}(3 \cos 3 x-4 \sin 3 x) \end{aligned} $

Solution

Let $y=\tan ^{-1} x$

Then,

$ \begin{aligned} & \frac{d y}{d x}=\frac{d}{d x}(\tan ^{-1} x)=\frac{1}{1+x^{2}} \\ & \therefore \frac{d^{2} y}{d x^{2}}=\frac{d}{d x}(\frac{1}{1+x^{2}})=\frac{d}{d x}(1+x^{2})^{-1}=(-1) \cdot(1+x^{2})^{-2} \cdot \frac{d}{d x}(1+x^{2}) \\ & \quad=\frac{-1}{(1+x^{2})^{2}} \times 2 x=\frac{-2 x}{(1+x^{2})^{2}} \end{aligned} $

Solution

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

Then,

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

Solution

Let $y=\sin (\log x)$

Then,

$ \begin{aligned} & \frac{d y}{d x}=\frac{d}{d x}[\sin (\log x)]=\cos (\log x) \cdot \frac{d}{d x}(\log x)=\frac{\cos (\log x)}{x} \\ & \begin{aligned} \therefore \frac{d^{2} y}{d x^{2}} & =\frac{d}{d x}[\frac{\cos (\log x)}{x}] \\ & =\frac{x \cdot \frac{d}{d x}[\cos (\log x)]-\cos (\log x) \cdot \frac{d}{d x}(x)}{x^{2}} \\ & =\frac{x \cdot[-\sin (\log x) \cdot \frac{d}{d x}(\log x)]-\cos (\log x) \cdot 1}{x^{2}} \\ & =\frac{-x \sin (\log x) \cdot \frac{1}{x}-\cos (\log x)}{x^{2}} \\ & =\frac{-[\sin (\log x)+\cos (\log x)]}{x^{2}} \end{aligned} \end{aligned} $

Solution

It is given that, $y=5 \cos x-3 \sin x$

Then,

$ \begin{aligned} & \frac{d y}{d x}=\frac{d}{d x}(5 \cos x)-\frac{d}{d x}(3 \sin x)=5 \frac{d}{d x}(\cos x)-3 \frac{d}{d x}(\sin x) \\ & =5(-\sin x)-3 \cos x=-(5 \sin x+3 \cos x) \\ & \therefore \frac{d^{2} y}{d x^{2}}=\frac{d}{d x}[-(5 \sin x+3 \cos x)] \\ & =-[5 \cdot \frac{d}{d x}(\sin x)+3 \cdot \frac{d}{d x}(\cos x)] \\ & =-[5 \cos x+3(-\sin x)] \\ & =-[5 \cos x-3 \sin x] \\ & =-y \\ & \therefore \frac{d^{2} y}{d x^{2}}+y=0 \end{aligned} $

Hence, proved.

Solution

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

Then,

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

Putting $x=\cos y$ in equation (i), we obtain

$\frac{d^{2} y}{d x^{2}}=\frac{-\cos y}{\sqrt{(1-\cos ^{2} y)^{3}}}$ $\Rightarrow \frac{d^{2} y}{d x^{2}}=\frac{-\cos y}{\sqrt{(\sin ^{2} y)^{3}}}$

$ =\frac{-\cos y}{\sin ^{3} y} $

$ =\frac{-\cos y}{\sin y} \times \frac{1}{\sin ^{2} y} $

$\Rightarrow \frac{d^{2} y}{d x^{2}}=-\cot y \cdot cosec^{2} y$

Solution

It is given that, $y=3 \cos (\log x)+4 \sin (\log x)$

Then,

$ \begin{aligned} & y_1=3 \cdot \frac{d}{d x}[\cos (\log x)]+4 \cdot \frac{d}{d x}[\sin (\log x)] \\ & =3 \cdot[-\sin (\log x) \cdot \frac{d}{d x}(\log x)]+4 \cdot[\cos (\log x) \cdot \frac{d}{d x}(\log x)] \\ & \therefore y_1=\frac{-3 \sin (\log x)}{x}+\frac{4 \cos (\log x)}{x}=\frac{4 \cos (\log x)-3 \sin (\log x)}{x} \\ & \therefore y_2=\frac{d}{d x}(\frac{4 \cos (\log x)-3 \sin (\log x)}{x}) \\ & =\frac{x{4 \cos (\log x)-3 \sin (\log x)}^{\prime}-{4 \cos (\log x)-3 \sin (\log x)}(x)^{\prime}}{x^{2}} \\ & =\frac{x[4{\cos (\log x)}^{\prime}-3{\sin (\log x)}^{\prime}]-{4 \cos (\log x)-3 \sin (\log x)} \cdot 1}{x^{2}} \\ & =\frac{x[-4 \sin (\log x) \cdot(\log x)^{\prime}-3 \cos (\log x) \cdot(\log x)^{\prime}]-4 \cos (\log x)+3 \sin (\log x)}{x^{2}} \\ & =\frac{x[-4 \sin (\log x) \cdot \frac{1}{x}-3 \cos (\log x) \cdot \frac{1}{x}]-4 \cos (\log x)+3 \sin (\log x)}{x^{2}} \\ & =\frac{-4 \sin (\log x)-3 \cos (\log x)-4 \cos (\log x)+3 \sin (\log x)}{x^{2}} \\ & =\frac{-\sin (\log x)-7 \cos (\log x)}{x^{2}} \\ & \therefore x^{2} y_2+x y_1+y \\ & =x^{2}(\frac{-\sin (\log x)-7 \cos (\log x)}{x^{2}})+x(\frac{4 \cos (\log x)-3 \sin (\log x)}{x})+3 \cos (\log x)+4 \sin (\log x) \\ & =-\sin (\log x)-7 \cos (\log x)+4 \cos (\log x)-3 \sin (\log x)+3 \cos (\log x)+4 \sin (\log x) \\ & =0 \end{aligned} $

Hence, proved.

Solution

It is given that, $y=A e^{m x}+B e^{n x}$

Then,

$ \begin{aligned} & \frac{d y}{d x}=A \cdot \frac{d}{d x}(e^{m x})+B \cdot \frac{d}{d x}(e^{n x})=A \cdot e^{m x} \cdot \frac{d}{d x}(m x)+B \cdot e^{n x} \cdot \frac{d}{d x}(n x)=A m e^{n x x}+B n e^{n x} \\ & \frac{d^{2} y}{d x^{2}}=\frac{d}{d x}(A m e^{m x}+B n e^{n x})=A m \cdot \frac{d}{d x}(e^{m x})+B n \cdot \frac{d}{d x}(e^{n x}) \\ & =A m \cdot e^{m x} \cdot \frac{d}{d x}(m x)+B n \cdot e^{n x} \cdot \frac{d}{d x}(n x)=A m^{2} e^{m x}+B n^{2} e^{n x} \\ & \therefore \frac{d^{2} y}{d x^{2}}-(m+n) \frac{d y}{d x}+m n y \\ & =A m^{2} e^{m x}+B n^{2} e^{n x}-(m+n) \cdot(A m e^{m x}+B n e^{n x})+m n(A e^{m x}+B e^{n x}) \\ & =A m^{2} e^{n x x}+B n^{2} e^{n x}-A m^{2} e^{n x x}-B m n e^{n x}-A m n e^{m x x}-B n^{2} e^{n x}+A m n e^{n x x}+B m n e^{n x} \\ & =0 \end{aligned} $

Hence, proved.

Solution

It is given that, $y=500 e^{7 x}+600 e^{-7 x}$

Then,

$ \begin{aligned} \frac{d y}{d x} & =500 \cdot \frac{d}{d x}(e^{7 x})+600 \cdot \frac{d}{d x}(e^{-7 x}) \\ & =500 \cdot e^{7 x} \cdot \frac{d}{d x}(7 x)+600 \cdot e^{-7 x} \cdot \frac{d}{d x}(-7 x) \\ & =3500 e^{7 x}-4200 e^{-7 x} \\ \therefore \frac{d^{2} y}{d x^{2}} & =3500 \cdot \frac{d}{d x}(e^{7 x})-4200 \cdot \frac{d}{d x}(e^{-7 x}) \\ & =3500 \cdot e^{7 x} \cdot \frac{d}{d x}(7 x)-4200 \cdot e^{-7 x} \cdot \frac{d}{d x}(-7 x) \\ & =7 \times 3500 \cdot e^{7 x}+7 \times 4200 \cdot e^{-7 x} \\ & =49 \times 500 e^{7 x}+49 \times 600 e^{-7 x} \\ & =49(500 e^{7 x}+600 e^{-7 x}) \\ & =49 y \end{aligned} $

Hence, proved.

Solution

The given relationship is $e^{y}(x+1)=1$

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

$\Rightarrow e^{y}=\frac{1}{x+1}$

Taking logarithm on both the sides, we obtain

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

Differentiating this relationship with respect to $x$, we obtain $\frac{d y}{d x}=(x+1) \frac{d}{d x}(\frac{1}{x+1})=(x+1) \cdot \frac{-1}{(x+1)^{2}}=\frac{-1}{x+1}$

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

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

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

Hence, proved.

Solution

The given relationship is $y=(\tan ^{-1} x)^{2}$

Then,

$y _1=2 \tan ^{-1} x \frac{d}{d x}(\tan ^{-1} x)$

$\Rightarrow y _1=2 \tan ^{-1} x \cdot \frac{1}{1+x^{2}}$

$\Rightarrow(1+x^{2}) y _1=2 \tan^{-1} x$

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

$(1+x^{2}) y _2+2 x y _1=2(\frac{1}{1+x^{2}})$

$\Rightarrow(1+x^{2})^{2} y _2+2 x(1+x^{2}) y _1=2$

Hence, proved.