Question

Find the second order derivatives of each of the following functions:

(i) x³ + tan x
(ii) sin (log x)
(iii) log (sin x)
(iv) e^x sin 5x
(v) e^6x cos 3x
(vi) x³ log x
(vii) tan⁻¹ x
(viii) x cos x
(ix) log (log x)

Answer 1

(i) We have,

$$\begin{gathered} y=x^3+\tan x \\ \mathrm{Differentiating}\mathrm{w}.\mathrm{r}.\mathrm{t}.\mathit{}x,\mathrm{we}\mathrm{get} \\ \frac{dy}{dx}=3x^2+{\sec }^{2}x \\ \mathrm{Differentiating}\mathrm{again}\mathrm{w}.\mathrm{r}.\mathrm{t}.\mathit{}\mathrm{x},\mathrm{we}\mathrm{get} \\ \frac{{\mathit{d}}^{\mathit{2}}y}{\mathit{d}{x}^2}=6x+2{\sec }^{2}x\tan x \end{gathered}$$

(ii) We have,

$$\begin{gathered} y=\sin \left(\log x\right) \\ \mathrm{Differentiating}\mathrm{w}.\mathrm{r}.\mathrm{t}.\mathit{}x,\mathrm{we}\mathrm{get} \\ \frac{\mathit{d}y}{\mathit{d}x}=\cos \left(\log x\right)\times \frac{1}{x} \\ \mathrm{Differentiating}\mathrm{again}\mathrm{w}.\mathrm{r}.\mathrm{t}.\mathit{}x,\mathrm{we}\mathrm{get} \\ \frac{{\mathit{d}}^{\mathit{2}}y}{\mathit{d}{x}^2}=-\sin \left(\log x\right)\frac{1}{x}\times \frac{1}{x}+\cos \left(\log x\right)\times \frac{-1}{x^2} \\ =\frac{-\left[\sin \left(\log x\right)+\cos \left(\log x\right)\right]}{x^2} \end{gathered}$$

(iii) We have,

$$\begin{gathered} y=\log \left(\sin x\right) \\ \mathrm{Differentiating}\mathrm{w}.\mathrm{r}.\mathrm{t}.\mathit{}x,\mathrm{we}\mathrm{get} \\ \frac{\mathit{d}y}{\mathit{d}x}=\frac{1}{\sin x}\times \cos x=\cot x \\ \mathrm{Differentiating}\mathrm{again}\mathrm{w}.\mathrm{r}.\mathrm{t}.\mathit{}x,\mathrm{we}\mathrm{get} \\ \frac{{\mathit{d}}^{\mathit{2}}y}{\mathit{d}{x}^2}=-{\mathrm{cosec}}^{2}x \end{gathered}$$

(iv) We have,

$$\begin{gathered} y=e^x\sin \left(5x\right) \\ \mathrm{Differentiating}\mathrm{w}.\mathrm{r}.\mathrm{t}.\mathit{}x,\mathrm{we}\mathrm{get} \\ \frac{\mathit{d}\mathit{y}}{\mathit{d}\mathit{x}}=e^x\sin 5x+e^x\cos 5x\times 5 \\ \mathrm{Differentiating}\mathrm{again}\mathrm{w}.\mathrm{r}.\mathrm{t}.\mathit{}x,\mathrm{we}\mathrm{get} \\ \frac{{\mathit{d}}^{\mathit{2}}y}{\mathit{d}{x}^2}=e^x\sin 5x+e^x\cos 5x\times 5+5e^x(-\sin 5x\times 5)+5e^x\cos 5x \\ =-24e^x\sin 5x+10e^x\cos 5x \\ =2e^x\left(5\cos 5x-12\sin 5x\right) \end{gathered}$$

(v) We have,
$$\begin{gathered} y=e^{6x}\cos 3x \\ \mathrm{Differentiating}\mathrm{w}.\mathrm{r}.\mathrm{t}.\mathit{}x,\mathrm{we}\mathrm{get} \\ \frac{\mathit{d}y}{\mathit{d}x}=e^{6x}\times 6\times \cos 3x+e^{6x}(-\sin 3x\times 3) \\ =6e^{6x}\cos 3x-3e^{6x}\sin 3x \\ \mathrm{Differentiating}\mathrm{again}\mathrm{w}.\mathrm{r}.\mathrm{t}.\mathit{}x,\mathrm{we}\mathrm{get} \\ \frac{{\mathit{d}}^{\mathit{2}}y}{\mathit{d}{x}^2}=6e^{6x}\cos 3x\times 6-6e^{6x}\sin 3x\times 3-3\times 6e^{6x}\sin 3x-3e^{6x}\times 3\cos 3x \\ =27e^{6x}\cos 3x-36e^{6x}\sin 3x \\ =9e^{6x}\left(3\cos 3x-4\sin 3x\right) \end{gathered}$$

(vi) We have,
$$\begin{gathered} y=x^3\log x \\ \mathrm{Differentiating}\mathrm{w}.\mathrm{r}.\mathrm{t}.\mathit{}x,\mathrm{we}\mathrm{get} \\ \frac{\mathit{d}\mathit{y}}{\mathit{d}\mathit{x}}=3x^2\log x+x^3\times \frac{1}{x} \\ =3x^2\log x+x^2 \\ \mathrm{Differentiating}\mathrm{again}\mathrm{w}.\mathrm{r}.\mathrm{t}.\mathit{}x,\mathrm{we}\mathrm{get} \\ \frac{{\mathit{d}}^{\mathit{2}}y}{\mathit{d}{x}^2}=6x\log x+3x^2\times \frac{1}{x}+2x \\ =6x\log x+5x \end{gathered}$$

(vii) We have,
$$\begin{gathered} y={\tan }^{-1}x \\ \mathrm{Differentiating}\mathrm{w}.\mathrm{r}.\mathrm{t}.\mathit{}x,\mathrm{we}\mathrm{get} \\ \frac{\mathit{d}y}{\mathit{d}x}=\frac{1}{1+x^2} \\ \mathrm{Differentiating}\mathrm{again}\mathrm{w}.\mathrm{r}.\mathrm{t}.\mathit{}x,\mathrm{we}\mathrm{get} \\ \frac{{\mathit{d}}^{\mathit{2}}y}{\mathit{d}{x}^2}=-\left(2x\right)\times \frac{1}{{\left(1+x^2\right)}^2}=\frac{-2x}{{\left(1+x^2\right)}^2} \end{gathered}$$


(viii) We have,
$$\begin{gathered} y=x\cos x \\ \mathrm{Differentiating}\mathrm{w}.\mathrm{r}.\mathrm{t}.\mathit{}x,\mathrm{we}\mathrm{get} \\ \frac{\mathit{d}\mathit{y}}{\mathit{d}\mathit{x}}=\cos x-x\sin x \\ \mathrm{Differentiating}\mathrm{again}\mathrm{w}.\mathrm{r}.\mathrm{t}.\mathit{}x,\mathrm{we}\mathrm{get} \\ \frac{{\mathit{d}}^{\mathit{2}}y}{\mathit{d}{x}^2}=-\sin x-\sin x-x\cos x \\ =-\left(2\sin x+x\cos x\right) \end{gathered}$$

(ix) We have,

$$\begin{gathered} y=\log \left(\log x\right) \\ \mathrm{Differentiating}\mathrm{w}.\mathrm{r}.\mathrm{t}.\mathit{}x,\mathrm{we}\mathrm{get} \\ \frac{\mathit{d}y}{\mathit{d}x}=\frac{1}{\log x}\times \frac{1}{x}=\frac{1}{x\log x} \\ \mathrm{Differentiating}\mathrm{again}\mathrm{w}.\mathrm{r}.\mathrm{t}.\mathit{}x,\mathrm{we}\mathrm{get} \\ \frac{{\mathit{d}}^{\mathit{2}}y}{\mathit{d}{x}^2}=\frac{0-\left(\log x+1\right)}{{\left(x\log x\right)}^2}=-\frac{\left(1+\log x\right)}{{\left(x\log x\right)}^2} \end{gathered}$$