Question

Differentiate

$x\sin 2x+5^x+k^k+{\left({\tan }^{2}x\right)}^3$

Answer 1

$\mathrm{Let}y=x\sin 2x+5^x+k^k+{\left({\tan }^{2}x\right)}^3$

Differentiate it with respect to x we get,

$$\begin{gathered} \frac{\mathit{d}\mathit{y}}{\mathit{d}\mathit{x}}=\frac{d}{dx}\left[x\sin 2x+5^x+k^k+\left({\tan }^{6}x\right)\right] \\ =\frac{d}{dx}\left(x\sin 2x\right)+\frac{d}{dx}\left(5^x\right)+\frac{d}{dx}\left(k^k\right)+\frac{d}{dx}\left({\tan }^{6}x\right) \\ =\left[x\frac{d}{dx}\left(\sin 2x\right)+\sin 2x\frac{d}{dx}\left(x\right)\right]+5^x{\log }_{e}5+0+6{\tan }^{5}x\times \frac{d}{dx}\left(\tan x\right)\left[\mathrm{Using}\mathrm{product}\mathrm{rule}\mathrm{and}\mathrm{chain}\mathrm{rule}\right] \\ =\left[x\cos 2x\frac{d}{dx}\left(2x\right)+\sin 2x\right]+5^x{\log }_{e}5+6{\tan }^{5}x{\sec }^{2}x \\ =2x\cos 2x+\sin 2x+5^x{\log }_{e}5+6{\tan }^{5}x{\sec }^{2}x \\ \mathrm{So},\frac{d}{dx}\left\{x\sin 2x+5^x+k^k+{\left({\tan }^{2}x\right)}^3\right\}=2x\cos 2x+\sin 2x+5^x{\log }_{e}5+6{\tan }^{5}x{\sec }^{2}x \end{gathered}$$