Question

Differentiate \[x \sin 2x + 5^x + k^k + \left( \tan^2 x \right)^3\] ?

Answer 1

\[\text{Let} y = x \sin2x + 5^x + k^k + \left( \tan^2 x \right)^3\] 

Differentiate it with respect to x we get,

\[\frac{d y}{d x} = \frac{d}{dx}\left[ x \sin2x + 5^x + k^k + \left( \tan^6 x \right) \right]\]

\[ = \frac{d}{dx}\left( x \sin2x \right) + \frac{d}{dx}\left( 5^x \right) + \frac{d}{dx}\left( k^k \right) + \frac{d}{dx}\left( \tan^6 x \right)\]

\[ = \left[ x\frac{d}{dx}\left( \sin2x \right) + \sin2x\frac{d}{dx}\left( x \right) \right] + 5^x \log_e 5 + 0 + 6 \tan^5 x \times \frac{d}{dx}\left( \tan x \right) \left[ \text{Using product rule and chain rule } \right]\]

\[ = \left[ x \cos2x\frac{d}{dx}\left( 2x \right) + \sin2x \right] + 5^x \log_e 5 + 6 \tan^5 x \sec^2 x\]

\[ = 2x \cos2x + \sin2x + 5^x \log_e 5 + 6 \tan^5 x \sec^2 x\]

\[So, \frac{d}{dx}\left\{ x \sin2x + 5^x + k^k + \left( \tan^2 x \right)^3 \right\} = 2x\cos2x + \sin2x + 5^x \log_e 5 + 6 \tan^5 x \sec^2 x\]