Question

Find the derivative of the following function:
f(x) = $(x+cosx)(x-tanx)$

Answer 1

Here $f\left(x\right)=(x+cosx)(x-tanx)$

$\therefore f^{\prime }\left(x\right)=\frac{d}{dx}\left[\right(x+cosx\left)\right(x-tanx\left)\right]$

$=(x+cosx)\frac{d}{dx}(x-tanx)+(x-tanx)\frac{d}{dx}(x+cosx)$

$=(x+cosx)(1-se{c}^{2}x)+(x-tanx)(1-sinx)$

$=-ta{n}^{2}x(x+cosx)+(x-tanx)(1-sinx)$ $[\because 1+ta{n}^{2}x=se{c}^{2}x]$