Differentiate:

To find: Differentiation of $e^{x} \cos x$

Formula used: (i) (uv)′ = u′v + uv′ (Leibnitz or product rule)

(ii)

$\frac{d e^{x}}{d x}=e^{x}$

(iii)

$\frac{d \cos x}{d x}=-\sin x$

Let us take $u=e^{x}$ and $v=\cos x$

$u^{\prime}=\frac{d u}{d x}=\frac{d e^{x}}{d x}=e^{x}$

$v^{\prime}=\frac{d v}{d x}=\frac{d \cos x}{d x}=-\sin x$

Putting the above obtained values in the formula:-

$(u v)^{\prime}=u^{\prime} v+u v^{\prime}$

$\left(e^{x} \cos x\right)^{\prime}=e^{x} x \cos x+e^{x} x-\sin x$

$=e^{x} \cos x-e^{x} \sin x$

$=e^{x}(\cos x-\sin x)$

Ans) $e^{x}(\cos x-\sin x)$