Question

differentiate
${e}^{x} cos x \:$
with respect to X
plz answer fast and step by step ​

Answer 1

AnswEr :

Given Expression,

$\sf \: y = e {}^{x} .cos(x)$

The above expression is of the form uv

Derivative of the above expression would be of the form uv' + v'u

Now,

$\sf \: y' = \dfrac{d( {e}^{x}) }{dx} \times cos (x) + {e}^{x} \times \dfrac{d(cos \: x)}{dx} \\ \\ \longrightarrow \: \sf \: y' = {e}^{x} cos(x) - {e}^{x} sin(x) \\ \\ \longrightarrow \: \boxed{\boxed{ \sf y' = {e}^{x} \bigg(cos \: x - sin \: x \bigg)}}$

• Derivative of cos x is - sin x

• Derivative of e^x is e^x

Derivative of the above expression is e^x(cos x - sin x)

Answer 2

Given term :-

$\sf{e^xcosx}$

Objective :-

To differentiate the given term with respect to x.

Solution :-

Let y = $\sf{e^xcosx}$.

$\displaystyle{\sf{\frac{dy}{dx}=\frac{d}{dx}(e^xcosx) }}$

$\sf{Applying\ product\ rule-}\\\\\displaystyle{\sf{\longrightarrow \frac{dy}{dx}=\frac{d}{dx}(e^x).cos(x)+e^x.\frac{d}{dx}(xosx) }}\\\\\displaystyle{\sf{\longrightarrow \frac{dy}{dx}= e^xcosx+e^x(-sinx)}}\\\\\displaystyle{\sf{\longrightarrow \frac{dy}{dx}= e^x cos x-e^xsinx}}\\\\\displaystyle{\sf{\longrightarrow \frac{dy}{dx}= -e^x(sinx-cosx)}}$

$\boxed{\boxed{\displaystyle{\sf{\color{lime}{\longrightarrow \frac{dy}{dx}= e^x(cosx-sinx)}}}}}$

$\rule{180}2$

Derivative formulas used :-

◙ Power rule :-

[a(x) × b(x)]' = a'(x).b(x) + a(x).b'(x)

◙ Differentiation of $\sf{e^x}$ is

◙ Differentiation of cosx = -sinx