Question

Find the derivative of the following function w.r.t.x:
$x\times \frac{5}{2} {e}^{x}$

Class 11th; Commerce
Chapter no. : 9th,, Differentiation


PLEASE HELP! I WANT IT URGENTLY T-T
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Answer 1

The question has been solved by using product rule.

Answer 2

SOLUTION:

Given That:

$\rm\longrightarrow f(x)=x\times\dfrac{5}{2}e^{x}$

Calculating the derivative:

$\rm\longrightarrow f'(x)=\dfrac{5}{2}\times\dfrac{d}{dx}(xe^{x})$

We know that:

$\rm\longrightarrow \dfrac{d}{dx}[f(x)\cdot g(x)]=f'(x)\cdot g(x)+f(x)\cdot g'(x)$

Therefore:

$\rm\longrightarrow f'(x)=\dfrac{5}{2}\times\bigg[e^{x}\cdot\dfrac{d}{dx}(x)+x\cdot\dfrac{d}{dx}e^{x}\bigg]$

$\rm\longrightarrow f'(x)=\dfrac{5}{2}\times[e^{x}+xe^{x}]$

$\rm\longrightarrow f'(x)=\dfrac{5}{2}e^{x}(1+x)$

Which is our required answer.

LEARN MORE:

$\begin{gathered}\boxed{\begin{array}{c|c}\bf f(x)&\bf\dfrac{d}{dx}f(x)\\ \\ \frac{\qquad\qquad}{}&\frac{\qquad\qquad}{}\\ \sf k&\sf0\\ \\ \sf sin(x)&\sf cos(x)\\ \\ \sf cos(x)&\sf-sin(x)\\ \\ \sf tan(x)&\sf{sec}^{2}(x)\\ \\ \sf cot(x)&\sf-{cosec}^{2}(x)\\ \\ \sf sec(x)&\sf sec(x)tan(x)\\ \\ \sf cosec(x)&\sf-cosec(x)cot(x)\\ \\ \sf\sqrt{x}&\sf\dfrac{1}{2\sqrt{x}}\\ \\ \sf log(x)&\sf\dfrac{1}{x}\\ \\ \sf{e}^{x}&\sf{e}^{x}\end{array}}\\ \end{gathered}$