Question

Find (e^x log x) d/ dx​

Answer 1

Solution:

To Determine: The derivative of the function.

Given That:

$\rm\longrightarrow y=e^{x}\ln(x)$

Differentiating both sides wrt x, we get:

$\rm\longrightarrow \dfrac{dy}{dx}=\dfrac{d}{dx}[e^{x}\ln(x)]$

We know that:

$\bigstar\:\underline{\boxed{\rm\dfrac{d}{dx}(fg)=f\cdot\dfrac{d}{dx}(g)+g\cdot\dfrac{d}{dx}f}}$

Using the above result, we get:

$\rm\longrightarrow \dfrac{dy}{dx}=e^{x}\cdot\dfrac{d}{dx}\ln(x)+ln(x)\cdot\dfrac{d}{dx}e^{x}$

We know that:

$\bigstar\:\underline{\boxed{\rm\dfrac{d}{dx}(e^{x})=e^{x}}}$

$\bigstar\:\underline{\boxed{\rm\dfrac{d}{dx}ln(x)=\dfrac{1}{x}}}$

Therefore, we get:

$\rm\longrightarrow \dfrac{dy}{dx}=\dfrac{e^{x}}{x}+ln(x)\cdot e^{x}$

$\rm\longrightarrow \dfrac{dy}{dx}=\dfrac{e^{x}+x e^{x}\ln(x)}{x}$

★ Which is our required answer.

Answer:

$\rm\hookrightarrow \dfrac{dy}{dx}=\dfrac{e^{x}+x e^{x}\ln(x)}{x}$

Learn More:

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