Question

y=ex. log x find
dy dx​

Answer 1

Given:

$\purple\bigstar\bf{y=e^x logx}$

To find :

$\blue\bigstar\rm{\dfrac{dy}{dx} =?}$

Differentation:

$\\ \red \bigstar \sf \: \frac{dy}{dx} = {e}^{x} = {e}^{x}$

$\\ \red \bigstar \sf \: \frac{dy}{dx} = logx = \frac{1}{x}$

Solution :

$\\ \implies \large \sf \: y = {e}^{x} \times logx \\ \\ \: \: \: \: \: \: \: \: \: \green\large \underline {\sf \: by \: using \: chain \: rule} \\ \\ \implies\large \sf \: \frac{ dy}{dx} = {e}^{x} \frac{dy}{dx} logx + \frac{dy}{dx} {e}^{x} logx \\ \\ \implies\large \sf \: \frac{dy}{dx} = \frac{ {e}^{x} }{x} \times 1 + {e}^{x} logx \\ \\ \implies\large \boxed{ \sf{ \frac{dy}{dx} = \frac{ {e}^{x} }{x} + {e}^{x} logx}}$

Additional information :

For solving differentiation we use product rule, quotient rule, and chain rule

Consider the value y=uv

➡ Product rule:

$\red\bigstar\boxed{\sf{\dfrac{dy}{dx} =u\dfrac{dv}{dx}v+\dfrac{du}{dx} u v}}$

Consider the value y=v/u

➡ Quotient Rule:

$\red\bigstar\boxed{\sf{\dfrac{dy}{dx} =\dfrac{u \dfrac{dv}{dx} v - \dfrac{du}{dx}uv}{u^2}}}$

Consider the value y=u²v²

➡ Chain rule :

$\red\bigstar\boxed{\sf{\dfrac{dy}{dx} =2u \dfrac{du}{dx}v² \times 2v \dfrac{dv}{dx}}}$

Answer 2

Step-by-step explanation:

I hope it help you

d logx/dX=1/x