Question

answes plz this question​

Answer 1

Given

• $\sf (ye)^x = e^x$

On Taking In both sides,we get

$\sf{yIn_e x = (x-y)In_e e}$

$\sf{yIn_e x = (x-y)}$

$\sf{y(1+In~x )=x \: \: \: o r \: \: \: y =\dfrac{x}{1+In~x}}$

On Differentiating both sides w.r.t x, we get

$\sf{\dfrac{dy}{dx}= \dfrac{(1+In~x) \dfrac{d}{dx}(x)-x\dfrac{d}{dx}(1+In~x)}{(1+In~x)^2}}$

we know

$\sf \Bigg[ \dfrac{d}{dx}\bigg(\dfrac{v}{u} \bigg)=\dfrac{u\dfrac{dv}{dx}-v\dfrac{du}{dx}}{u²}\Bigg]$

$\sf{\dfrac{dy}{dx}= \dfrac{1+In~x - \dfrac{1}{x}(x)}{(1+In~x)^2}}$

$\sf{\dfrac{dy}{dx}= \dfrac{ \cancel{1}+In~x -\cancel1}{(1+In~x)^2}}$

$\boxed{\therefore\mathcal{\dfrac{dy}{dx}= \dfrac{In~x }{(1+In~x)^2}}}$

Answer 2

Step-by-step explanation:

hope this helps you better