Question

If $y={(2x+3)}^{(3x-5)},$ find $\frac{dy}{dx}$.

Answer 1

Solution:-

Given:- $y={(2x+3)}^{(3x-5)}$

To find:- $\frac{dy}{dx}=?$

$y={(2x+3)}^{(3x-5)}$

Taking log both sides, we have

$\log y=\log \left({(2x+3)}^{(3x-5)}\right)$

$\log y=(3x-5)\log (2x+3)$

Now, differentiating both sides w.r.t. x, we have

$\frac{d}{dx}\left(\log y\right)=(3x-5).\frac{d}{dx}\left(\log (2x+3)\right)+\left(\log (2x+3)\right).\frac{d}{dx}(3x-5)$

$\frac{1}{y}\frac{dy}{dx}=(3x-5)(\frac{1}{2x+3}.2)+\left(\log (2x+3)\right)\left(3\right)$

$\frac{dy}{dx}=y(\frac{2(3x-5)}{2x+3}+3\log (2x+3))$

$\Rightarrow \frac{dy}{dx}=\left({(2x+3)}^{(3x-5)}\right)(\frac{(6x-10)}{2x+3}+3\log (2x+3))$