Question

differentiate x^nlogax​

Answer 1

Answer:

$\sf{x^{n-1}[n.ln(ax)+1]}$

Step-by-step explanation:

To differentiate $\sf{x^nln(ax)}$.

Let y = $\sf{x^nln(ax)}$.

$\sf{\dfrac{dy}{dx}=\dfrac{d}{dx}[x^nln(ax)] }$

Applying product rule :-

$\displaystyle \sf{ =\frac{d}{dx} (x^n).ln(ax)+x^n.\frac{d}{dx}[ln(x)] }$

$\displaystyle \sf{=nx^{n-1}ln(ax)+x^n.\frac{1}{ax}.\frac{d}{dx}(ax) }$

$\displaystyle \sf{ =nx^{n-2}ln(ax)+\frac{a.\frac{d}{dx}(x).x^{n-1} }{a} }$

$\displaystyle \sf{ =nx^{n-1}ln(ax)+1x^{n-1}}$

$\sf{=nx^{n-1}ln(ax)+x^{n-1}}$

Simplifying :-

$\sf{x^{n-1}[n.ln(ax)+1]}$

Therefore, the answer.

Answer 2

REFER TO THE ATTACHMENT

HOPE IT HELPS :)