Question

For any constant $a$, find the derivative of $\frac{x^n-a^n}{x-a}$.

Answer 1

$\frac{d}{dx}\left(\frac{x^n-a^n}{x-a}\right)$ where a is arbitary constant

we can use the chain rule to differentiate

$d(I.II)=Id\left(II\right)+IId\left(I\right)$

$\Rightarrow \frac{d}{dx}\left(\frac{x^n-a^n}{x-a}\right)=$

$\{\because \frac{d}{dx}(x^n)=n{x}^{n-1}\}$

$=(x^n-a^n)\frac{d}{dx}\left(\frac{1}{x-a}\right)+\frac{1}{(x-a)}\frac{d}{dx}(x^n-a^n)$

$=(x^n-a^n)[-\frac{1}{(x-a{)}^2}]+\frac{1}{(x-a)}[n{x}^{n-1}=0]$

$\Rightarrow -\frac{(x^n-a^n)}{(x-a{)}^2}+\frac{n{x}^{n-1}}{(x-a)}$

$\Rightarrow \frac{n{x}^{n-1}}{(x-a)}=\frac{(x^n-a^n)}{(x-a{)}^2}$.