Question

Differentiate the following using quotient rule :
$f\left(x\right)=\frac{x^4}{{\log }_{a}x}$

Answer 1

Given that,

$f\left(x\right)=\frac{x^4}{{\log }_{a}x}$

On differentiating w.r.t $x$, we get

$f^{\prime }\left(x\right)=\frac{d}{dx}\left(\frac{x^4}{{\log }_{a}x}\right)$

Since, $\frac{d}{dx}\left(\frac{u}{v}\right)=\frac{v\frac{d\left(u\right)}{dx}-u\frac{d\left(v\right)}{dx}}{v^2}$

Therefore,

$f^{\prime }\left(x\right)=\frac{{\log }_{a}x\left(4x^3\right)-x^4\left(\frac{1}{x{\log }_{e}a}\right)}{{\left({\log }_{a}x\right)}^2}$

$f^{\prime }\left(x\right)=\frac{4x^3{\log }_{a}x-\left(\frac{x^3}{{\log }_{e}a}\right)}{{\left({\log }_{a}x\right)}^2}$

$f^{\prime }\left(x\right)=\frac{4x^3{\log }_{a}x}{{\left({\log }_{a}x\right)}^2}-\frac{\frac{x^3}{{\log }_{e}a}}{{\left({\log }_{a}x\right)}^2}$

$f^{\prime }\left(x\right)=\frac{4x^3}{\left({\log }_{a}x\right)}-\frac{x^3}{{\log }_{e}a{\left({\log }_{a}x\right)}^2}$

$f^{\prime }\left(x\right)=\frac{4x^3}{\left({\log }_{a}x\right)}-\frac{x^3}{{\log }_{e}a{\left(\frac{{\log }_{e}x}{{\log }_{e}a}\right)}^2}$

$f^{\prime }\left(x\right)=\frac{4x^3}{\left({\log }_{a}x\right)}-\frac{x^3{\log }_{e}a}{{\left({\log }_{e}x\right)}^2}$

Hence, this is the answer.