Question

What is the derivative of `f(x)=(x^3)/(9 (3lnx-1))`?

Answer 1

`f'(x)=(x^2(3lnx-2))/(3(3lnx-1)^2)`

Explanation:

First, note that this can be simplified as

`f(x)=x^3/(27lnx-9)`

To differentiate this, we can use the quotient rule, which states that

`d/dx[(g(x))/(h(x))]=(g'(x)h(x)-h'(x)g(x))/[h(x)]^2`

Applying this to the function at hand, we see that

`f'(x)=((27lnx-9)d/dx[x^3]-x^3d/dx[27lnx-9])/(27lnx-9)^2`

These derivatives are fairly simple to find. The one thing that may be tricky is remembering that `d/dx[lnx]=1/x`, so `d/dx[27lnx]=27/x`.

`f'(x)=((27lnx-9)(3x^2)-x^3(27/x))/(27lnx-9)^2`

Simplify.

`f'(x)=(81x^2lnx-27x^2-27x^2)/(27lnx-9)^2`

`f'(x)=(27x^2(3lnx-2))/(81(3lnx-1)^2)`

Note that while it appears a `9` was factored from the denominator, it really was `9^2` since the `9` was factored from two terms, since the denominator is squared.

`f'(x)=(x^2(3lnx-2))/(3(3lnx-1)^2)`