Question

How do you differentiate `f(x)= ln(x^2/(3-x))`?

Answer 1

`frac(x-6)(x(x-3))`

Explanation:

According to the Chain Rule, `d/(dx)[lnu]=(u')/u`.
Therefore, `f'(x)=frac(d/(dx)[x^2/(3-x)])(x^2/(3-x))`.

We will have to use the Quotient Rule.
`d/(dx)[x^2/(3-x)]=(d/(dx)[x^2]*(3-x)-x^2*d/(dx)[3-x])/(3-x)^2`
`=(2x(3-x)-x^2(-1))/(3-x)^2=(-x^2+6x)/(3-x)^2`

`f'(x)=((-x^2+6x)/(3-x)^2)/(x^2/(3-x))=(-x^2+6x)/(3-x)^2((3-x)/x^2)=(x(6-x))/(x^2(3-x))`
`color(blue)(f'(x)=frac(6-x)(x(3-x))=frac(x-6)(x(x-3))`