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

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Answer 1 · 2015-11-16T02:22:46

$frac(x-6)(x(x-3))$

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.
$d/(dx)[x^2/(3-x)]=(d/(dx)[x^2]*(3-x)-x^2*d/(dx)[3-x])/(3-x)^2$
$=(2x(3-x)-x^2(-1))/(3-x)^2=(-x^2+6x)/(3-x)^2$

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

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