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

Answer 1

#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))#

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Answer 2

Ezra Adams

To differentiate ( f(x) = \ln\left(\frac{x^2}{3-x}\right) ), you can use the chain rule and the quotient rule. The derivative is ( f'(x) = \frac{2x - x^2}{x(3-x)} ).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.