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

Answer 1

Charles Arocha

#(sec^2x-2x)/(tanx-x^2)#

Set #y=lnu, u=tanx-x^2#

#dy/(du)=1/u, (du)/(dx)=sec^2x-2x#

#(dy)/(dx)=(dy)/(du)*(du)/(dx)#

#=1/(tanx-x^2)*sec^2x-2x#

#=(sec^2x-2x)/(tanx-x^2)#

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Answer 2

Evelyn Agard

To differentiate ( f(x) = \ln(\tan(x) - x^2) ), you would use the chain rule and the derivative rules for logarithmic and trigonometric functions:

Differentiate the outer function (\ln(u)) with respect to (u), where (u = \tan(x) - x^2), to get (\frac{1}{u}).
Multiply by the derivative of the inner function, which is obtained by applying the chain rule. The derivative of (\tan(x)) is (\sec^2(x)), and the derivative of (x^2) is (2x).
Therefore, the derivative of (f(x)) is:

[ f'(x) = \frac{1}{\tan(x) - x^2} \left(\sec^2(x) - 2x\right) ]

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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.