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How do you differentiate #f(x)= x^2*tan^-1 x#?

Answer 1

The answer is #f'(x)=x^2/(1+x^2)+2x tan^{-1}(x)#

This follows from the Product Rule #d/dx(g(x)h(x))=g(x)h'(x)+g'(x)h(x)# (with #g(x)=x^2# and #h(x)=tan^{-1}(x)#), the Power Rule #d/dx(x^(n))=nx^(n-1)#, and the derivative of inverse tangent #d/dx(tan^{-1}(x))=1/(1+x^2)#.

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Answer 2

Luke Alvarez

To differentiate the function ( f(x) = x^2 \tan^{-1}(x) ), you can use the product rule, which states that the derivative of a product of two functions is the derivative of the first function times the second function plus the first function times the derivative of the second function.

Let ( u(x) = x^2 ) and ( v(x) = \tan^{-1}(x) ). Then, ( u'(x) = 2x ) and ( v'(x) = \frac{1}{1+x^2} ).

Applying the product rule:

[ f'(x) = u'(x)v(x) + u(x)v'(x) ]

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

[ = 2x \tan^{-1}(x) + \frac{x^2}{1+x^2} ]

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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.