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What is the derivative of this function #y=tan^-1(x^3)-x#?

Answer 1

Emilia Aguilar

#dy/dx = (3x^2)/(x^6+1)-1#

#y = arctan(x^3)-x#

#dy/dx= 1/((x^3)^2+1) * d/dx(x^3) -d/dx x# (Standard differential and chain rule)

#= 1/(x^6+1) * 3x^2 -1# (Power rule)

#= (3x^2)/(x^6+1)-1#

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Answer 2

Ezra Adams

To find the derivative of the function ( y = \tan^{-1}(x^3) - x ), use the chain rule and the derivative of ( \tan^{-1}(x) ), which is ( \frac{1}{1+x^2} ).

First, find the derivative of ( \tan^{-1}(x^3) ): [ \frac{d}{dx}(\tan^{-1}(x^3)) = \frac{1}{1+(x^3)^2} \cdot 3x^2 ]

Then, find the derivative of ( -x ): [ \frac{d}{dx}(-x) = -1 ]

Combining these, the derivative of the function is: [ \frac{d}{dx}(y) = \frac{3x^2}{1+x^6} - 1 ]

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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.