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What is the derivative of tan inverse of #(x^2 y^5)#?

1 Answer

Shwetank Mauria

May 26, 2016

#d/dx(tan^(-1)(x^2y^5))=1/(1+x^4y^10)[2xy^5+5x^2y^4(dy)/(dx)]#

Explanation:

We have to find derivative of #tan^(-1)(x^2y^5)#. For this we use the concept of chain rule for implicit differentiation as well as product rule..

Hence #d/dx(tan^(-1)(x^2y^5))#

= #1/(1+(x^2y^5)^2)[2x xxy^5+x^2xx5y^4xx(dy)/(dx)]#

= #1/(1+x^4y^10)[2xy^5+5x^2y^4(dy)/(dx)]#

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