How do you take the derivative of #tan^-1(x^2)#?

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Answer 1 · 2015-08-26T10:26:54

#y^' = (2x)/(1 + x^4)#

You can differentiate a function #y = tan^(-1)(x^2)# by using implicit differentiation.

So, if you have a function #y = tan^(-1)(x^2)#, then you know that you can write

#tan(y) = x^2#

Differentiate both sides with respect to #x# to get

#d/(dy)(tany) * (dy)/dx = d/dx(x^2)#

#sec^2y * (dy)/dx = 2x#

This is equivalent to saying that

#(dy)/dx = (2x)/sec^2y#

Remember that you have

#color(blue)(sec^2x = 1 + tan^2x)#

which means that you get

#(dy)/dx = (2x)/(1 + tan^2y)#

Finally, replace #tan^2y# with #x^2# to get

#(dy)/dx = color(green)((2x)/(1 + x^4))#