How do you find the derivative of #y = arctan(x^2)#?

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Answer 1 · 2016-09-16T03:24:22

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

Note that #d/dxarctan(x)=1/(1+x^2)#. Thus, according to the chain rule:

#d/dxarctan(f(x))=1/(1+f(x)^2)*f'(x)#

Thus:

#dy/dx=d/dxarctan(x^2)=1/(1+(x^2)^2)*d/dxx^2=(2x)/(1+x^4)#

Answer 2 · 2016-09-16T03:28:14

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

Rearrange the equation:

#tan(y)=x^2#

Differentiate both sides. Recall to use the chain rule on the left hand side.

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

Note that #sec^2(y)=tan^2(y)+1#:

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

Since #tan(y)=x^2#:

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

Solving for #dy/dx#:

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