# Given that #tany= x^2#, what is the value of #dy/dx#?

##### 1 Answer

Dec 7, 2016

#### Explanation:

We write

#siny/cosy = x^2#

#(cosy(cosy)(dy/dx) - (-siny xx siny)dy/dx)/(cosy)^2 = 2x#

#(cos^2y(dy/dx) + sin^2y(dy/dx))/cos^2y = 2x#

We use the identity

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

Use the identity

#dy/dxsec^2y = 2x#

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

Hopefully this helps!