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!