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

Question

2809 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-12-07T03:28:26

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

We write #tany# as #siny/cosy# and differentiate using the quotient rule (with respect to #x#) .

#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 #sin^2beta + cos^2beta = 1# to simplify further...

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

Use the identity #sectheta = 1/costheta#...

#dy/dxsec^2y = 2x#

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

Hopefully this helps!