How do you differentiate given y=(secx^3)sqrt(sin2x)y=(secx3)sin2x?

1 Answer
May 16, 2018

dy/dx=secx^3((cos2x)/sqrt(sin2x)+3x^2tanx^3sqrt(sin2x))dydx=secx3(cos2xsin2x+3x2tanx3sin2x)

Explanation:

We have y=uvy=uv where uu and vv are both functions of xx.

dy/dx=uv'+vu'

u=secx^3
u'=3x^2secx^3tanx^3

v=(sin2x)^(1/2)
v'=(sin2x)^(-1/2)/2*d/dx[sin2x]=(sin2x)^(-1/2)/2*2cos2x=(cos2x)/sqrt(sin2x)

dy/dx=(secx^3cos2x)/sqrt(sin2x)+3x^2secx^3tanx^3sqrt(sin2x)

dy/dx=secx^3((cos2x)/sqrt(sin2x)+3x^2tanx^3sqrt(sin2x))