Question

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

Answer 1

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

Explanation:

We have `y=uv` where `u` and `v` are both functions of `x`.

`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))`