Question

I have a calculus I question about derivatives:1) Find the derivative of y=cos^-1(2x)

Answer 1

`y=\cos^(-1)(2x)`

1. Let u=cos(2x)

`\begin{gathered} y=u^(-1) \\ \\ Chain\text{ }rule: \\ (d)/(dx)f(u)=(d)/(du)(f(u))*(d)/(dx)(u) \\ \\ (dy)/(dx)=(d)/(du)u^(-1)*(d)/(dx)cos(2x) \\ \\ (dy)/(dx)=-1u^(-2)*(-2sin(2x)) \\ \\ u=cos(2x) \\ \\ (dy)/(dx)=-cos^(-2)(2x)*(-2sin(2x)) \end{gathered}`

Simplify:

`\begin{gathered} (dy)/(dx)=-(1)/(cos^2(2x))*(-2sin(2x)) \\ \\ (dy)/(dx)=(2sin(2x))/(cos(2x)cos2x)) \\ \\ (dy)/(dx)=2*(sin(2x))/(cos(2x))*(1)/(cos(2x)) \\ \\ (sinx)/(cosx)=tanx \\ \\ (1)/(cosx)=secx \\ \\ \\ (dy)/(dx)=2tan(2x)sec(2x) \end{gathered}`

Then, the derivate of the given function is:

2tan(2x)sec(2x)