Question

What is the derivative of `f(x)=tan(ln(cosx))`?

Answer 1

`dy/dx = -sec^2(ln(cos(x)))tan(x)`

Explanation:

So we have

`y = tan(ln(cos(x)))`

Let's say that `ln(cos(x)) = u`, so we have

`y = tan(u)`

`dy/dx = d/(du)tan(u)(du)/dx`
`dy/dx = sec^2(u)d/dx(ln(cos(x))`

Let's say that `cos(x) = v`, so we have

`dy/dx = sec^2(u)d/(dv)ln(v)(dv)/dx`
`dy/dx = sec^2(u)/vd/dxcos(x)`
`dy/dx = -(sec^2(u)sin(x))/v`

Putting everything in terms of `x`

`dy/dx = -sec^2(ln(cos(x)))sin(x)/cos(x)`
`dy/dx = -sec^2(ln(cos(x)))tan(x)`