Question

How do you differentiate `f(x)=(2x-cos^3x)^2-(lnx)^2` using the chain rule?

Answer 1

`dy/dx = (4+6cos^2(x)sin(x))(2x-cos^3(x)) - (2ln(x))/x`

Explanation:

So we have

`y = (2x - cos^3(x))^2 - ln^2(x)`

Derivating it

`dy/dx = d/dx(2x - cos^3(x))^2 - d/dx(ln^2(x))`

Let's solve the log part first, let's call `ln(x) = u`

`dy/dx = d/dx(2x-cos^3(x))^2 - d/(du)u^2(du)/dx`
`dy/dx = d/dx(2x-cos^3(x))^2 - 2u*1/x`

Let's say that `(2x-cos^3(x)) = v`

`dy/dx = d/(dv)v^2(dv)/dx - 2u*1/x`
`dy/dx = 2v*d/dx(2x-cos^3(x)) - 2u*1/x`

We break that derivative in two parts and apply the distributive

`dy/dx = 2v*d/dx(2x) - 2v*d/dx(cos^3(x)) - 2u*1/x`
`dy/dx = 4v - 2v*d/dx(cos^3(x)) - 2u*1/x`

Let's say that `w = cos(x)`

`dy/dx = 4v - 2v*d/(dw)w^3(dw)/dx - 2u*1/x`
`dy/dx = 4v - 6vw^2*d/dx(cos(x)) - 2u*1/x`
`dy/dx = 4v + 6vw^2sin(x) - 2u*1/x`

Put everything in terms of `x`

`dy/dx = (4+6w^2sin(x))v-(2u)/x`
`dy/dx = (4+6cos^2(x)sin(x))(2x-cos^3(x)) - (2ln(x))/x`