Question

How do you differentiate `f(x)=tan((1/cos(7x))^2)` using the chain rule?

Answer 1

`dy/dx = 14sec^2(sec^2(7x))tan(7x)sec^2(7x)`

Explanation:

Okay, so, we have

`y = tan(sec^2(7x))`

Call `sec^2(7x) = u`

`dy/dx = d/(du)tan(u)(du)/dx = sec^2(u)(d)/dxsec^2(7x)`

Call `sec(7x) = v`

`dy/dx = sec^2(u)d/(dv)v^2(dv)/dx = 2v*sec^2(u)(d)/dx(sec(7x))`

Call `7x = w`

`dy/dx = 2v*sec^2(u)d/(dw)sec(w)(dw)/dx`
`dy/dx = 2v*sec^2(u)*tan(w)sec(w)d/dx(7x)`
`dy/dx = 14v*sec^2(u)*tan(w)sec(w)`

Substituting `w` back to `x`

`dy/dx = 14v*sec^2(u)*tan(7x)sec(7x)`

Substituting `v` back to `x`

`dy/dx = 14*sec^2(u)*tan(7x)sec^2(7x)`

Substituing `u` back to `x`

`dy/dx = 14sec^2(sec^2(7x))tan(7x)sec^2(7x)`