Question

How do you differentiate `f(x)=x^2lnsec(x^2)` using the chain rule?

Answer 1

`dy/dx =2x(x^2*tan(x^2) + ln(sec(x^2)))`

Explanation:

So we have

`y = x^2ln(sec(x^2))`

Derivating and using the product rule we have

`dy/dx = x^2d/dx(ln(sec(x^2)) + ln(sec(x^2))d/dxx^2`

`dy/dx = x^2d/dx(ln(sec(x^2)) + 2xln(sec(x^2))`

Let's say `sec(x^2) = u` then, by the chain rule we have

`dy/dx = x^2d/dx(ln(u)) + 2xln(sec(x^2))`
`dy/dx = x^2d/(du)(ln(u))(du)/dx + 2xln(sec(x^2))`
`dy/dx = x^2/u*d/dx(sec(x^2)) + 2xln(sec(x^2))`

Let's say `x^2 = v` then we have

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

`dy/dx = x^2/u*tan(v)*sec(v)*2x + 2xln(sec(x^2))`

Putting it all in terms of `x` we have

`dy/dx =2x(x^2/sec(x^2)*tan(x^2)*sec(x^2) + ln(sec(x^2)))`

`dy/dx =2x(x^2*tan(x^2) + ln(sec(x^2)))`