Question

How do you differentiate `f(x)=sec(1/sqrt(3x) )` using the chain rule?

Answer 1

`f'(x)=-(3(3x)^(-3/2))/2sec((3x)^(-1/2))tan((3x)^(-1/2))=-(3sec(1/sqrt(3x))tan(1/sqrt(3x)))/(2sqrt(3x)^3)`

Explanation:

The equation given can be simplified as `f(x)=sec((3x)^(-1/2))`

If `f(x)=sec(g(x))`, then `f'(x)=g'(x)sec(g(x))tan(g(x))`

If `g(x)=(h(x))^n`, then `g'(x)=h(x)*n(h(x))^(n-1)`

The derivative of `3x` is`3`.

`-1/2-1=-3/2`

`g'(x)=3*-1/2(3x)^(-3/2)=-(3(3x)^(-3/2))/2`

So, `f'(x)=-(3(3x)^(-3/2))/2sec((3x)^(-1/2))tan((3x)^(-1/2))=-(3sec(1/sqrt(3x))tan(1/sqrt(3x)))/(2sqrt(3x)^3)`