Question

How do you differentiate `f(x)=(sec^5 (1/x))^(1/3)` using the chain rule?

Answer 1

`d/dx(sec^5(1/x))^(1/3)=-5/(3x^2)sec^(5/3)(1/x)tan(1/x)`

Explanation:

`f(x)=(sec^5(1/x))^(1/3)=sec^(5/3)(1/x)`

Here, we have `f(x)=g(x)^(5/3)`, `g(x)=sec(h(x))` and `h(x)=1/x`

Hence according to chin rule

`(df)/(dx)=(df)/(dg)xx(dg)/(dh)xx(dh)/(dx)`

= `5/3(g(x))^(2/3)xxsec(h(x))tan(h(x))xx(-1)/x^2`

= `5/3sec^(2/3)(1/x) xxsec(1/x)tan(1/x)xx(-1)/x^2`

= `-5/(3x^2)sec^(5/3)(1/x)tan(1/x)`