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

1 Answer
May 14, 2016

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)