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

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Answer 1 · 2016-05-14T04:55:32

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

#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)#