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

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

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

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

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

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