Question

How do you differentiate `f(x) = 4/sqrt(tan^3(1/x)` using the chain rule?

Answer 1

`f'(x)=(6sec^2(1/x))/(x^2sqrt(tan^5(1/x)))`

Explanation:

By rewriting a bit,

`f(x)=4(tan^3(x^(-1)))^(-1/2)`

By applying Power Rule & Chain Rule repeatedly,

`f'(x)=4 cdot [-1/2(tan^3(x^(-1)))^(-3/2)cdot(tan^3(x^(-1)))']`

`=-2(tan^3(x^(-1)))^(-3/2)cdot 3tan^2(x^(-1))cdot[tan(x^(-1))]'`

`=-6(tan^3(x^(-1)))^(-3/2)cdot tan^2(x^(-1))cdot sec^2(x^(-1))cdot(x^(-1))'`

`=-6(tan^3(x^(-1)))^(-3/2)cdot tan^2(x^(-1))cdot sec^2(x^(-1))cdot(-x^(-2))`

By cleaning up a bit,

#=(6tan^2(1/x)sec^2(1/x))/(x^2sqrt(tan^9(1/x))) =(6tan^2(1/x)sec^2(1/x))/(x^2tan^2(1/x)sqrt(tan^5(1/x))) =(6sec^2(1/x))/(x^2sqrt(tan^5(1/x)))#

I hope that this was clear.