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

Apr 13, 2017

$f ' \left(x\right) = \frac{6 {\sec}^{2} \left(\frac{1}{x}\right)}{{x}^{2} \sqrt{{\tan}^{5} \left(\frac{1}{x}\right)}}$

#### Explanation:

By rewriting a bit,

$f \left(x\right) = 4 {\left({\tan}^{3} \left({x}^{- 1}\right)\right)}^{- \frac{1}{2}}$

By applying Power Rule & Chain Rule repeatedly,

$f ' \left(x\right) = 4 \cdot \left[- \frac{1}{2} {\left({\tan}^{3} \left({x}^{- 1}\right)\right)}^{- \frac{3}{2}} \cdot \left({\tan}^{3} \left({x}^{- 1}\right)\right) '\right]$

$= - 2 {\left({\tan}^{3} \left({x}^{- 1}\right)\right)}^{- \frac{3}{2}} \cdot 3 {\tan}^{2} \left({x}^{- 1}\right) \cdot \left[\tan \left({x}^{- 1}\right)\right] '$

$= - 6 {\left({\tan}^{3} \left({x}^{- 1}\right)\right)}^{- \frac{3}{2}} \cdot {\tan}^{2} \left({x}^{- 1}\right) \cdot {\sec}^{2} \left({x}^{- 1}\right) \cdot \left({x}^{- 1}\right) '$

$= - 6 {\left({\tan}^{3} \left({x}^{- 1}\right)\right)}^{- \frac{3}{2}} \cdot {\tan}^{2} \left({x}^{- 1}\right) \cdot {\sec}^{2} \left({x}^{- 1}\right) \cdot \left(- {x}^{- 2}\right)$

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.