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

1 Answer
Apr 13, 2017

Answer:

#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.