How do you find the derivative of #((x^3)-(7/x))^-2 #?

1 Answer
marfre
Feb 12, 2017

#(-2x(3x^4+7))/(x^4-7)^3#

Explanation:

Rewrite & then use a combination of the power rule & chain rule :
#y = (x^3-7x^-1)^-2#

#y' = -2(x^3-7x^-1)^-3(3x^2+7x^-2)#

Simplify by making all exponents positive & find common denominators:

#y' = (-2(3x^2+7/x^2))/(x^3-7/x)^3 = (-2((3x^4+7)/x^2))/((x^4-7)/x)^3#

Simplify by cubing both numerator & denominator of the denominator term and multiplying reciprocals:
#y' = -2((3x^4+7)/x^2)(x^3)/(x^4-7)^3#

Simplify and rearrange to get the final answer:
#(-2x(3x^4+7))/(x^4-7)^3#

Related questions
See all questions in Chain Rule
Impact of this question
1768 views around the world
You can reuse this answer
Creative Commons License