How do you differentiate #y=((x^2+1)/(x^2-1))^3#?

Question

Emily Ammerman · Accepted Answer

# y'=-12x(((x^2+1)^2)/(x^2-1)^4)# #y=((x^2+1)/(x^2-1))^3# #y'=3((x^2+1)/(x^2-1))^2*((x^2+1)/(x^2-1))'# # ((x^2+1)/(x^2-1))'=((2x(x^2-1)-(x^2+1)2x)/(x^2-1)^2)# #=((2x((x^2-1)-(x^2+1)))/(x^2-1)^2)# #=((2x((x^2-1-x^2-1)))/(x^2-1)^2)# #=((2x((-2)))/(x^2-1)^2)# #=((-4x)/(x^2-1)^2)# #y'=3((x^2+1)/(x^2-1))^2*((x^2+1)/(x^2-1))'# #=> y'=3((x^2+1)/(x^2-1))^2*((-4x)/(x^2-1)^2)# #=> y'=3((x^2+1)^2/(x^2-1)^2)*((-4x)/(x^2-1)^2)# #=> y'=-12x(((x^2+1)^2)/(x^2-1)^4)#

Christopher Agresta · Answer

undefined To differentiate the function $ y = \left( \frac{x^2+1}{x^2-1} \right)^3 $, you can use the chain rule. First, find the derivative of the inner function $ \frac{x^2+1}{x^2-1} $, and then multiply it by the derivative of $ x^3 $.

The derivative of $ \frac{x^2+1}{x^2-1} $ can be found using the quotient rule, which states that for functions $ u(x) $ and $ v(x) $, $ \left( \frac{u}{v} \right)' = \frac{u'v - uv'}{v^2} $.

Let $ u(x) = x^2+1 $ and $ v(x) = x^2-1 $. Then $ u'(x) = 2x $ and $ v'(x) = 2x $.

Using the quotient rule, $ \left( \frac{u}{v} \right)' = \frac{(2x)(x^2-1) - (x^2+1)(2x)}{(x^2-1)^2} $.

Simplify the expression.

After finding $ \frac{du}{dv} $, multiply it by the derivative of $ x^3 $, which is $ 3x^2 $.

So, $ \frac{dy}{dx} = 3x^2 \left( \frac{(2x)(x^2-1) - (x^2+1)(2x)}{(x^2-1)^2} \right) $.