Find the derivative of the function $(y^2-1)^2/(y^2+1)$

Question

How do I solve this? I tried using the quotient and the chain rule but I can't seem to get the correct answer.

Answer 1

You can set $\;f(y)=\dfrac{(y^2-1)^2}{y^2+1}$ and use logarithmic differentiation: \begin{align} \frac{f'(y)}{f(y)}&=2\,\frac{(y^2-1)'}{y^2-1}-\frac{(y^2+1)'}{y2+1}=2\,\frac{2y}{y^2-1}-\frac{2y}{y^2+1}=2y\biggl(\frac{2}{y^2-1}-\frac{1}{y^2+1}\biggr)=\frac{2y(y^2+5)}{(y^2-1)(y^2+1)} \end{align} whence $$ f'(y)=\frac{f'(y)}{f(y)}\,f(y)=\frac{2y(y^2+5)(y^2-1)}{(y^2+1)^2}. $$

Answer 2

Let: $$p=(y^2-1)^2\to p'=2(2y)(y^2-1)=4y(y^2-1) \text{ [chain rule]}$$ $$q=y^2+1 \to q'=2y \text{ [implicit]}$$ Then the quotient rule: $$\bigg(\frac pq\bigg)'=\frac{p'q-q'p}{q^2}$$ See if you can take this forward

Answer 3

HINT:

You can simplify the chain rule using $$(y^2-1)^2=\{(y^2+1)-2\}^2=?$$