Answer
$$y'=5\Bigg(\frac{(t+1)(t-1)}{(t-2)(t+3)}\Bigg)^5\Big(\frac{1}{t+1}+\frac{1}{t-1}-\frac{1}{t-2}-\frac{1}{t+3}\Big)$$
Work Step by Step
$$y=\Bigg(\frac{(t+1)(t-1)}{(t-2)(t+3)}\Bigg)^5$$
Take the natural logarithm of both sides: $$\ln y=\ln\Bigg(\frac{(t+1)(t-1)}{(t-2)(t+3)}\Bigg)^5=5\ln\frac{(t+1)(t-1)}{(t-2)(t+3)}$$ $$\ln y =5\Big(\ln(t+1)(t-1)-\ln(t-2)(t+3)\Big)$$ $$\ln y=5\Big(\ln(t+1)+\ln(t-1)-\ln(t-2)-\ln(t+3)\Big)$$
Now take the derivative of both sides, and remember that $(\ln x)'=1/x$:
$$\frac{1}{y}\times y'=5\Big(\frac{(t+1)'}{t+1}+\frac{(t-1)'}{t-1}-\frac{(t-2)'}{t-2}-\frac{(t+3)'}{t+3}\Big)$$ $$\frac{y'}{y}=5\Big(\frac{1}{t+1}+\frac{1}{t-1}-\frac{1}{t-2}-\frac{1}{t+3}\Big)$$
Find $y'$: $$y'=5y\Big(\frac{1}{t+1}+\frac{1}{t-1}-\frac{1}{t-2}-\frac{1}{t+3}\Big)$$ $$y'=5\Bigg(\frac{(t+1)(t-1)}{(t-2)(t+3)}\Bigg)^5\Big(\frac{1}{t+1}+\frac{1}{t-1}-\frac{1}{t-2}-\frac{1}{t+3}\Big)$$
