# Taking the derivative of a product, which is made up of two quotients

I'm working on a calculus question, where we're asked to find $g'(z)$ for $$g(z)=\frac{z^2-2z-8}{z-3}\cdot\frac{z^2-9}{z-4}$$ So I was thinking that first you have to use the quotient rule on each side individually and then multiply them, but I seem to keep getting stuck and it's becoming pretty frustrating.

• Product rule quotient rule combo deal. – Randall Jan 31 '18 at 4:20
• Have you tried canceling anything first? – Randall Jan 31 '18 at 4:20
• ah, apply quotient rule and to each and then product rule? I hadn't tried cancelling anything either actually. – millartime Jan 31 '18 at 4:22
• Well, cancelling is going to be easier... – Randall Jan 31 '18 at 4:22
• I'll give that a go, thanks – millartime Jan 31 '18 at 4:23

Make your problem simpler by hand first before bringing in the machinery. $$g(z)=\frac{z^2-2z-8}{z-3}\cdot\frac{z^2-9}{z-4}=\frac{(z-4)(z+2)}{z-3}\cdot\frac{(z+3)(z-3)}{z-4}=(z+2)(z+3)=z^2+5z+6$$ $$g'(z)=2z+5$$
First,let's factor and simplify: $$\frac{(z-4)(z+2)}{(z-3)}\cdot\frac{(z+3)(z-3)}{(z-4)}=(z+2)(z+3)=z^2+5z+6 \implies g'(z)=2z+5$$