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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.

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  • $\begingroup$ Product rule quotient rule combo deal. $\endgroup$ – Randall Jan 31 '18 at 4:20
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    $\begingroup$ Have you tried canceling anything first? $\endgroup$ – Randall Jan 31 '18 at 4:20
  • $\begingroup$ ah, apply quotient rule and to each and then product rule? I hadn't tried cancelling anything either actually. $\endgroup$ – millartime Jan 31 '18 at 4:22
  • $\begingroup$ Well, cancelling is going to be easier... $\endgroup$ – Randall Jan 31 '18 at 4:22
  • $\begingroup$ I'll give that a go, thanks $\endgroup$ – millartime Jan 31 '18 at 4:23
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Make your problem simpler by hand first before bringing in the machinery. $$g(z)=\frac{z^2-2z-8}{z-3}\cdot\frac{z^2-9}{z-4}=\frac{(z-4)(z+2)}{z-3}\cdot\frac{(z+3)(z-3)}{z-4}=(z+2)(z+3)=z^2+5z+6$$ $$g'(z)=2z+5$$

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  • $\begingroup$ now that's helpful. Thank you! I wouldn't have realized that it could have been solved so easily. $\endgroup$ – millartime Jan 31 '18 at 4:29
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First,let's factor and simplify: $$\frac{(z-4)(z+2)}{(z-3)}\cdot\frac{(z+3)(z-3)}{(z-4)}=(z+2)(z+3)=z^2+5z+6 \implies g'(z)=2z+5$$

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