How do you differentiate $g(z)=(z^2+1)/(x^3-5)$ using the quotient rule?

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Answer 1 · 2017-09-20T21:12:26

$(-z^4 - 3z^2 -10z)/(z^6 - 10z^3 +25)$

The quotient rule can be stated as;

$d/dz f(z)/g(z) = (f'(z)g(z) - f(z)g'(z))/g^2(z)$

We can choose our $f(z)$ and $g(z)$ and take each derivative separately. We only need the power rule here.

$f(z) = z^2+1$
$f'(z) = 2z$

$g(z) = z^3-5$
$g'(z) = 3z^2$

Now we have all of the pieces we need, we can plug them into our power rule function.

$(2z(z^3-5) - 3z^2(z^2+1))/(z^3-5)^2$

Now we can simplify our terms.

$(2z^4 - 10z - 3z^4 - 3z^2)/(z^6-10z^3 +25)$

$(-z^4 - 3z^2 -10z)/(z^6 - 10z^3 +25)$

How do you differentiate g(z)=(z^2+1)/(x^3-5)g(z)=(z^2+1)/(x^3-5) using the quotient rule?