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

Question

Aurora Alvarado · Accepted Answer

#(-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)#

Luke Alvarez · Answer

undefined To differentiate $ g(z) = \frac{{z^2 + 1}}{{z^3 - 5}} $ using the quotient rule:

1. Identify $ f(z) = z^2 + 1 $ and $ h(z) = z^3 - 5 $.
2. Apply the quotient rule:

$$ g'(z) = \frac{{h(z) \cdot f'(z) - f(z) \cdot h'(z)}}{{[h(z)]^2}} $$

3. Find the derivatives of $ f(z) $ and $ h(z) $:

$$ f'(z) = 2z $$
$$ h'(z) = 3z^2 $$

4. Plug the derivatives and original functions into the quotient rule formula:

$$ g'(z) = \frac{{(z^3 - 5)(2z) - (z^2 + 1)(3z^2)}}{{(z^3 - 5)^2}} $$