# How do you find the derivative of 6(z^2+z-1)^-1?

Aug 30, 2017

$- \frac{12 z + 6}{{z}^{2} + z - 1} ^ 2$

#### Explanation:

$\text{differentiate using the "color(blue)"chain rule}$

$\text{given "y=f(g(x)" then}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = f ' \left(g \left(x\right)\right) \times g ' \left(x\right) \leftarrow \text{ chain rule}$

$\frac{d}{\mathrm{dz}} \left(6 {\left({z}^{2} + z - 1\right)}^{-} 1\right)$

$= - 6 {\left({z}^{2} + z - 1\right)}^{-} 2 \times \frac{d}{\mathrm{dz}} \left({z}^{2} + z - 1\right)$

$= \frac{- 6 \left(2 z + 1\right)}{{z}^{2} + z - 1} ^ 2 = - \frac{12 z + 6}{{z}^{2} + z - 1} ^ 2$

Aug 30, 2017

Recall the power rule: $\frac{d}{\mathrm{dx}} {x}^{n} = n {x}^{n - 1}$. When we have a function to a power, we still use the power rule to differentiate it, but we do so as well as using the chain rule.

Combining the chain rule with the power rule for some function $u$ gives us: $\frac{d}{\mathrm{dx}} {u}^{n} = n {u}^{n - 1} \frac{\mathrm{du}}{\mathrm{dx}}$

Thus:

$\frac{d}{\mathrm{dz}} 6 {\left({z}^{2} + z - 1\right)}^{-} 1 = 6 \left(- 1 {\left({z}^{2} + z - 1\right)}^{-} 2\right) \frac{d}{\mathrm{dz}} \left({z}^{2} + z - 1\right)$

And we can use the power rule to find the derivative of ${z}^{2} + z - 1$:

$\frac{d}{\mathrm{dz}} 6 {\left({z}^{2} + z - 1\right)}^{-} 1 = - 6 {\left({z}^{2} + z - 1\right)}^{-} 2 \left(2 z + 1\right)$

=color(blue)((-6(2z+1))/(z^2+z-1)^2