# How do you differentiate h(y)=1/(y^3+2y+1) using the quotient rule?

Aug 25, 2016

$h ' \left(y\right) = \frac{- 3 {y}^{2} - 2}{{y}^{3} + 2 y + 1} ^ 2$

#### Explanation:

The quotient rule states that for some function in the form $h \left(y\right) = f \frac{y}{g} \left(y\right)$, the function's derivative equals $h ' \left(y\right) = \frac{f ' \left(y\right) g \left(y\right) - f \left(y\right) g ' \left(y\right)}{g \left(y\right)} ^ 2$.

Here, for $h \left(y\right) = \frac{1}{{y}^{3} + 2 y + 1}$, our two functions being divided are:

$\left\{\begin{matrix}f \left(y\right) = 1 \\ g \left(y\right) = {y}^{3} + 2 y + 1\end{matrix}\right.$

We will need to know their derivatives in a moment:

$\left\{\begin{matrix}f ' \left(y\right) = 0 \\ g ' \left(y\right) = 3 {y}^{2} + 2\end{matrix}\right.$

Plugging these into the quotient rule formula gives us:

$h ' \left(y\right) = \frac{\left(0\right) \left({y}^{3} + 2 y + 1\right) - \left(1\right) \left(3 {y}^{2} + 2\right)}{{y}^{3} + 2 y + 1} ^ 2$

Simplifying:

$h ' \left(y\right) = \frac{- 3 {y}^{2} - 2}{{y}^{3} + 2 y + 1} ^ 2$