# How do you use the Product Rule to find the derivative of y(u)=(u^-2 + u^-3)(u^5 - 2u^2)e^(2u)?

Aug 9, 2015

${y}^{'} = {e}^{2 u} \cdot \left(2 {u}^{3} + 5 {u}^{2} + 2 u - 4 - 4 {u}^{- 1} + 2 {u}^{- 2}\right)$

#### Explanation:

You're dealing with a product of three functions

$y \left(u\right) = f \left(u\right) \cdot g \left(u\right) \cdot h \left(u\right)$

which means that the formula for differentiating using the product rule will take the form

$\textcolor{b l u e}{\frac{d}{\mathrm{du}} \left(y\right) = \left[\frac{d}{\mathrm{du}} \left(f \left(u\right)\right)\right] \cdot g \left(u\right) \cdot h \left(u\right) + f \left(u\right) \cdot \left[\frac{d}{\mathrm{du}} \left(g \left(u\right)\right)\right] \cdot h \left(u\right) + f \left(u\right) \cdot g \left(u\right) \cdot \frac{d}{\mathrm{du}} \left(h \left(u\right)\right)}$

$f \left(u\right) = {u}^{- 2} + {u}^{- 3}$

$g \left(u\right) = {u}^{5} - 2 {u}^{2}$

$h \left(u\right) = {e}^{2 u}$

The derivative of $y \left(u\right)$ will thus be

$\frac{d}{\mathrm{du}} \left(y \left(u\right)\right) = \left[\frac{d}{\mathrm{du}} \left({u}^{- 2} + {u}^{- 3}\right)\right] \cdot \left({u}^{5} - 2 {u}^{2}\right) \cdot {e}^{2 u} + \left({u}^{- 2} + {u}^{- 3}\right) \cdot \left[\frac{d}{\mathrm{du}} \left({u}^{5} - 2 {u}^{2}\right)\right] \cdot {e}^{2 u} + \left({u}^{- 2} + {u}^{- 3}\right) \cdot \left({u}^{5} - 2 {u}^{2}\right) \cdot \frac{d}{\mathrm{dx}} \left({e}^{2 u}\right)$

To make the calculations easier to follow, I'll take each of those three derivatives separately.

$\frac{d}{\mathrm{du}} \left({u}^{- 2} + {u}^{- 3}\right) = \frac{d}{\mathrm{du}} {u}^{- 2} + \frac{d}{\mathrm{du}} {u}^{- 3}$

$\frac{d}{\mathrm{du}} \left({u}^{- 2} + {u}^{- 3}\right) = - 2 {u}^{- 3} + \left(- 3 {u}^{- 4}\right)$

$\frac{d}{\mathrm{du}} \left({u}^{- 2} + {u}^{- 3}\right) = - 2 {u}^{- 3} - 3 {u}^{- 4}$

Next, you have

$\frac{d}{\mathrm{du}} \left({u}^{5} - 2 {u}^{2}\right) = \frac{d}{\mathrm{du}} {u}^{5} - 2 \frac{d}{\mathrm{du}} {u}^{2}$

$\frac{d}{\mathrm{du}} \left({u}^{5} - 2 {u}^{2}\right) = 5 {u}^{4} - 4 u$

To differentiate ${e}^{2 u}$, use the chain rule for ${e}^{t}$, with $t = 2 u$. This will get you

$\frac{d}{\mathrm{du}} \left({e}^{t}\right) = \frac{d}{\mathrm{dt}} \left({e}^{t}\right) \cdot \frac{d}{\mathrm{du}} \left(t\right)$

$\frac{d}{\mathrm{du}} \left({e}^{t}\right) = {e}^{t} \cdot 2$

$\frac{d}{\mathrm{du}} \left({e}^{2 u}\right) = 2 {e}^{2 u}$

I'll break up the target derivative further to write the three products separately. The first product is

$\left[\frac{d}{\mathrm{du}} \left({u}^{-} 2 + {u}^{-} 3\right)\right] \cdot \left({u}^{5} - 2 {u}^{2}\right) \cdot {e}^{2 u}$

$\left(- 2 {u}^{- 3} - 3 {u}^{-} 4\right) \cdot \left({u}^{5} - 2 {u}^{2}\right) \cdot {e}^{2 u}$

$\left(- 2 {u}^{2} + 4 {u}^{-} 1 - 3 u + 6 {u}^{-} 2\right) \cdot {e}^{2 u} \to \textcolor{b l u e}{\left({P}_{1}\right)}$

The second product is

$\left({u}^{-} 2 + {u}^{-} 3\right) \cdot \left[\frac{d}{\mathrm{du}} \left({u}^{5} - 2 {u}^{2}\right)\right] \cdot {e}^{2 u}$

$\left({u}^{-} 2 + {u}^{-} 3\right) \cdot \left(5 {u}^{4} - 4 u\right) \cdot {e}^{2 u}$

$\left(5 {u}^{2} - 4 {u}^{-} 1 + 5 u - 4 {u}^{-} 2\right) \cdot {e}^{2 u} \to \textcolor{b l u e}{\left({P}_{2}\right)}$

The third product is

$\left({u}^{-} 2 + {u}^{-} 3\right) \cdot \left({u}^{5} - 2 {u}^{2}\right) \cdot \frac{d}{\mathrm{du}} {e}^{2 u}$

$2 \cdot \left({u}^{3} - 2 + {u}^{-} 2 - 2 {u}^{-} 1\right) \cdot {e}^{2 u} \to \textcolor{b l u e}{\left({P}_{3}\right)}$

The target derivative will thus be

${y}^{'} = \textcolor{b l u e}{{P}_{1}} + \textcolor{b l u e}{{P}_{2}} + \textcolor{b l u e}{{P}_{3}}$

${y}^{'} = {e}^{2 u} \cdot \left(- 2 {u}^{-} 2 + 4 {u}^{-} 1 - 3 u + 6 {u}^{-} 2 + 5 {u}^{2} - 4 {u}^{-} 1 + 5 u - 4 {u}^{-} 2 + 2 {u}^{3} - 4 + 2 {u}^{-} 2 - 4 {u}^{-} 1\right)$

This can be simplified to

${y}^{'} = \textcolor{g r e e n}{{e}^{2 u} \cdot \left(2 {u}^{3} + 5 {u}^{2} + 2 u - 4 - 4 {u}^{- 1} + 2 {u}^{- 2}\right)}$