# How do you use the product rule to differentiate y= (2+x) /( 2-3x)#?

Dec 9, 2016

It is (perhaps) more obvious to use the quotient rule, but we can use the product rule (and the chain rule).

#### Explanation:

To differentiate using the product rule, we must first write the quotient as a product

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

I use the product rule in the following order:

$\frac{d}{\mathrm{dx}} \left(u v\right) = u ' v + u v '$.

$\frac{\mathrm{dy}}{\mathrm{dx}} = {\overbrace{\left(1\right)}}^{u '} {\overbrace{{\left(2 - 3 x\right)}^{-} 1}}^{v} + {\overbrace{\left(2 + x\right)}}^{u} {\overbrace{\left[- 1 {\left(2 - 3 x\right)}^{-} 2 \cdot \frac{d}{\mathrm{dx}} \left(2 - 3 x\right)\right]}}^{v '}$

$= {\left(2 - 3 x\right)}^{-} 1 + \left(2 + x\right) \left[- {\left(2 - 3 x\right)}^{-} 2 \left(- 3\right)\right]$

$= {\left(2 - 3 x\right)}^{-} 1 + 3 \left(2 + x\right) {\left(2 - 3 x\right)}^{-} 2$

We're finished with the calculus, but we can do some algebra:

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

$= \frac{2 - 3 x + 6 + 3 x}{2 - 3 x} ^ 2$

$= \frac{8}{2 - 3 x} ^ 2$