# How do you differentiate y = (x + 7)^10 (x^2 + 2)^7?

Jul 6, 2015

$y ' = \left(10 \left({x}^{2} + 2\right) + 14 x \left(x + 7\right)\right) {\left(x + 7\right)}^{9} {\left({x}^{2} + 2\right)}^{6}$

$= \left(24 {x}^{2} + 98 x + 20\right) {\left(x + 7\right)}^{9} {\left({x}^{2} + 2\right)}^{6}$

#### Explanation:

$y = {\left(x + 7\right)}^{10} {\left({x}^{2} + 2\right)}^{7}$ is of the form:

$y = U \left(x\right) V \left(x\right)$

An equation of this form is differentaited like this:

$y ' = U ' \left(x\right) V \left(x\right) + U \left(x\right) V ' \left(x\right)$

$U \left(x\right)$ and $V \left(x\right)$ are both of the form:

$U \left(x\right) = g \left(f \left(x\right)\right)$

An equation of this form is differentiated like this:

$U ' \left(x\right) = f ' \left(x\right) g ' \left(f \left(x\right)\right)$

$\rightarrow U ' \left(x\right) = \frac{d \left(x + 7\right)}{\mathrm{dx}} \frac{d \left({\left(x + 7\right)}^{10}\right)}{d \left(x + 7\right)} = 1 \cdot 10 {\left(x + 7\right)}^{9}$
$= 10 {\left(x + 7\right)}^{9}$

$\rightarrow V ' \left(x\right) = \frac{d \left({x}^{2} + 2\right)}{\mathrm{dx}} \frac{d \left({\left({x}^{2} + 2\right)}^{7}\right)}{d \left({x}^{2} + 2\right)} = 2 x \cdot 7 {\left({x}^{2} + 2\right)}^{6}$
$= 14 x {\left({x}^{2} + 2\right)}^{6}$

Therefore:

$y ' = 10 {\left(x + 7\right)}^{9} {\left({x}^{2} + 2\right)}^{7} + 14 x {\left(x + 7\right)}^{10} {\left({x}^{2} + 2\right)}^{6}$

$= \left(10 \left({x}^{2} + 2\right) + 14 x \left(x + 7\right)\right) {\left(x + 7\right)}^{9} {\left({x}^{2} + 2\right)}^{6}$
$= \left(24 {x}^{2} + 98 x + 20\right) {\left(x + 7\right)}^{9} {\left({x}^{2} + 2\right)}^{6}$