# How do you differentiate f(x)=x^2(x+7)^3 using the product rule?

Nov 7, 2016

$f ' \left(x\right) = {\left(x + 7\right)}^{2} \left(x\right) \left(5 x + 14\right)$

#### Explanation:

$f \left(x\right) = {\left(x\right)}^{2} {\left(x + 7\right)}^{3}$

Product rule: if $f \left(x\right) = A B$, then $f ' \left(x\right) = A B ' + A ' B$

In this case, too differentiate terms "A" and "B", we have to use the chain rule.

$f ' \left(x\right) = {\left(x\right)}^{2} \left(3\right) {\left(x + 7\right)}^{2} + \left(2 x\right) {\left(x + 7\right)}^{3}$

Now simplify by factoring:
$f ' \left(x\right) = {\left(x + 7\right)}^{2} \left(x\right) \left[3 x + 2 \left(x + 7\right)\right]$
$f ' \left(x\right) = {\left(x + 7\right)}^{2} \left(x\right) \left(3 x + 2 x + 14\right)$
$f ' \left(x\right) = {\left(x + 7\right)}^{2} \left(x\right) \left(5 x + 14\right)$

Nov 7, 2016

$\therefore f ' \left(x\right) = x {\left(x + 7\right)}^{2} \left(5 x + 14\right\}$

#### Explanation:

If you are studying maths, then you should learn the Product Rule for Differentiation, and practice how to use it:

$\frac{d}{\mathrm{dx}} \left(u v\right) = u \frac{\mathrm{dv}}{\mathrm{dx}} + v \frac{\mathrm{du}}{\mathrm{dx}}$, or, $\left(u v\right) ' = \left(\mathrm{du}\right) v + u \left(\mathrm{dv}\right)$

I was taught to remember the rule in words; "The first times the derivative of the second plus the second times the derivative of the first ".

So with $f \left(x\right) = {x}^{2} {\left(x + 7\right)}^{3}$ Then

$\left\{\begin{matrix}\text{Let "u=x^2 & => & (du)/dx=2x \\ "And "v=(x+7)^3 & => & (dv)/dx=3(x+7)^2 " (by chain rule)}\end{matrix}\right.$

$\therefore \frac{d}{\mathrm{dx}} \left(u v\right) = u \frac{\mathrm{dv}}{\mathrm{dx}} + v \frac{\mathrm{du}}{\mathrm{dx}}$
$\therefore f ' \left(x\right) = \left({x}^{2}\right) \left(3 {\left(x + 7\right)}^{2}\right) + \left({\left(x + 7\right)}^{3}\right) \left(2 x\right)$
$\therefore f ' \left(x\right) = 3 {x}^{2} {\left(x + 7\right)}^{2} + 2 x {\left(x + 7\right)}^{3}$
$\therefore f ' \left(x\right) = x {\left(x + 7\right)}^{2} \left\{3 x + 2 \left(x + 7\right)\right\}$
$\therefore f ' \left(x\right) = x {\left(x + 7\right)}^{2} \left(3 x + 2 x + 14\right\}$
$\therefore f ' \left(x\right) = x {\left(x + 7\right)}^{2} \left(5 x + 14\right\}$