# How do you differentiate f(x)= (x + 7)^10 (x^2 + 2)^7  using the product rule?

Dec 14, 2015

Product Rule:
If we have two functions $f \left(x\right)$ and $g \left(x\right)$, then
$\frac{d}{\mathrm{dx}}$ $f \left(x\right) g \left(x\right)$ = ${f}^{'} \left(x\right) g \left(x\right) + f \left(x\right) {g}^{'} \left(x\right)$

Chain Rule:
If we have two functions, $f \left(x\right)$ and $g \left(x\right)$, then
$\frac{d}{\mathrm{dx}}$ $f \left(g \left(x\right)\right)$=g^'(x)•f^'(g(x))

Original Equation:
$f \left(x\right) = {\left(x + 7\right)}^{10} {\left({x}^{2} + 2\right)}^{7}$

Use the chain rule to find the derivative of the first equation:
${\left(x + 7\right)}^{10}$
(1)•(10)(x+7)^9
$10 {\left(x + 7\right)}^{9}$

Now multiply by the second function:
$10 {\left(x + 7\right)}^{9} {\left({x}^{2} + 2\right)}^{7}$

Next, use the chain rule to find the derivative of the second equation:
${\left({x}^{2} + 2\right)}^{7}$
$\left(2 x\right) \left(7\right) {\left({x}^{2} + 2\right)}^{6}$
$\left(14 x\right) {\left({x}^{2} + 2\right)}^{6}$

Multiply by the first function:
${\left(x + 7\right)}^{10} \left(14 x\right) {\left({x}^{2} + 2\right)}^{6}$

Add together the two multiplied functions:
$\left(10 {\left(x + 7\right)}^{9} {\left({x}^{2} + 2\right)}^{7}\right)$+$\left({\left(x + 7\right)}^{10} \left(14 x\right) {\left({x}^{2} + 2\right)}^{6}\right)$

Factor out ${\left(x + 7\right)}^{9}$ and ${\left({x}^{2} + 2\right)}^{6}$:
$\left\{{\left(x + 7\right)}^{9} {\left({x}^{2} + 2\right)}^{6}\right\} 10 \left({x}^{2} + 2\right) + \left(x + 7\right) \left(14 x\right)$

Multiply and combine like terms for the factored section:
$10 \left({x}^{2} + 2\right) + \left(x + 7\right) \left(14 x\right)$
$10 {x}^{2} + 20 + 14 {x}^{2} + 98 x$
$24 {x}^{2} + 98 x + 20$

Put back the factors.
${f}^{'} \left(x\right) = \left\{{\left(x + 7\right)}^{9} {\left({x}^{2} + 2\right)}^{6}\right\} \left(24 {x}^{2} + 98 x + 20\right)$