How do you integrate f(x)=x(x^7+15)^3 using the product rule?

Jan 13, 2018

$f ' \left(x\right) = {\left({x}^{7} + 15\right)}^{3} + 21 {x}^{7} {\left({x}^{7} + 15\right)}^{2}$

Explanation:

I'm assuming that you meant differentiate, because we cannot integrate using the product rule.

$f \left(x\right) = x {\left({x}^{7} + 15\right)}^{3}$

Use the product rule, which states that

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

Let $u = x$, $v = {\left({x}^{7} + 15\right)}^{3}$

We now need to differentiate $v = {\left({x}^{7} + 15\right)}^{3}$, so we are going to use the chain rule, which states that

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\mathrm{dy}}{\mathrm{du}} \cdot \frac{\mathrm{du}}{\mathrm{dx}}$

Since we already have a $u$ for the product rule, and we are trying to differentiate $v$, let's rewrite the rule as

$\frac{\mathrm{dv}}{\mathrm{dx}} = \frac{\mathrm{dv}}{\mathrm{da}} \cdot \frac{\mathrm{da}}{\mathrm{dx}}$

Let $a = {x}^{7} + 15$, so $\frac{\mathrm{da}}{\mathrm{dx}} = 7 {x}^{6}$, $v = {a}^{3}$, $\frac{\mathrm{dv}}{\mathrm{da}} = 3 {a}^{2}$

Combining, we get $\frac{\mathrm{dv}}{\mathrm{dx}} = 7 {x}^{6} \cdot 3 {a}^{2}$

$\frac{\mathrm{dv}}{\mathrm{dx}} = 21 {x}^{6} {a}^{2}$

Putting back $a = {x}^{7} + 15$, we get

$\frac{\mathrm{dv}}{\mathrm{dx}} = 21 {x}^{6} {\left({x}^{7} + 15\right)}^{2}$

$v ' = 21 {x}^{6} {\left({x}^{7} + 15\right)}^{2}$

Now that we have differentiated $v$, let's put it in the product rule

$\frac{d}{\mathrm{dx}} \left(f \left(x\right)\right) = 1 \cdot {\left({x}^{7} + 15\right)}^{3} + x \left(21 {x}^{6} {\left({x}^{7} + 15\right)}^{2}\right)$

$f ' \left(x\right) = {\left({x}^{7} + 15\right)}^{3} + 21 {x}^{7} {\left({x}^{7} + 15\right)}^{2}$