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

Apr 17, 2016

Use the product rule, and the power and chain rules.

Explanation:

$y = u v$ $\text{ }$ $\Rightarrow$ $\text{ }$ $y ' = u ' v + u v '$

In this case $u = {\left(x + 7\right)}^{10}$,

so $u ' = 10 {\left(x + 7\right)}^{9} \left[\frac{d}{\mathrm{dx}} \left(x + 7\right)\right] = 10 {\left(x + 7\right)}^{9} \left[1\right] = 10 {\left(x + 7\right)}^{9}$.

And $v = {\left({x}^{2} + 2\right)}^{7}$

So $v ' = 7 {\left({x}^{2} + 2\right)}^{6} \left[\frac{d}{\mathrm{dx}} \left({x}^{2} + 2\right)\right] = 7 {\left({x}^{2} + 2\right)}^{6} \cdot 2 x$

With practice, it will not be necessary to write all of the above.

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

We've finished the calculus, but we can clear up the answer using algebra:

We really have two terms (Things that are added are called "terms".)

$y ' = {\underbrace{10 {\left(x + 7\right)}^{9} {\left({x}^{2} + 2\right)}^{7}}}_{\text{First Term" + underbrace((x+7)^10*7(x^2+2)^6 * 2x)_"Second Term}}$

Usually, we'll clean up both terms, but the first term here is fine, so we'll just clean up the second.

$y ' = {\underbrace{10 {\left(x + 7\right)}^{9} {\left({x}^{2} + 2\right)}^{7}}}_{\text{First Term" + underbrace(14x (x+7)^10 (x^2+2)^6)_"Second Term}}$

Notice the common factors. Let's factor them out:

$y ' = 2 {\left(x + 7\right)}^{9} {\left({x}^{2} + 2\right)}^{6} \left[5 \left({x}^{2} + 2\right) + x \left(x + 7\right)\right]$

Now we can finish by simplifying the expression in the square brackets.

$y ' = 2 {\left(x + 7\right)}^{9} {\left({x}^{2} + 2\right)}^{6} \left[6 {x}^{2} + 17\right]$

You can change those square brackets to parentheses and move that factor to right after the $2$ if you think it looks nicer.

$y ' = 2 \left(6 {x}^{2} + 17\right) {\left(x + 7\right)}^{9} {\left({x}^{2} + 2\right)}^{6}$