Question

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

Answer 1

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

Explanation:

`y = uv` `" "` `rArr` `" "` `y' = u'v+uv'`

In this case `u = (x+7)^10`,

so `u'=10(x+7)^9[d/dx(x+7)]=10(x+7)^9[1] =10(x+7)^9`.

And `v = (x^2+2)^7`

So `v'=7(x^2+2)^6[d/dx(x^2+2)] = 7(x^2+2)^6*2x`

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

`y' = 10(x+7)^9(x^2+2)^7 + (x+7)^10*7(x^2+2)^6 * 2x`

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(x+7)^9(x^2+2)^7)_"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(x+7)^9(x^2+2)^7)_"First Term" + underbrace(14x (x+7)^10 (x^2+2)^6)_"Second Term"`

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

`y' = 2(x+7)^9(x^2+2)^6[5(x^2+2)+x(x+7)]`

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

`y' = 2(x+7)^9(x^2+2)^6[6x^2+17]`

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(6x^2+17)(x+7)^9(x^2+2)^6`