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

1 Answer
Apr 17, 2016

Answer:

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#