# Differentiating this Expression

• Jul 24th 2011, 12:32 PM
YoungMarbleGiant
Differentiating this Expression
Can anyone help me with this? The question asks to differentiate the following...

(x-5)^8 * (x+3)^6

The way I attempted it was to try using the product rule, but I don't get the same answer as stated in the book.

I used the chain rule to differentiate each of the two arguments, getting 8(x-5)^7 and 6(x+3)^5 respectively. At that point I attempted to use the product rule, getting:

8(x-5)^7*(x+3)^6 + (x-5)^8*6(x+3)^5

but this is not the correct answer, unless there's some algebra I'm not seeing with the answer above - I suspect I'm doing something amiss. Any help is appreciated!
• Jul 24th 2011, 12:40 PM
Plato
Re: Differentiating this Expression
Quote:

Originally Posted by YoungMarbleGiant
differentiate the following...
\$\displaystyle (x-5)^8 * (x+3)^6\$
I used the chain rule to differentiate each of the two arguments, getting 8(x-5)^7 and 6(x+3)^5 respectively. At that point I attempted to use the product rule, getting:
\$\displaystyle 8(x-5)^7*(x+3)^6 + (x-5)^8*6(x+3)^5\$
but this is not the correct answer

That answer is certainly correct. Why do you doubt it?
It can be factored: \$\displaystyle (x-5)^7(x+3)^5[8(x+3)+6(x-5)]\$
• Jul 24th 2011, 12:51 PM
YoungMarbleGiant
Re: Differentiating this Expression
Quote:

Originally Posted by Plato
That answer is certainly correct. Why do you doubt it?
It can be factored: \$\displaystyle (x-5)^7(x+3)^5[8(x+3)+6(x-5)]\$

I was sure the method was correct, but the answer in the exercise book is quite different. It gives the answer as \$\displaystyle 2(7x-3)(x-5)^7(x+3)^5\$.

I looked at ways of simplifying/factoring, but the \$\displaystyle (7x-3)\$ was throwing me off. Do you have any idea how to get to this answer from the point where I reached? Thank you.
• Jul 24th 2011, 12:55 PM
Siron
Re: Differentiating this Expression
That's exactly what Plato said, because \$\displaystyle [8(x+3)+6(x-5)]=(8x+24+6x-30)=14x-6=2(7x-3)\$
• Jul 24th 2011, 01:06 PM
YoungMarbleGiant
Re: Differentiating this Expression
Alright, thanks to you both.