find y' when y=(x^2+2x)(x^4+5x^2+3)^5

[havenoclu]

I'm running into trouble with this one. I'm not sure if my answer is correct or if I'm going about it the wrong way. The whole (...)^5 is messing me up. I don't know if I should take the derivative of everything inside the paren's..or just of the (...)^5...if that makes any sense..anyways, here's what I came up with (i also tried to use a difference quotient to solve...but it turned out long..and ugly):

y'=(2x-2)5(x^4+5x^2+3)^4

=(10x-10)(x^4+5x^2+3)^4 <--is that correct?

 

[stapel]

Has your class not covered the Product Rule yet?

Thank you.

Eliz.

 

[havenoclu]

I believe it was gone over briefly...Is that the

(f+g)' or (f-g)'

rule? My professor called it a theorem...i'm not sure if he referred to it as the "product rule"...so are you saying my answer is wrong, and i should use the product rule or something?

 

[skeeter]

product rule ...

\(\displaystyle \L \frac{d}{dx}[fg] = fg' + gf'\)

for your "product" (which will also involve the chain rule, btw) ...

\(\displaystyle \L y=(x^2+2x)(x^4+5x^2+3)^5\)

\(\displaystyle \L \frac{dy}{dx} = (x^2+2x)*5(x^4+5x^2+3)^4*(4x^3+10x) + (x^4+5x^2+3)^5*(2x+2)\)

I leave it to you to perform the standard algebraic "clean-up".