Derivatives again: y = (x^2 - 6)(x^3 + 2)
- Thread starter scrum
- Start date
Re: Derivatives again.
\(\displaystyle \L y = (x^2 - 6)(x^3 + 2)\)
Use the product rule
\(\displaystyle \L y' = \frac{d}{dx}(x^2 - 6)(x^3 + 2) \,+\, (x^2 - 6)\frac{d}{dx}(x^3 + 2)\)
\(\displaystyle \L y' = (2x)(x^3 + 2) \,+\, (x^2 - 6)(3x^2)\)
Simplify if you want to.
John
scrum said:Okay I have to get the derivative of this.
I multiply it out and get
x^6 -6x^3 +2x^2 -12
and differentiate to
6x^5 -18^2 +4x
and it says i'm wrong, but i don't see where i'm doing anything wrong. i've done it over 3 times and i'm still wrong.
\(\displaystyle \L y = (x^2 - 6)(x^3 + 2)\)
Use the product rule
\(\displaystyle \L y' = \frac{d}{dx}(x^2 - 6)(x^3 + 2) \,+\, (x^2 - 6)\frac{d}{dx}(x^3 + 2)\)
\(\displaystyle \L y' = (2x)(x^3 + 2) \,+\, (x^2 - 6)(3x^2)\)
Simplify if you want to.
John
