Differentiate
g(x)=(x^3+1)(3x^2-1)
so I use g'(x) X h(x) + g(x) X h'(x)
so the derivative of (x^3+1)=3x^2
and (3x^3-1)=6x
So (3x^2)(x^3+1) + (x^3+1)(6x) Is this correct so far. If so this is where I am stuck. I am having trouble simplifing.
You made a mistake applying the "derivative of a product" formula.
g'(x) = (3x^2)(3x^2-1) + (x^3+1)*6x
= 12x^4 -3x^2 +6x
= 3x(4x^3 -x^2 +2)
Actually I figure it out.
In google type: calc101
When you see list of resultc click on:
Calc101com Automatic Calculus and Algebra Help
When page be open clik option: derivatives
When this page be open in rectacangle type:
(x^3+1)(3x^2-1)
and click options DO IT
You will see solution step-by-step
By the way on this site you can practice any kind of derivation.
Thank You!!
Yes, you are on the right track with using the product rule for differentiation. Let's simplify the expression further.
You correctly found the derivative of g(x) with respect to x as:
g'(x) = (3x^2)(3x^2-1) + (x^3+1)(6x)
To simplify this expression, we can distribute and combine like terms:
g'(x) = 3x^4 - 3x^2 + 6x^4 + 6x
Now, let's combine the terms with the same power of x:
g'(x) = (3x^4 + 6x^4) + (-3x^2 + 6x)
Combining the monomials with the same power of x, we get:
g'(x) = 9x^4 + 3x
Therefore, the derivative of g(x) = (x^3+1)(3x^2-1) is g'(x) = 9x^4 + 3x.