Differentiate.
g(x)=(x^3+1)(3x^2-1)
I know that the answer is
15x^4-3x^2+6x but I donot know how to get to this. I was using
g'(x) X h(x) + g(x) X h'(x) and I was using the derivatives 3x^2 and 6x but I guess this is wrong. Please help!! Thank you.
g '(x) = 6x(x^3+1) + 3x^2(3x^2-1)
= 6x^4 + 6x + 9x^4 - 3x^2
= 15x^4 - 3x^2 + 6x
Thank You
To differentiate the function g(x) = (x^3 + 1)(3x^2 - 1), you need to apply the product rule of differentiation. The product rule states that the derivative of the product of two functions is equal to the derivative of the first function times the second function, plus the first function times the derivative of the second function.
Let's break it down step by step:
Step 1: Identify the two functions for the product rule.
In this case, the two functions are f(x) = x^3 + 1 and g(x) = 3x^2 - 1.
Step 2: Find the derivatives of the two functions.
The derivative of f(x) = x^3 + 1 is f'(x) = 3x^2.
The derivative of g(x) = 3x^2 - 1 is g'(x) = 6x.
Step 3: Apply the product rule.
According to the product rule, the derivative of the product of two functions, f(x) and g(x), is given by:
(f(x)g(x))' = f'(x)g(x) + f(x)g'(x).
Plugging in the values, we get:
(g(x))' = (x^3 + 1)g'(x) + (3x^2 - 1)f'(x)
Step 4: Simplify.
(g(x))' = (x^3 + 1)(6x) + (3x^2 - 1)(3x^2)
= 6x^4 + 6x + 9x^4 - 3x^2
= 15x^4 - 3x^2 + 6x
Therefore, the derivative of g(x) = (x^3 + 1)(3x^2 - 1) is 15x^4 - 3x^2 + 6x.
So, your answer is correct. You just needed to include the term (3x^2 - 1)(3x^2) in your calculation.