Differential calculus using product rule pls with detailed explanation (2x+3)^3(4x^2-1)^2
Did you not look at replies to the same question you posted yesterday?
http://www.jiskha.com/display.cgi?id=1300049644
You have stated a function but not a question. Do you want the derivative of that function? If so, say so.
Reiny answered your question yesterday
To find the derivative of the function (2x+3)^3(4x^2-1)^2 using the product rule, you need to apply the following steps:
Step 1: Identify the functions
In this case, we have two functions: f(x) = (2x+3)^3 and g(x) = (4x^2-1)^2.
Step 2: Determine the derivative of each function
To find the derivatives of f(x) and g(x), we'll need to use the chain rule as well as the power rule.
Derivative of f(x) = (2x+3)^3:
To find the derivative of this function, consider it as the composition of two functions: u(x) = 2x+3 and v(u) = u^3.
Derivative of u(x) = 2
Derivative of v(u) = 3u^2
Using the chain rule, we have:
[Derivative of f(x)] = [Derivative of v(u)] * [Derivative of u(x)]
= 3(2x+3)^2 * 2
Simplifying, we get:
[Derivative of f(x)] = 6(2x+3)^2
Derivative of g(x) = (4x^2-1)^2:
To find the derivative of this function, we use the power rule.
[Derivative of g(x)] = 2(4x^2-1) * (8x)
Simplifying further, we have:
[Derivative of g(x)] = 16x(4x^2-1)
Step 3: Apply the product rule
The product rule states that 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)
Applying the product rule to our functions f(x) = (2x+3)^3 and g(x) = (4x^2-1)^2, we get:
[(2x+3)^3 * (4x^2-1)^2]' = (6(2x+3)^2 * (4x^2-1)^2) + ((2x+3)^3 * 16x(4x^2-1))
Now, simplify the expression:
[(2x+3)^3 * (4x^2-1)^2]' = 6(2x+3)^2 * (4x^2-1)^2 + 16x(2x+3)^3 * (4x^2-1)
This is the derivative of the function (2x+3)^3 * (4x^2-1)^2 using the product rule.