F(x)= ln((3x+6)^4 (x^3 -2)^5 )
F'(x)=
To find the derivative of F(x) = ln((3x+6)^4 (x^3 -2)^5), we can use the chain rule.
The chain rule states that if we have a function g(f(x)), where g(x) is another function and f(x) is the inner function, then the derivative of g(f(x)) is given by g'(f(x)) * f'(x).
In this case, g(x) = ln(x), and f(x) = (3x+6)^4 (x^3 -2)^5.
Now, let's find the derivative step by step:
1. Find the derivative of the function inside the natural logarithm ln((3x+6)^4 (x^3 -2)^5):
Using the product rule, the derivative of (3x+6)^4 (x^3 -2)^5 with respect to x is:
d/dx [(3x+6)^4 (x^3 -2)^5]
= [(3x+6)^4] * d/dx [(x^3 -2)^5] + [(x^3 -2)^5] * d/dx [(3x+6)^4]
2. Find the derivative of each term separately:
a. d/dx [(3x+6)^4]:
Using the chain rule, let u = 3x+6. Then, d/dx [(3x+6)^4] = 4u^3 * d/dx (u)
= 4(3x+6)^3 * d/dx (3x+6)
= 4(3x+6)^3 * 3
= 12(3x+6)^3
b. d/dx [(x^3 -2)^5]:
Using the chain rule, let v = x^3 - 2. Then, d/dx [(x^3 -2)^5] = 5v^4 * d/dx (v)
= 5(x^3 - 2)^4 * d/dx (x^3 - 2)
= 5(x^3 - 2)^4 * 3x^2
= 15x^2 (x^3 - 2)^4
3. Substitute the derivatives back into the original equation:
= [(3x+6)^4] * 15x^2 (x^3 - 2)^4 + [(x^3 -2)^5] * 12(3x+6)^3
So the derivative of F(x) = ln((3x+6)^4 (x^3 -2)^5) is:
F'(x) = [(3x+6)^4] * 15x^2 (x^3 - 2)^4 + [(x^3 -2)^5] * 12(3x+6)^3