if f(x)=ln(x^3 +1) find f '(3), correct to 2 decimal places
To find the derivative of the function f(x) = ln(x^3 + 1), you can use the chain rule.
The chain rule states that if you have a composite function, f(g(x)), then the derivative of that composite function is given by f'(g(x)) * g'(x).
In this case, the outer function f(x) is the natural logarithm, and the inner function g(x) is x^3 + 1.
First, find the derivative of the outer function, f'(x), which is simply 1/x.
Next, find the derivative of the inner function, g'(x), which is 3x^2.
Now, apply the chain rule:
f '(x) = f'(g(x)) * g'(x)
= (1 / g(x)) * g'(x)
= (1 / (x^3 + 1)) * (3x^2)
To find f'(3), substitute x = 3 into the derivative expression:
f'(3) = (1 / (3^3 + 1)) * (3 * 3^2)
= (1 / 28) * 27
= 27 / 28
≈ 0.96429 (rounded to 5 decimal places)
Therefore, f'(3) ≈ 0.96 (correct to 2 decimal places).