find f"(x) if f(x)= cube root(x^2-2x+1)
Note that x^2 - 2x + 1 = (x-1)^2
f(x) = (x-1)^2/3
f' = 2/3 (x-1)^-1/3
f'' = -2/9 (x-1)^-4/3
To find the second derivative of the function f(x) = ∛(x^2 - 2x + 1), we need to follow these steps:
Step 1: Find the first derivative of f(x). Let's call it f'(x).
Step 2: Find the second derivative of f(x) by taking the derivative of f'(x).
Step 1: Find f'(x)
To find the first derivative of f(x), we can use the power rule combined with the chain rule.
Let's start by rewriting f(x) as (x^2 - 2x + 1)^(1/3).
Using the chain rule, the first derivative f'(x) can be found as follows:
f'(x) = (1/3)(x^2 - 2x + 1)^(-2/3) * (2x - 2)
Step 2: Find f"(x)
Now that we have f'(x), we can find the second derivative, f"(x).
To do this, we need to take the derivative of f'(x) with respect to x.
Differentiating f'(x), we get:
f"(x) = d/dx [f'(x)]
= d/dx [(1/3)(x^2 - 2x + 1)^(-2/3) * (2x - 2)]
To simplify this expression, let's focus on differentiating each term separately.
For the first term (1/3)(x^2 - 2x + 1)^(-2/3):
Using the chain rule, we differentiate the outer function (-2/3) times the derivative of the inner function (x^2 - 2x + 1).
d/dx [(x^2 - 2x + 1)^(-2/3)]
= (-2/3)(x^2 - 2x + 1)^(-2/3 - 1) * (2x - 2)
For the second term, we differentiate (2x - 2):
d/dx (2x - 2) = 2
So, the second derivative f"(x) is given by:
f"(x) = (-2/3)(x^2 - 2x + 1)^(-2/3 - 1) * (2x - 2) + 2
Simplifying further, we get the final answer:
f"(x) = (-2/3)(2x - 2)/(x^2 - 2x + 1)^(5/3) + 2
Therefore, the second derivative of f(x) is (-2/3)(2x - 2)/(x^2 - 2x + 1)^(5/3) + 2.