f'(x) = (x^2-1)/x
sorry, i meant what is the antiderivative of
f'(x) = (x^2-1)/x
Write f(x) as
f(x) = x - 1/x, and integrate it one term at a time.
The integral (or "antiderivative", if you insist on using that term) of x is (x^2)/2. The integral of -1/x is -ln x. An arbitrary constant term can be added to the integral.
To find the derivative of f(x) = (x^2 - 1) / x, you can use the quotient rule. The quotient rule states that if you have a function of the form f(x) = g(x) / h(x), then its derivative is given by:
f'(x) = (h(x) * g'(x) - g(x) * h'(x)) / (h(x))^2
In this case, g(x) = x^2 - 1 and h(x) = x. Now, let's calculate the derivatives of g(x) and h(x).
1. Derivative of g(x) = x^2 - 1:
To find the derivative of x^2 - 1, you can simply apply the power rule. The power rule states that if you have a function of the form f(x) = x^n, then its derivative is given by:
f'(x) = n * x^(n-1)
Applying the power rule to g(x) = x^2 - 1, we get:
g'(x) = 2 * x^(2-1) = 2x
2. Derivative of h(x) = x:
To find the derivative of x, you can simply apply the power rule with n = 1:
h'(x) = 1 * x^(1-1) = 1
Now, we have g'(x) = 2x and h'(x) = 1. Let's substitute these values into the quotient rule formula:
f'(x) = (h(x) * g'(x) - g(x) * h'(x)) / (h(x))^2
= (x * 2x - (x^2 - 1) * 1) / (x^2)^2
= (2x^2 - x^2 + 1) / x^4
= (x^2 + 1) / x^4
Therefore, the derivative of f(x) = (x^2 - 1) / x is f'(x) = (x^2 + 1) / x^4.