Question

Find the derivative of the function \(\mathrm{y}=\int_{0}^{\mathrm{x}} \frac{1-\mathrm{t}+\mathrm{t}^{2}}{1+\mathrm{t}+\mathrm{t}^{2}} \mathrm{dt}\) at \(\mathrm{x}=1\).

Short answer

Answer: The derivative of the function y at x=1 is \(\frac{1}{3}\).

Step-by-step solution

Apply the Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus states that if a function F(x) is defined as the integral of a function f(t) from a constant a to x, then the derivative of F(x) with respect to x is f(x). In our problem, we have: \(y = \int_{0}^{x} \frac{1-t+t^2}{1+t+t^2} dt\) According to the Fundamental Theorem of Calculus, the derivative of y with respect to x is: \(y' = \frac{1-x+x^2}{1+x+x^2}\)

Evaluate y'(1)

Now, we will evaluate the derivative of y at x=1: \(y'(1) = \frac{1-1+1^2}{1+1+1^2} = \frac{1}{3}\) So, the derivative of the function y at x=1 is \(\frac{1}{3}\).