Question

Find the derivative of \(\mathrm{y}=\int_{1-3 \times}^{1} \frac{\mathrm{u}^{3}}{1+\mathrm{u}^{2}} \mathrm{du}\).

Short answer

Answer: The derivative of y with respect to x is \(y'(x) = \frac{3(1-3x)^3}{1+(1-3x)^2}\).

Step-by-step solution

Identify the limits of integration and the function in the given integral

The given integral is y = \(\int_{1-3x}^{1}\frac{u^{3}}{1+u^{2}} du\). The limits of integration are a = 1 - 3x and b = 1, and the function inside the integral is \(\frac{u^3}{1+u^2}\).

Apply the Leibniz Rule for differentiation of an integral

According to the Leibniz Rule, if y = \(\int_{a(x)}^{b(x)} f(u, x) du\), then the derivative y' is given by: \(y'(x) = f(b(x), x)b'(x) - f(a(x), x)a'(x)\) In this case, \(f(u, x) = \frac{u^{3}}{1+u^{2}}\), \(a(x) = 1-3x\), \(b(x) = 1\), \(a'(x) = -3\), and \(b'(x) = 0\) since b is a constant.

Substitute the values of the limits and their derivatives into the Leibniz Rule

We now substitute the given values into the Leibniz Rule: \(y'(x) = \frac{1^3}{1+1^2}(0) - \frac{(1-3x)^3}{1+(1-3x)^2}(-3)\) Since \(b'(x) = 0\), the first term in the equation becomes zero, and we are left with: \(y'(x) = 3\frac{(1-3x)^3}{1+(1-3x)^2}\)

Simplify the expression for the derivative

Finally, we simplify the expression to obtain the derivative of y with respect to x: \(y'(x) = \frac{3(1-3x)^3}{1+(1-3x)^2}\) This is the derivative of y with respect to x.