Question

Find both first partial derivatives. \(f(x, y)=\int_{x}^{y}(2 t+1) d t+\int_{y}^{x}(2 t-1) d t\)

Short answer

The first partial derivatives of the given function are \( f_{x} = -2 \) and \( f_{y} = 2 \).

Step-by-step solution

Calculate \(f_{x}\)

First, compute the partial derivative of \(f\) with respect to \(x\). The derivative of an integral from a constant to a variable (in this case \(x\)) of a function is that function evaluated at the variable, and the derivative of an integral from one independent variable to another independent variable is the negative of the derivative of the first. The derivative of the integral from \(y\) to \(x\) of \(2t - 1\) is just \(-1 \times (2x - 1)\). Therefore, \[f_{x} = (2x - 1) - (2x + 1) = -2 .\]

Calculate \(f_{y}\)

Next, compute the partial derivative of \(f\) with respect to \(y\). The derivative of the integral from \(x\) to \(y\) of \(2t + 1\) is just \(2y + 1\). The derivative of the integral from \(y\) to \(x\) of \(2t - 1\) is the negative of \((2y - 1)\) as we have to reverse the sign because of the integration limits. Therefore, \[f_{y} = (2y + 1) - (2y - 1) = 2 .\]