Question

Find both first partial derivatives. \(h(x, y)=e^{-\left(x^{2}+y^{2}\right)}\)

Short answer

\(\frac{\partial h}{\partial x} = -2x \cdot e^{-\left(x^{2}+y^{2}\right)}\), \(\frac{\partial h}{\partial y} = -2y \cdot e^{-\left(x^{2}+y^{2}\right)}\)

Step-by-step solution

Calculating The Partial Derivative With Respect To \(x\)

Let's start by calculating the first partial derivative with respect to \(x\). The function is \(h(x, y) = e^{-\left(x^{2}+y^{2}\right)}\). The chain rule states that the derivative of a composite function is the derivative of the outside function times the derivative of the inside function. In this case, the 'outside' function is the exponential function, and the 'inside' function is the negation of the sum of squares. The derivative of the outside function is simply itself since the derivative of \(e^x\) is \(e^x\), and the derivative of the inside function with respect to \(x\) is \(-2x\). Thus, the first partial derivative with respect to \(x\) is: \(\frac{\partial h}{\partial x}= e^{-\left(x^{2}+y^{2}\right)} \cdot -2x = -2x \cdot e^{-\left(x^{2}+y^{2}\right)}\).

Calculating The Partial Derivative With Respect To \(y\)

Now, let's find the first partial derivative with respect to \(y\). By the same logic as in the previous step, the outside function is the exponential function and the inside function is the negation of the sum of squares. The derivative of the outside function is \(e^{-\left(x^{2}+y^{2}\right)}\), and the derivative of the inside function with respect to \(y\) is \(-2y\). Thus, the first partial derivative with respect to \(y\) is: \(\frac{\partial h}{\partial y} = e^{-\left(x^{2}+y^{2}\right)} \cdot -2y = -2y \cdot e^{-\left(x^{2}+y^{2}\right)}\).