Problem 15 - Calculus Early Transcendentals

Question

Find the first partial derivatives of the following functions. $$f(x, y)=e^{x^{2} y}$$

Step-by-Step Solution

Answer

Question: Determine the first partial derivatives of the function \(f(x, y) = e^{x^2 y}\) with respect to x and y. Answer: The first partial derivatives of the given function are: \(f_x(x, y) = e^{x^2 y} (2xy)\) \(f_y(x, y) = e^{x^2 y} (x^2)\)

Step 1: Write down the given function

The function we're working with is: $$f(x, y) = e^{x^2 y}$$

Step 2: Find the first partial derivative with respect to x

To find the first partial derivative with respect to x (denoted as \(f_x\) or \(\frac{\partial f}{\partial x}\)), we'll differentiate \(f(x, y)\) with respect to x, treating y as a constant: $$f_x(x, y) = \frac{\partial}{\partial x}(e^{x^2 y})$$ Now, we'll use the chain rule, which states that the derivative of a composite function is the derivative of the outer function times the derivative of the inner function. Using the chain rule, we get: $$f_x(x, y) = e^{x^2 y} \frac{\partial}{\partial x}(x^2 y)$$ The derivative of \(x^2 y\) with respect to x is: $$\frac{\partial}{\partial x}(x^2 y) = 2xy$$ So, the first partial derivative with respect to x is: $$f_x(x, y) = e^{x^2 y} (2xy)$$

Step 3: Find the first partial derivative with respect to y

To find the first partial derivative with respect to y (denoted as \(f_y\) or \(\frac{\partial f}{\partial y}\)), we'll differentiate \(f(x, y)\) with respect to y, treating x as a constant: $$f_y(x, y) = \frac{\partial}{\partial y}(e^{x^2 y})$$ We'll use the chain rule again: $$f_y(x, y) = e^{x^2 y} \frac{\partial}{\partial y}(x^2 y)$$ The derivative of \(x^2 y\) with respect to y is: $$\frac{\partial}{\partial y}(x^2 y) = x^2$$ So, the first partial derivative with respect to y is: $$f_y(x, y) = e^{x^2 y} (x^2)$$

Step 4: Write down the first partial derivatives

The first partial derivatives of the function \(f(x, y) = e^{x^2 y}\) are: $$f_x(x, y) = e^{x^2 y} (2xy)$$ $$f_y(x, y) = e^{x^2 y} (x^2)$$