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Chapter 13: Problem 60

Compute the first partial derivatives of the following functions. $$f(x, y)=\ln \left(1+e^{-x y}\right)$$

Short Answer

Expert verified
Question: Find the first partial derivatives of the function \(f(x, y) = \ln(1 + e^{-xy})\). Answer: The first partial derivatives are: 1. With respect to x: \(\frac{\partial f}{\partial x} = \frac{-ye^{-xy}}{1 + e^{-xy}}\) 2. With respect to y: \(\frac{\partial f}{\partial y} = \frac{-xe^{-xy}}{1 + e^{-xy}}\)

Step by step solution

01

Find the partial derivative with respect to x

First, we'll compute the partial derivative with respect to x. Remember that while taking the partial derivative, we'll treat y as a constant. Using the chain rule, we have: $$\frac{\partial f}{\partial x} = \frac{\partial}{\partial x}\left(\ln(1 + e^{-xy})\right) = \frac{1}{1 + e^{-xy}}\frac{\partial}{\partial x}\left(1 + e^{-xy}\right)$$ Now, differentiate the term inside the parenthesis with respect to x, using the chain rule again: $$\frac{\partial}{\partial x}\left(1 + e^{-xy}\right)= -ye^{-xy}$$ Now, substituting this back into the original expression: $$\frac{\partial f}{\partial x} = \frac{1}{1 + e^{-xy}}(-ye^{-xy})$$
02

Find the partial derivative with respect to y

Next, we'll compute the partial derivative with respect to y. Remember that while taking the partial derivative, we'll treat x as a constant. Using the chain rule, we have: $$\frac{\partial f}{\partial y} = \frac{\partial}{\partial y}\left(\ln(1 + e^{-xy})\right) = \frac{1}{1 + e^{-xy}}\frac{\partial}{\partial y}\left(1 + e^{-xy}\right)$$ Now, differentiate the term inside the parenthesis with respect to y, using the chain rule again: $$\frac{\partial}{\partial y}\left(1+e^{-xy}\right)= -xe^{-xy}$$ Now, substituting this back into the original expression: $$\frac{\partial f}{\partial y} = \frac{1}{1 + e^{-xy}}(-xe^{-xy})$$
03

Write down the final solution

After finding both partial derivatives, we have: 1. Partial derivative with respect to x: $$\frac{\partial f}{\partial x} = \frac{-ye^{-xy}}{1 + e^{-xy}}$$ 2. Partial derivative with respect to y: $$\frac{\partial f}{\partial y} = \frac{-xe^{-xy}}{1 + e^{-xy}}$$

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