Problem 60
Question
Compute the first partial derivatives of the following functions. $$f(x, y)=\ln \left(1+e^{-x y}\right)$$
Step-by-Step Solution
Verified Answer
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: \(\frac{\partial f}{\partial y} = \frac{-xe^{-xy}}{1 + e^{-xy}}\)
1Step 1: 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:
$$\frac{\partial f}{\partial x} = \frac{1}{1 + e^{-xy}}(-ye^{-xy})$$
2Step 2: 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:
$$\frac{\partial f}{\partial y} = \frac{1}{1 + e^{-xy}}(-xe^{-xy})$$
3Step 3: 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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