Find the first partial derivatives of the function: $f(x, y) = \int_{y}^{x} \cos(e^t) \, dt$

Neetesh Kumar | December 3, 2024

Calculus Homework Help

This is the solution to Math 1D
Assignment: 14.3 Question Number 16
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Step-by-step solution:

We are given the function: $f(x, y) = \int_{y}^{x} \cos(e^t) \, dt$

This is a definite integral with variable limits. To find the partial derivatives with respect to $x$ and $y$, we apply the Leibniz rule for differentiation of integrals with variable limits.

Step 1: Find $f_x(x, y)$

To find $f_x(x, y)$, we differentiate $f(x, y)$ with respect to the upper limit $x$. By the Leibniz rule:

$\frac{\partial}{\partial x} \int_{y}^{x} \cos(e^t) \, dt = \cos(e^x).$

Thus:

$f_x(x, y) = \cos(e^x).$

Step 2: Find $f_y(x, y)$

To find $f_y(x, y)$, we differentiate $f(x, y)$ with respect to the lower limit $y$. By the Leibniz rule, when differentiating with respect to the lower limit, we introduce a negative sign:

$\frac{\partial}{\partial y} \int_{y}^{x} \cos(e^t) \, dt = -\cos(e^y).$

Thus:

$f_y(x, y) = -\cos(e^y).$

Final Answer:

$f_x(x, y) = \boxed{\cos(e^x)}$

$f_y(x, y) = \boxed{-\cos(e^y)}$

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