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

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Question :

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

Find the first partial derivatives of the function:
$f(x, y) = \int_{y}^{x} \co | Doubtlet.com

Solution:

Neetesh Kumar

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)=yxcos(et)dtf(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 xx and yy, we apply the Leibniz rule for differentiation of integrals with variable limits.

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

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

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

Thus:

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

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

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

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

Thus:

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

Final Answer:

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

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


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