To find the partial derivative of the function $u = e^x (x \cos y - y \sin y)$ with respect to $x$, we apply the product rule of differentiation.

Derivative Calculation

Given: $u = e^x (x \cos y - y \sin y),$

we differentiate with respect to $x$:

$\frac{\partial u}{\partial x} = \frac{\partial}{\partial x} \left( e^x (x \cos y - y \sin y) \right).$

Applying the product rule, we get:

$\frac{\partial u}{\partial x} = \frac{\partial}{\partial x}(e^x) (x \cos y - y \sin y) + e^x \frac{\partial}{\partial x} (x \cos y - y \sin y).$

Calculating the derivatives:

1. $\frac{\partial}{\partial x}(e^x) = e^x,$
2. $\frac{\partial}{\partial x}(x \cos y - y \sin y) = \cos y.$

Therefore:

$\frac{\partial u}{\partial x} = e^x (x \cos y - y \sin y) + e^x \cos y.$

Combining terms:

$\frac{\partial u}{\partial x} = e^x (x \cos y - y \sin y + \cos y).$

Thus, the partial derivative of $u$ with respect to $x$ is:

$\frac{\partial u}{\partial x} = e^x (x \cos y - y \sin y + \cos y).$

Would you like further explanation on any step or more details?

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Relative Questions:

1. What is the product rule in differentiation?
2. How does the chain rule apply in partial differentiation?
3. What is the importance of partial derivatives in multivariable calculus?
4. How can you interpret the partial derivative of a function in a physical context?
5. What are the differences between partial derivatives and total derivatives?
6. How can partial derivatives be used to find the slope of a surface at a point?
7. Can partial derivatives help in finding critical points of multivariable functions?
8. What role do partial derivatives play in the Taylor series expansion of functions?

Tip:

When calculating partial derivatives, remember to treat all other variables as constants except the one you are differentiating with respect to.