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Is the partial derivative $f_x$ of $\cos(y)e^x = \cos(y)e^x$? where $\cos(y)$ is a constant $k$?

What about $f_y$?

Is it $-\sin(y)e^x$ where $e^x$ is a constant just like $e^2$ is?

2 Answers 2

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Yes. Essentially, you have something like $f(x,y)=g(x)h(y)$, so whenever you take a derivative with respect to $x$, $h(y)$ is "like a constant", and when you take a derivative with respect to $y$, $g(x)$ is "like a constant".

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$\begin{align} \frac{\partial(\cos y\cdot e^x)}{\partial x} &= \cos y\cdot e^x \\ \frac{\partial(\cos y\cdot e^x)}{\partial y} &= -\sin y\cdot e^x. \end{align}$ When differentiating with respect to $x$, then $y$ is regarded as constant, hence also $\cos y$.