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$\frac{d}{d x}\left(e^{x} \cos x\right)$

To find the derivative of the function $e^x \cos x$ with respect to $x$, we need to use the product rule. The product rule states that if you have a function $h(x) = f(x)g(x)$, then its derivative $h'(x)$ is given by $f'(x)g(x) + f(x)g'(x)$.

In this case, let $f(x) = e^x$ and $g(x) = \cos x$.

First, find the derivative of $f(x)$: $f'(x) = \frac{d}{dx}(e^x) = e^x$.

Next, find the derivative of $g(x)$: $g'(x) = \frac{d}{dx}(\cos x) = -\sin x$.

Now, apply the product rule: $\frac{d}{d x}\left(e^{x} \cos x\right) = f'(x)g(x) + f(x)g'(x)$ Substitute the functions and their derivatives into the formula: $= (e^x)(\cos x) + (e^x)(-\sin x)$ $= e^x \cos x - e^x \sin x$ We can factor out $e^x$ from both terms: $= e^x (\cos x - \sin x)$

The final answer is $\boxed{e^x (\cos x - \sin x)}$.