Differentiate the following with respect to $x$: $\cos (\log x + e^x)$,where $x > 0$.

Let $y = \cos (\log x + e^x)$.
By using the chain rule,we differentiate with respect to $x$:
$\frac{dy}{dx} = \frac{d}{dx} [\cos (\log x + e^x)]$
Using the derivative of $\cos(u)$ which is $-\sin(u) \cdot \frac{du}{dx}$,we get:
$\frac{dy}{dx} = -\sin (\log x + e^x) \cdot \frac{d}{dx} (\log x + e^x)$
Now,differentiate the terms inside the bracket:
$\frac{d}{dx} (\log x) = \frac{1}{x}$ and $\frac{d}{dx} (e^x) = e^x$.
Substituting these back into the expression:
$\frac{dy}{dx} = -\sin (\log x + e^x) \cdot (\frac{1}{x} + e^x)$
Therefore,the final derivative is:
$\frac{dy}{dx} = -(\frac{1}{x} + e^x) \sin (\log x + e^x)$ for $x > 0$.