Question

Differentiate $\frac{\sin x+e^x}{\cos x+e^x}$ with respect to $x$.

Answer 1

$\frac{d}{dx}\left(\frac{\sin x+\log x}{\cos x+e^x}\right)$
$=\frac{(\cos x+e^x)\frac{d}{dx}(\sin x+e^x)-(\sin x+e^x)\frac{d}{dx}(\cos x+e^x)}{(\cos x+e^x{)}^2}$
$=\frac{(\cos x+e^x)(\cos x+e^x)-(\sin x+e^x)(-\sin x+e^x)}{(\cos x+e^x{)}^2}$
$=\frac{{(\cos x+e^x)}^2-(-{\sin }^{2}x+e^{2x})}{(\cos x+e^x{)}^2}$
$=\frac{{\cos }^{2}x+2e^x\cos x+e^{2x}+{\sin }^{2}x-e^{2x}}{(\cos x+e^x{)}^2}$
$=\frac{2e^x\cos x+1}{(\cos x+e^x{)}^2}$