<s>[INST] The derivative of $f(x)=\frac{3 \sin x}{2+\cos x}$ My solution:
The background
$$\begin{align} &3\frac{d}{dx}\left(\frac{\sin(x)}{2+\cos(x)}\right)\\
&=3\frac{\frac{d}{dx}(\sin(x))(2+\cos(x))-\frac{d}{dx}(2+\cos(x))\sin(x)}{(2+\cos(x))^2}\\
&=3\frac{\cos(x)(2+\cos(x))-(-\sin(x))\sin(x)}{(2+\cos(x))^2}\\
&=\frac{3+6\cos(x)}{(2+\cos(x))^2}\end{align}$$
However, I plugged $f(x)=\dfrac{3 \sin x}{2+\cos x}$ into the derivative calculator in wolfram alpha and received the following calculation:
$$\frac{3\left(\sin^2(x)+\cos^2(x)+2\cos(x)\right)}{(2+\cos(x))^2}$$
Is my solution incorrect and the one from wolfram alpha correct? If so, where did I go wrong?
 [/INST] Your answer is correct. Use trig. identity $\sin^2(x)+\cos^2(x)=1$
$$\therefore \frac{3(\sin^2(x)+\cos^2(x)+2\cos(x))}{(2+\cos(x))^2}=\frac{3(1+2\cos(x))}{(2+\cos(x))^2}=\frac{3+6\cos(x)}{(2+\cos(x))^2}$$
Above matches your solution
</s>