Answer

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**Hint:**In this problem we need to calculate the derivative of the given function. We can observe that the given function is a fraction with numerator $\sin x$, denominator $1+\cos x$. In differentiation we have the formula $\dfrac{d}{dx}\left( \dfrac{u}{v} \right)=\dfrac{{{u}^{'}}v-u{{v}^{'}}}{{{v}^{2}}}$. So, we will compare the given function with $\dfrac{u}{v}$ and write the values of $u$, $v$. After knowing the values of $u$, $v$ we will differentiate the both the values to get the values of ${{u}^{'}}$, ${{v}^{'}}$. After knowing the value of ${{u}^{'}}$, ${{v}^{'}}$ we will use the division’s derivative formula and simplify the equation to get the required result.

**Complete step by step solution:**

Given that, $\dfrac{\sin x}{1+\cos x}$.

Comparing the above fraction with $\dfrac{u}{v}$, then we will get

$u=\sin x$, $v=1+\cos x$.

Considering the value $u=\sin x$.

Differentiating the above equation with respect to $x$, then we will get

$\Rightarrow \dfrac{du}{dx}=\dfrac{d}{dx}\left( \sin x \right)$

We have the differentiation formula $\dfrac{d}{dx}\left( \sin x \right)=\cos x$, then we will have

$\Rightarrow {{u}^{'}}=\cos x$

Considering the value $v=1+\cos x$.

Differentiating the above equation with respect to $x$, then we will get

$\Rightarrow \dfrac{dv}{dx}=\dfrac{d}{dx}\left( 1+\cos x \right)$

We know that the differentiation of constant is zero and $\dfrac{d}{dx}\left( \cos x \right)=-\sin x$, then we will have

$\Rightarrow {{v}^{'}}=-\sin x$

Now the derivative of the given fraction is given by

$\begin{align}

& \dfrac{d}{dx}\left( \dfrac{\sin x}{1+\cos x} \right)=\dfrac{{{u}^{'}}v-u{{v}^{'}}}{{{v}^{2}}} \\

& \Rightarrow \dfrac{d}{dx}\left( \dfrac{\sin x}{1+\cos x} \right)=\dfrac{\cos x\left( 1+\cos x \right)-\sin x\left( -\sin x \right)}{{{\left( 1+\cos x \right)}^{2}}} \\

\end{align}$

Simplifying the above equation, then we will have

$\Rightarrow \dfrac{d}{dx}\left( \dfrac{\sin x}{1+\cos x} \right)=\dfrac{\cos x+{{\cos }^{2}}x+{{\sin }^{2}}x}{{{\left( 1+\cos x \right)}^{2}}}$

We have the trigonometric identity ${{\sin }^{2}}x+{{\cos }^{2}}x=1$, then the above equation is modified as

$\begin{align}

& \Rightarrow \dfrac{d}{dx}\left( \dfrac{\sin x}{1+\cos x} \right)=\dfrac{1+\cos x}{{{\left( 1+\cos x \right)}^{2}}} \\

& \Rightarrow \dfrac{d}{dx}\left( \dfrac{\sin x}{1+\cos x} \right)=\dfrac{1}{1+\cos x} \\

\end{align}$

**Hence the derivative of the given equation $\dfrac{\sin x}{1+\cos x}$ is $\dfrac{1}{1+\cos x}$.**

**Note:**In this problem we can observe that ${{v}^{'}}=-\sin x=-u$. For this type of equation calculation of the integration is also simple. We have the integration formula $\int{\dfrac{{{f}^{'}}\left( x \right)}{f\left( x \right)}dx}=\log \left| f\left( x \right) \right|+C$. We will use this formula and simplify it to get the integration value.

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