How do you differentiate $f\left( x \right)=\dfrac{\cot x}{1+\cot x}$?

Hint: In this problem we need to calculate the derivative of the given function. We can observe that the given function is in the form of fraction. So, we will consider the numerator and denominator separately. Now we will calculate the differentiation of both numerator and denominator individually. But in the problem, they have asked to calculate the derivative of the fraction. So, we will use the differentiation formula $\dfrac{d}{dx}\left( \dfrac{u}{v} \right)=\dfrac{v\dfrac{du}{dx}-u\dfrac{dv}{dx}}{{{v}^{2}}}$. We will use the calculated values in the above equation and simplify the obtained equation, then we will get the required result.

Complete step by step answer:
Given that, $f\left( x \right)=\dfrac{\cot x}{1+\cot x}$.
The above function is in the form of fraction and the numerator in the above equation is $\cot x$, denominator in the above equation is $1+\cot x$.
Let us assume that $u=\cot x$, $v=1+\cot x$.
Considering the equation $u=\cot x$.
Differentiating the above equation with respect to $x$, then we will have
$\dfrac{du}{dx}=\dfrac{d}{dx}\left( \cot x \right)$
We have the differentiation formula $\dfrac{d}{dx}\left( \cot x \right)=-{{\csc }^{2}}x$, then we will have
$\Rightarrow \dfrac{du}{dx}=-{{\csc }^{2}}x$
Considering the equation $v=1+\cot x$.
Differentiating the above equation with respect to $x$, then we will get
$\dfrac{dv}{dx}=\dfrac{d}{dx}\left( 1+\cot x \right)$
Applying differentiation for each term individually, then we will get
$\dfrac{dv}{dx}=\dfrac{d}{dx}\left( 1 \right)+\dfrac{d}{dx}\left( \cot x \right)$
We have the differentiation value of constant is always zero and $\dfrac{d}{dx}\left( \cot x \right)=-{{\csc }^{2}}x$. Applying both the formulas in the above equation, then we will get
$\Rightarrow \dfrac{dv}{dx}=-{{\csc }^{2}}x$
Now differentiating the given equation with respect to $x$, then we will have
${{f}^{'}}\left( x \right)=\dfrac{d}{dx}\left( \dfrac{\cot x}{1+\cot x} \right)$
The above equation is in the form of $\dfrac{d}{dx}\left( \dfrac{u}{v} \right)$ which is given by $\dfrac{d}{dx}\left( \dfrac{u}{v} \right)=\dfrac{v\dfrac{du}{dx}-u\dfrac{dv}{dx}}{{{v}^{2}}}$. Now the derivative of the given function will be
${{f}^{'}}\left( x \right)=\dfrac{\left( 1+\cot x \right)\dfrac{d}{dx}\left( \cot x \right)-\cot x\dfrac{d}{dx}\left( 1+\cot x \right)}{{{\left( 1+\cot x \right)}^{2}}}$
Substituting the value, we have in the above equation, then we will get
$\begin{align}
  & {{f}^{'}}\left( x \right)=\dfrac{\left( 1+\cot x \right)\left( -{{\csc }^{2}}x \right)-\cot x\left( -{{\csc }^{2}}x \right)}{{{\left( 1+\cot x \right)}^{2}}} \\
 & \Rightarrow {{f}^{'}}\left( x \right)=\dfrac{-{{\csc }^{2}}x\left( 1+\cot x \right)+{{\csc }^{2}}x\cot x}{{{\left( 1+\cot x \right)}^{2}}} \\
\end{align}$
Taking $-{{\csc }^{2}}x$ as common in the numerator, then we will have
$\begin{align}
  & \Rightarrow {{f}^{'}}\left( x \right)=\dfrac{-{{\csc }^{2}}x\left( 1+\cot x-\cot x \right)}{{{\left( 1+\cot x \right)}^{2}}} \\
 & \Rightarrow {{f}^{'}}\left( x \right)=-\dfrac{{{\csc }^{2}}x}{{{\left( 1+\cot x \right)}^{2}}} \\
\end{align}$

Hence the derivative of the given equation $f\left( x \right)=\dfrac{\cot x}{1+\cot x}$ is $-\dfrac{{{\csc }^{2}}x}{{{\left( 1+\cot x \right)}^{2}}}$.

Note: We can also simplify the obtained equation by converting the whole equation in terms of $\sin x$, $\cos x$ by substituting the known as $\csc x=\dfrac{1}{\sin x}$, $\cot x=\dfrac{\cos x}{\sin x}$ and we will simplify the obtained equation and then we will apply differentiation to get the required result.