Question Statement

Find the derivative of the function $y=\frac{\cot x}{x+1}$ with respect to $x$.

---

Background and Explanation

This problem involves differentiating a quotient of two functions, which requires the quotient rule from differential calculus. You should be familiar with the derivatives of basic trigonometric functions, specifically that the derivative of $\cot x$ is $-{\mathrm{cosec}}^{2}x$.

---

Solution

We are given the function: $y=\frac{\cot x}{x+1}$

This is a quotient where the numerator is $u=\cot x$ and the denominator is $v=x+1$. To find $\frac{dy}{dx}$, we apply the quotient rule: $\frac{d}{dx}\left(\frac{u}{v}\right)=\frac{v\cdot \frac{du}{dx}-u\cdot \frac{dv}{dx}}{v^2}$

First, we determine the derivatives of the individual components:

• The derivative of the numerator: $\frac{d}{dx}(\cot x)=-{\mathrm{cosec}}^{2}x$
• The derivative of the denominator: $\frac{d}{dx}(x+1)=1$

Substituting these into the quotient rule formula:

$\begin{array}{rl}\frac{dy}{dx}& =\frac{(x+1)\frac{d}{dx}(\cot x)-\cot x\frac{d}{dx}(x+1)}{(x+1{)}^2}\\ & =\frac{(x+1)\left(-{\mathrm{cosec}}^{2}x\right)-\cot x(1+0)}{(x+1{)}^2}\\ & =\frac{-(x+1){\mathrm{cosec}}^{2}x-\cot x}{(x+1{)}^2}\end{array}$

Thus, the final derivative is: $\frac{dy}{dx}=\frac{-(x+1){\mathrm{cosec}}^{2}x-\cot x}{(x+1{)}^2}$

This can also be written as: $\frac{dy}{dx}=-\frac{(x+1){\mathrm{cosec}}^{2}x+\cot x}{(x+1{)}^2}$

---

Key Formulas or Methods Used

• Quotient Rule: $\frac{d}{dx}\left(\frac{u}{v}\right)=\frac{v\frac{du}{dx}-u\frac{dv}{dx}}{v^2}$
• Derivative of Cotangent: $\frac{d}{dx}(\cot x)=-{\mathrm{cosec}}^{2}x$ (also written as $-{\csc }^{2}x$)
• Power Rule for Linear Terms: $\frac{d}{dx}(x+1)=1$

---

Summary of Steps

1. Identify the function as a quotient with $u=\cot x$ and $v=x+1$
2. Compute derivatives of numerator and denominator separately: $u^{\prime }=-{\mathrm{cosec}}^{2}x$ and $v^{\prime }=1$
3. Apply the quotient rule formula $\frac{v{u}^{\prime }-u{v}^{\prime }}{v^2}$
4. Substitute the calculated derivatives into the formula
5. Simplify the expression by distributing the negative sign through the first term in the numerator