## Task Condition

Find the derivative.

$$
y=\frac{\operatorname{ctg} x+x}{1-x \cdot \operatorname{ctg} x}
$$
## Solution

$y^{\prime}=\left(\frac{\operatorname{ctg} x+x}{1-x \cdot \operatorname{ctg} x}\right)^{\prime}=\frac{(\operatorname{ctg} x+x)^{\prime} \cdot(1-x \cdot \operatorname{ctg} x)-(\operatorname{ctg} x+x) \cdot(1-x \cdot \operatorname{ctg} x)^{\prime}}{(1-x \cdot \operatorname{ctg} x)^{2}}=$

$$
\begin{aligned}
& =\frac{\left(-\frac{1}{\sin ^{2} x}+1\right) \cdot(1-x \cdot \operatorname{ctg} x)-(\operatorname{ctg} x+x) \cdot\left(-\operatorname{ctg} x+\frac{x}{\sin ^{2} x}\right)}{(1-x \cdot \operatorname{ctg} x)^{2}}= \\
& =\frac{-\frac{1}{\sin ^{2} x}+\frac{x \cdot \operatorname{ctg} x}{\sin ^{2} x}+1-x \cdot \operatorname{ctg} x+\operatorname{ctg}^{2} x-\frac{x \cdot \operatorname{ctg} x}{\sin ^{2} x}+x \cdot \operatorname{ctg} x-\frac{x^{2}}{\sin ^{2} x}}{(1-x \cdot \operatorname{ctg} x)^{2}}= \\
& =\frac{-\frac{1}{\sin ^{2} x}+1+\operatorname{ctg}^{2} x-\frac{x^{2}}{\sin ^{2} x}}{(1-x \cdot \operatorname{ctg} x)^{2}}=\frac{-\frac{x^{2}}{\sin ^{2} x}}{(1-x \cdot \operatorname{ctg} x)^{2}}=-\frac{x^{2}}{\sin ^{2} x(1-x \cdot \operatorname{ctg} x)^{2}}= \\
& =-\frac{x^{2}}{(\sin x-x \cdot \cos x)^{2}}
\end{aligned}
$$

## Problem Kuznetsov Differentiation 15-12