## Condition of the problem

Find the differential $d y$.

$y=\sqrt{\operatorname{ctg} x}-\frac{\sqrt{\operatorname{tg}^{3} x}}{3}$
## Solution

$d y=y^{\prime} \cdot d x=\left(\sqrt{\operatorname{ctg} x}-\frac{\sqrt{\operatorname{tg}^{3} x}}{3}\right)^{\prime} d x=\left(\frac{1}{2 \sqrt{\operatorname{ctg} x}} \cdot \frac{-1}{\sin ^{2} x}-\frac{1}{3} \cdot \frac{1}{2 \sqrt{\operatorname{tg}^{3} x}} \cdot 3 \operatorname{tg}^{2} x \cdot \frac{1}{\cos ^{2} x}\right) d x=$ $=\left(\frac{1}{2 \sqrt{\operatorname{ctg} x}} \cdot \frac{-1}{\sin ^{2} x}-\frac{1}{2 \sqrt{\operatorname{tg} x}} \cdot \operatorname{tg} x \cdot \frac{1}{\cos ^{2} x}\right) d x=\left(-\frac{1}{\sin x \sqrt{4 \sin x \cdot \cos x}}-\frac{1}{\cos x \sqrt{4 \cos x \cdot \sin x}} \cdot \operatorname{tg} x\right) d x=$ $=-\frac{1}{\sqrt{2 \sin 2 x}}\left(\frac{1}{\sin x}+\frac{1}{\cos x} \cdot \operatorname{tg} x\right) d x=-\frac{1}{\sqrt{2 \sin 2 x}}\left(\frac{\cos ^{2} x}{\sin x \cdot \cos ^{2} x}+\frac{\sin x}{\cos ^{2} x}\right) d x=$ $=-\frac{1}{\sin x \cdot \cos ^{2} x \cdot \sqrt{2 \sin 2 x}} \cdot d x=-\frac{\sqrt{2}}{\cos x \cdot \sqrt{\sin ^{3} 2 x}} \cdot d x$

## Problem Kuznetsov Differentiation 4-16