Question

Найти дифференциал dy

Answer 1

Ответ:

$\displaystyle y=ln\, tg\dfrac{x}{2}-\frac{x}{sinx}\ \ \ ,\ \ \ \ \ \ \boxed{\ dy=y'(x)\cdot dx\ }\\\\\\y'=\frac{1}{tg\frac{x}{2}}\cdot \frac{1}{cos^2\frac{x}{2}}\cdot \frac{1}{2}-\frac{sinx-x\cdot cosx}{sin^2x}=\frac{1}{sin\frac{x}{2}\cdot cos\frac{x}{2}}\cdot \frac{1}{2}-\frac{sinx-x\cdot cosx}{sin^2x}\\\\\\=\frac{1}{sinx}-\frac{sinx-x\cdot cosx}{sin^2x}\\\\\\\\dy=\Big(\frac{1}{sinx}-\frac{sinx-x\cdot cosx}{sin^2x}\Big)\, dx$