Question

вычислите вторую производную y=-lnctg2x

Answer 1

$y=-ln(ctg2x)\\\\y'=-\frac{1}{ctg2x}\cdot (ctg2x)'=-\frac{1}{ctg2x}\cdot \frac{-1}{sin^22x}\cdot (2x)'=\\\\=\frac{1}{ctg2x}\cdot \frac{1}{sin^22x}\cdot 2$

$y''=(\frac{2}{ctg2x\cdot sin^22x})'=\frac{2'\cdot ctg2x\cdot sin^22x-2\cdot ((ctg2x)'\, sin^22x+ctg2x\cdot (sin^22x)')}{ctg^22x\cdot sin^42x}=\\\\=\frac{0-2\cdot (-\frac{2}{sin^22x}\cdot sin^22x+ctg2x\cdot 2\, sin2x\cdot cos2x\cdot 2)}{ctg^22x\cdot sin^42x}=\frac{4-4ctg2x\cdot sin4x}{ctg^22x\cdot sin^42x}$

$2^\star )\; \; \; y'= \frac{2}{ctg2x\cdot sin^22x}=\frac{2}{\frac{cos2x}{sin2x}\cdot sin^22x}=\frac{2}{sin2x\cdot cos2x}=\frac{4}{sin4x}\\\\y''=\frac{0-4\cdot cos4x\cdot 4}{sin^24x}=-\frac{16\cdot cos4x}{sin^24x}=-\frac{4\cdot ctg4x}{sin4x}$