Question

Помогите найти производную неявной функции с подробным объяснением:
y=tg(x+y)
(d^2)y/d(x^2) -?

Answer 1

$y = tan(x+y)\\y' = \frac{1+y'}{cos^2(x+y)} \\y'(1 - \frac{1}{cos^2(x+y)}) = \frac{1}{cos^2(x+y)}\\y'(\frac{cos^2(x+y)-1}{cos^2(x+y)}) = \frac{1}{cos^2(x+y)}\\-tan^2(x+y)*y' = \frac{1}{cos^2(x+y)} \\y' = -\frac{1}{cos^2(x+y)*tan^2(x+y)} = -\frac{1}{sin^2(x+y)} \\y'' = (y')' = \frac{2}{sin^3(x+y)}*cos(x+y)*(1+y') = \frac{2cos(x+y)}{sin^3(x+y)}(1 -\frac{1}{sin^2(x+y)}) = -\frac{2cos(x+y)}{sin^3(x+y)} * \frac{cos^2(x+y)}{sin^2(x+y)} = -\frac{2}{tan^3(x+y)sin^2(x+y)} = -\frac{2}{y^3sin^2(x+y)}$

$\frac{d^2y}{dx^2} = -\frac{2}{y^3sin^2(x+y)}$