Question

Счастными производными! вычислите значения производной сложной функции z=z(x; y), где x=x(t), y=y(t) при

Answer 1

$z=z(x,y)\; ,\; \; x=x(t)\; ,\; \; y=y(t)\\\\ \frac{dz}{dt} = \frac{\partial z}{\partial x} \cdot \frac{dx}{dt}+ \frac{\partial z}{\partial y} \cdot \frac{dy}{dt} \\\\\\z= \frac{x}{y} -\frac{y}{x} \; ,\; \; x=sin2t\; ,\; \; y=tg^2t\; ,\; \; t_0= \frac{\pi}{4} \; ;\\\\ \frac{\partial z}{\partial x} = \frac{1}{y} +\frac{y}{x^2} \; ,\; \; \frac{\partial z}{\partial y}=-\frac{x}{y^2}-\frac{1}{x} \; ;\\\\ \frac{dx}{dt}=2\, cos2t\; ,\; \; \frac{dy}{dt}= \frac{2tgt}{cos^2t} \; ;$

$\frac{dz}{dt}=(\frac{1}{y} +\frac{x}{y^2})\cdot 2cos2t+(-\frac{x}{y^2}-\frac{1}{x})\cdot \frac{2tgt}{cos^2t} \\\\t_0= \frac{\pi }{4}\; \; \Rightarrow \; \; x_0=sin(2\cdot \frac{\pi}{4})=sin\frac{\pi}{2}=1\; ,\; y_0=tg^2\frac{\pi}{4}=1^2=1\; ,\\\\ \frac{dz(t_0)}{dt} =(1+1)\cdot 2\underbrace {cos\frac{\pi}{2}}_{0}+(-1-1)\cdot \frac{2tg\frac{\pi}{4}}{cos^2\frac{\pi}{4}} =-2\cdot \frac{2\cdot 1}{(\frac{\sqrt2}{2})^2} = -\frac{4\cdot 4}{2} =-8$