Question

Найдите производную dz/dt функции z=arccos(x+y), x=3t, y=4t^3. Решите плизз))

Answer 1

$x=3t\; ,\; \; y=4t^3\\\\z=arccos(x+y)=arccos(3t+4t^3)\\\\z'_{t}=\frac{dz}{dt}=-\frac{1}{\sqrt{1-(3t+4t^3)^2}}\cdot (3+12t^2)\\\\ili\\\\\frac{dz}{dt}=\frac{\partial z}{\partial x}\cdot \frac{dx}{dt}+\frac{\partial z}{\partial y}\cdot \frac{dy}{dt}\\\\\frac{dz}{dt}=-\frac{1}{\sqrt{1-(x+y)^2}}\cdot 3-\frac{1}{\sqrt{1-(x+y)^2}}\cdot 12t^2=-\frac{1}{\sqrt{1-(x+y)^2}}\cdot (3+12t^2)=\\\\=-\frac{1}{\sqrt{1-(3t+4t^3)^2}}\cdot (3+12t^2)$