Question

Найти dz/dx и dz/dy функции заданой неявно yz=arctg (xz)

Answer 1

$yz=\mathrm{arctg} (xz)$

1.

$(yz)'_x=(\mathrm{arctg} (xz))'_x$

$yz'_x=\dfrac{1}{1+(xz)^2}\cdot (xz)'_x$

$yz'_x=\dfrac{1}{1+x^2z^2}\cdot (x'_xz+xz'_x)$

$yz'_x=\dfrac{1}{1+x^2z^2}\cdot (z+xz'_x)$

$y(1+x^2z^2)z'_x=z+xz'_x$

$y(1+x^2z^2)z'_x-xz'_x=z$

$(y(1+x^2z^2)-x)z'_x=z$

$z'_x=\dfrac{z}{y(1+x^2z^2)-x}$

$\boxed{\dfrac{\partial z}{\partial x} =\dfrac{z}{y(1+x^2z^2)-x}}$

2.

$(yz)'_y=(\mathrm{arctg} (xz))'_y$

$y'_yz+yz'_y=\dfrac{1}{1+(xz)^2}\cdot(xz)'_y$

$z+yz'_y=\dfrac{1}{1+x^2z^2}\cdot xz'_y$

$\dfrac{1}{1+x^2z^2}\cdot xz'_y-yz'_y=z$

$\left(\dfrac{x}{1+x^2z^2}-y\right)z'_y=z$

$z'_y=\dfrac{z}{\dfrac{x}{1+x^2z^2}-y}=\dfrac{z(1+x^2z^2)}{x-y(1+x^2z^2)}$

$\boxed{\dfrac{\partial z}{\partial y} =\dfrac{z(1+x^2z^2)}{x-y(1+x^2z^2)}}$