1. Вычислить значения частных производных f’x (M0), f’y (M0), f’z (M0) для данной функции...

3.9.
$f(x,y,z) = arcsin( \frac{x^2}{y} -z); M_0(2,5,0) \\ \\ \frac{df}{dx} = \frac{( \frac{x^2}{y} -z)_x^'}{ \sqrt{1-(\frac{x^2}{y} -z)^2} } = \frac{ \frac{2x}{y}}{ \sqrt{1-(\frac{x^2}{y} -z)^2} } = \frac{ 2x}{y \sqrt{1-(\frac{x^2}{y} -z)^2} } = \\ \\ = \frac{ 2*2}{ 5\sqrt{1-(\frac{2^2}{5} -0)^2} } } = \frac{4}{5* \sqrt{1- \frac{16}{25} } } = \frac{4}{3 }$

$\frac{df}{dy} = \frac{( \frac{x^2}{y} -z)_y^'}{ \sqrt{1-(\frac{x^2}{y} -z)^2} } = \frac{ -\frac{x^2}{y^2}}{ \sqrt{1-(\frac{x^2}{y} -z)^2} } =- \frac{ x^2}{y^2 \sqrt{1-(\frac{x^2}{y} -z)^2} } = \\ \\ = \frac{ 2^2}{ 5^2 \sqrt{1-(\frac{2^2}{5} -0)^2} } } = \frac{4}{5* \sqrt{1- \frac{16}{25} } } =- \frac{4}{15 }$

$\frac{df}{dz} = \frac{( \frac{x^2}{y} -z)_z^'}{ \sqrt{1-(\frac{x^2}{y} -z)^2} } = \frac{ -1}{ \sqrt{1-(\frac{x^2}{y} -z)^2} } = \\ \\ = -\frac{ 1}{ \sqrt{1-(\frac{2^2}{5} -0)^2} } } = -\frac{1}{ \sqrt{1- \frac{16}{25} } } =- \frac{5}{3 }$