Question

100 БАЛЛОВ!! СРОЧНО!! ПОМОГИТЕ ПОЖАЛУЙСТА!!!

Answer 1

Ответ:

$1) v = \sqrt[x]{ {e}^{y} } = {e}^{ \frac{y}{x} }$

$\frac{dv}{dx} = {e}^{ \frac{y}{x} } \times y \times ( {x}^{ - 1} )' = \\ = {e}^{ \frac{y}{x} } \times y \times ( - {x}^{ - 2} ) = \\ = - \frac{y \sqrt[x]{ {e}^{y} } }{ {x}^{2} }$

$\frac{dv}{dy} = {e}^{ \frac{y}{x} } \times \frac{1}{x} = \frac{ \sqrt[x]{ {e}^{y} } }{x}$

$2)z = \frac{ {x}^{2} + x + 1 }{ ln(y) }$

$\frac{dz}{dx} = \frac{1}{ ln(y) } \times (2x + 1) = \frac{2x + 1}{ ln(y) }$

$\frac{dz}{dy} = ( {x}^{2} + x + 1) \times ( - {ln}^{ - 2} (y)) \times \frac{1}{y} = \\ = - \frac{ {x}^{2} + x + 1 }{y {ln}^{2} y}$

Answer 2

Ответ:

$1)\ \ V=\sqrt[x]{e^{y}}=e^{\frac{y}{x}}\\\\\dfrac{\partial V}{\partial x}=V'_{x}=e^{\frac{y}{x}}\cdot \dfrac{-y}{x^2}\ \ \ ,\ \ \ \ \dfrac{\partial V}{\partial y}=V'_{y}=e^{\frac{y}{x}}\cdot \dfrac{1}{x}\\\\\\2)\ \ Z=\dfrac{x^2+x+1}{lny}\\\\\\\dfrac{\partial Z}{\partial x}=Z'_{x}=\dfrac{1}{lny}\cdot (2x+1)\ \ \ ,\\\\\\\dfrac{\partial Z}{\partial y}=Z'_{y}=(x^2+x+1)\cdot \dfrac{-1}{y\cdot ln^2y}=-\dfrac{x^2+x+1}{y\cdot ln^2y}$