Question

Дифференцирования составных функций:

Answer 1

10.18

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

$\frac{dz}{dx} = z'_x + z'_y \times y'_x \\$

$z'_y = \frac{1}{ {x}^{2} + {y}^{2} } \times 2y \\$

$y'_x = {e}^{x}$

$\frac{dz}{dx} = \frac{2x}{ {x}^{2} + {y}^{2} } + \frac{2y \times {e}^{x} }{ {x}^{2} + {y}^{2} } = \\ = \frac{2x + 2y {e}^{x} }{ {x}^{2} + {y}^{2} }$

10.19

$\frac{dz}{dx} = \frac{dz}{du} \times \frac{du}{dx} + \frac{dz}{dv} \times \frac{dv}{dx} \\$

(в правой части закруглённые d. Внизу, соответственно, тоже)

$\frac{dz}{du} = \frac{1}{2} \times \frac{1}{ \frac{u}{v} } \times \frac{1}{v} = \frac{v}{2u} \times \frac{1}{v} = \frac{1}{2u} \\ \\ \frac{du}{dx} = 2tgx \times \frac{1}{ \cos {}^{2} (x) } \\ \\ \frac{dz}{dv} = \frac{v}{2u} \times ( - uv {}^{ - 2} ) = \frac{v}{2u} \times ( - \frac{u}{v {}^{2} } ) = \\ = - \frac{1}{2v} \\ \\ \frac{dv}{dx} = 2ctgx \times ( - \frac{1}{ \sin {}^{2} (x) } ) \\ \\ \frac{dz}{dx} = \frac{1}{2u} \times \frac{2tgx}{ \cos {}^{2} (x) } - \frac{1}{2v} \times ( - \frac{2ctgx}{ \sin {}^{2} (x) } ) = \\ = \frac{tgx}{u \cos {}^{2} (x) } + \frac{ctgx}{v \sin {}^{2} (x) }$