Question

Найти производные от функций, заданных неявно
1) 3x⁴y⁵+$e^{7x-4y}$ = 4x⁵+2y⁴
2) y sin x = cos(x-y)

Answer 1

Ответ:

1.

$3 {x}^{4} {y}^{5} + {e}^{7x - 4y} = 4 {x}^{5} + 2 {y}^{4} \\ (3 {x}^{4} )' \times {y}^{5} + ( {y}^{5} )' \times 3 {x}^{4} + {e}^{7x - 4y} \times (7x - 4y)' = 20 {x}^{4} + 8 {y}^{3} y'\\ 12 {x}^{3} {y}^{5} + 5 {y}^{4} y' \times 3 {x}^{4} + {e}^{7x - 4y} \times (7 - 4y') = 20 {x}^{4} + 8 {y}^{3} y' \\ 15 {x}^{4} {y}^{4} y'- 4 {e}^{7x - 4y} y'- 8 {y}^{3} y'= 20 {x}^{4} - 12 {x}^{3} {y}^{5} - 7 {e}^{7x - 4y} \\ y'(15 {x}^{4} {y}^{4} - 4 {e}^{7x - 4y} - 8 {y}^{3} ) = 20 {x}^{4} - 12 {x}^{3} {y}^{5} - 7 {e}^{7x - 4y} \\ y '= \frac{20 {x}^{4} - 12 {x}^{3} {y}^{5} - 7 {e}^{7x - 4y} }{15 {x}^{4} {y}^{4} - 4 {e}^{7x - 4y} - 8 {y}^{3} }$

2.

$y \sin(x) = \cos(x - y) \\ (y)' \sin(x) + ( \sin(x)) '\times y = - \sin(x - y) \times (x - y)'\\ y' \sin(x) + y\cos(x) = - \sin(x - y) \times (1 - y') \\ y'\sin(x) + y\cos(x) = - \sin(x - y) + y' \sin(x - y) \\ y' \sin(x) - y' \sin(x - y) = - y \cos(x) - \sin(x - y) \\ y'( \sin(x) - \sin(x - y) ) = - (y \cos(x) + \sin(x - y) ) \\ y' = - \frac{y \cos(x) + \sin(x - y) }{ \sin(x) - \sin(x - y) }$