Question

Помогите найти производную неявной функции.

Answer 1

Ответ:

$8 {x}^{2} \times {e}^{4x + {y}^{2} } = 5y + ln( {x}^{3} + y ) \\ (8 {x}^{2} ) '{e}^{4x + {y}^{2} } + ( {e}^{4x + {y}^{2} } )' \times 8 {x}^{2} = 5y' + \frac{1}{ {x}^{3} + y } \times (3 {x}^{2} + y') \\ 16x {e}^{4x + {y}^{2} } + {e}^{4x + {y}^{2} } \times (4 + 2yy')8 {x}^{2} = 5y'+ \frac{3}{ {x}^{3} + y } + \frac{ y'} { {x}^{3} + y} \\ 16x {e}^{4x + {y}^{2} } + 32 {x}^{2} {e}^{4x + {y}^{2} } + 16yy' {x}^{2} {e}^{4x + {y}^{2} } - 5y' - \frac{y'}{ {x}^{3} + y } = \frac{3}{ {x}^{3} + y } \\ y'(16y {x}^{2} {e}^{4x + {y}^{2} } - 5 - \frac{1}{ {x}^ {3} + y } ) = \frac{3}{ {x}^{3} + y } - 16x {e}^{4x + {y}^{2} } - 32 {x}^{2} {e}^{4x + {y}^{2} } \\ y '\times \frac{(16 {x}^{2} y {e}^{4x + {y}^{2} })( {x}^{3} + y) - 1 }{ {x}^{3} + y } = \frac{3 - {e}^{4x - {y}^{2} }( {x}^{3} + y)(16x + 32 {x}^{2}) }{ {x}^{3} + y } \\ y' = \frac{3 - {e}^{4x - {y}^{2} }( {x}^{3} + y)(16x + 32 {x}^{2}) }{ {x}^{3} + y } \times \frac{ {x}^{3} + y}{(16 {x}^{2} y {e}^{4x + {y}^{2} })( {x}^{3} + y) - 1} \\ y '= \frac{3 - {e}^{4x - {y}^{2} }( {x}^{3} + y)(16x + 32 {x}^{2}) }{ (16 {x}^{2} y {e}^{4x + {y}^{2} })( {x}^{3} + y) - 1}$