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Helena737
@Helena737
July 2022
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Вычислить производную от функции заданной неявно
[tex] \frac{1}{ \sqrt{2x+y^2} }+ln(cosx)=0 [/tex]
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oganesbagoyan
Verified answer
;
обозначение : sqrt(2x+y²) ⇔√2x+y² !
===============================
((2x+y²) ^ (-1/2)) ' +(1/cosx)*(cosx) ' = 0;
-1/2*(2x+y²)^(-3/2)*(2x+y²)' -tqx=0 ;
-1/2*(2x+y²*)^(-3/2)*(2+2y*y') -tqx=0;
-1/(2x+y2)^(3/2)*(1+y*y') -tqx=0 ;
(1+y*y')/(2x+y²)^3/2 +tqx=0 ;
1+y*y' = -tqx*(2x+y²)^3/2;
y*y '= -tqx*(2x+y²)^3/2 -1;
y ' = -1/y*(tqx*(2x+y²)^3/2 -1) .
====================================
1/√2x+y² = -Ln(cosx) ;
√2x+y² = -1/Ln(cosx); [ Ln(cosx) <Ln1=0 ] ;
2x+y² = 1/Ln²(cosx) ;
y²= 1/Ln²(cosx) -2x ;
y = (+/-√(1/Ln²(cosx) -2x ) .
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Verified answer
;обозначение : sqrt(2x+y²) ⇔√2x+y² !
===============================
((2x+y²) ^ (-1/2)) ' +(1/cosx)*(cosx) ' = 0;
-1/2*(2x+y²)^(-3/2)*(2x+y²)' -tqx=0 ;
-1/2*(2x+y²*)^(-3/2)*(2+2y*y') -tqx=0;
-1/(2x+y2)^(3/2)*(1+y*y') -tqx=0 ;
(1+y*y')/(2x+y²)^3/2 +tqx=0 ;
1+y*y' = -tqx*(2x+y²)^3/2;
y*y '= -tqx*(2x+y²)^3/2 -1;
y ' = -1/y*(tqx*(2x+y²)^3/2 -1) .
====================================
1/√2x+y² = -Ln(cosx) ;
√2x+y² = -1/Ln(cosx); [ Ln(cosx) <Ln1=0 ] ;
2x+y² = 1/Ln²(cosx) ;
y²= 1/Ln²(cosx) -2x ;
y = (+/-√(1/Ln²(cosx) -2x ) .