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graciewyatt471
08.01.2020 •
Mathematics
Find the first, second and third implicit derivatives of x^2+2y^2=16
i'm pretty sure the first derivative is -x/(2y), and the second is -(2y^2+x^2)/(4y^3)
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Ответ:
dy/dx = -x/(2y)
d²y/dx² = (-2y² − x²) / (4y³)
Step-by-step explanation:
x² + 2y² = 16
Take implicit derivative once:
2x + 4y dy/dx = 0
4y dy/dx = -2x
dy/dx = -x/(2y)
Take derivative a second time:
d²y/dx² = [ (2y) (-1) − (-x) (2 dy/dx) ] / (2y)²
d²y/dx² = (-2y + 2x dy/dx) / (4y²)
d²y/dx² = (-2y + 2x (-x/(2y))) / (4y²)
d²y/dx² = (-2y − x²/y) / (4y²)
d²y/dx² = (-2y² − x²) / (4y³)
Ответ:
y''' = 3(x² + 2y²)/(4y³)
Step-by-step explanation:
x² +2y² =16
2x + 4y(y') = 0
4y(y') = -2x
2y(y') = -x
y' = -x/2y
2y(y') = -1
2[y'×y' + y×y"] = -1
2(y')² + 2y(y") = -1
2(-x/2y)² + 2y(y") = -1
2(x²/4y²) + 2y(y") = -1
2y(y") = -1 - 2(x²/4y²)
8y³(y") = -4y² - 2x²
4y³(y") = -2y² - x²
y" = [-x² - 2y²] ÷ (4y³)
y" = -(x² + 2y²)/(4y³)
4y³(y") = -2y² - x²
4y³(y"') + 12y²(y')(y") = -4y(y') - 2x
4y³(y''') + 12y³[-(x² + 2y²)/(4y³)] =
-4y(-x/2y) - 2x
4y³(y''') + 3(-x²-2y²) = 2x - 2x
4y³(y''') = 3(x² + 2y²)
y''' = 3(x² + 2y²)/(4y³)
Ответ:
Hope this helps!