Calculus Implicit Differentiation: solving for y' (x^4)+(x^2y^2)+(y^3)=5 Answer is: -[2x(2x^2+y^2)]÷[y(2x^2+3y)] What are the steps?
回答 (2)
2016-07-04 6:20 pm
x⁴ + x²y² + y³ = 5
(x⁴ + x²y² + y³)' = 0
(x⁴)' + (x²y²)' + (y³)' = 0
4x³ + x²(y²)' + y²(x²)' + 3y² = 0
4x³ + x²(2yy') + y²(2x) + 3y²y' = 0
4x³ + 2x²yy' + 2xy² + 3y²y' = 0
2x²yy' + 3y²y' = -4x³ - 2xy²
y(2x² + 3y)y' = -2x(2x² + y²)
y' = [-2x(2x² + y²)] / [y(2x² + 3y)]
(x⁴ + x²y² + y³)' = 0
(x⁴)' + (x²y²)' + (y³)' = 0
4x³ + x²(y²)' + y²(x²)' + 3y² = 0
4x³ + x²(2yy') + y²(2x) + 3y²y' = 0
4x³ + 2x²yy' + 2xy² + 3y²y' = 0
2x²yy' + 3y²y' = -4x³ - 2xy²
y(2x² + 3y)y' = -2x(2x² + y²)
y' = [-2x(2x² + y²)] / [y(2x² + 3y)]
2016-07-04 6:15 pm
x^4 + x^2y^2 + y^3 = 5
Differentiate both sides with respect to x:
4x^3 + 2xy^2 + 2x^2y dy/dx + 3y^2 dy/dx = 0
(2x^2y + 3y^2) dy/dx = −(4x^3+2xy^2)
dy/dx = −(4x^3+2xy^2) / (2x^2y+3y^2)
dy/dx = −2x(2x^2+y^2) / (y(2x^2+3y))
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