Asked by olivia
find dy/dx for y^2+2xy-x^3=12
so far, i have (2y)(dy/dx)+(2)(dy/dx)=3x^2
so far, i have (2y)(dy/dx)+(2)(dy/dx)=3x^2
Answers
Answered by
Reiny
no, for the 2xy term you have to use the product rule
2y dy/dx + (2x)(dy/dx) + 2y(1) - 3x^2 = 0
keep all the dy/dx terms on the left, everybody else goes to the right
2y dy/dx + 2x dy/dx = 3x^2 - 2y
factor out the dy/dx
dy/dx (2y + 2x) = (3x^2 - 2y)
divide
dy/dx = (3x^2 - 2y)/(2y + 2x)
2y dy/dx + (2x)(dy/dx) + 2y(1) - 3x^2 = 0
keep all the dy/dx terms on the left, everybody else goes to the right
2y dy/dx + 2x dy/dx = 3x^2 - 2y
factor out the dy/dx
dy/dx (2y + 2x) = (3x^2 - 2y)
divide
dy/dx = (3x^2 - 2y)/(2y + 2x)
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