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Find dy/dx for the relation? (Please check my answer)

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2 Answers

2xy^2+3*ln(y)=x^2-3y^3
Differentiate both sdies with respect to x,
2(y^2 + 2xyy') + 3y'/y = 2x-9y^2y'
Plug in (3, 1),
2(1+6y')+3y' = 6-9y'
Solve for y',
24y' = 4
y' = 1/6 <==Answer
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Attn: To get it done simpler, plug in (3,1) before you solve y' as a function of x.
This is an implicit differentiation problem.
 
2*x*y^2 + 3*LN(y) = x^2 - 3*y^3
 
differentiating using the product rule and chain rule for the first half of the left side, and the chain rule for the second half of the left side yields
 
4xyy' + 2y^2 + (3/y)y'
 
for the first half of the right side we differentiate as usual and for the last half we use the chain rule, which yields
 
2x - (9y^2)y'
 
the whole thing being
 
4xyy' + 2y^2 + (3/y)y' = 2x - (9y^2)y'
 
grouping like terms and solving for y' (dy/dx) yields
 
dy/dx = (2x -2y^2) / (4xy + (3/y) + 9y^2)
 
Plugging in or point at (3,1) yields
 
dy/dx = (6 - 2) / (12 + 3 + 9) = 4/24 = 1/6

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