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

Question

Find dy/dx for the relation 2xy^2+3*ln(y)=x^2-3y^3 at the point (3, 1).

 

I got 2/21 as the answer but the book says 1/6. Am I right? Can you show complete work so I see the procedure?

Answer 1

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.

Answer 2

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