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Question

A curve is given by x⁵-x⁴y⁴+y⁵=-1507

If y(-3)=2 evaluate dy/dx

Answer 1

I gave the answer that was posted a downvote because it said that (- 3)^5 was equal to - 729 when in fact it is - 243. The problem is consistent because (- 243) + (- 1296) + 32 does equal - 1507.

Differentiating implicitly gives 5x⁴ - (4x³ * y⁴ + x⁴ * 4y³ * y') + 5y⁴ * y' = 0. Solving for y', we have y' = (5x⁴ - 4x³y⁴) ÷ (4y³x⁴ - 5y⁴) = (405 - (4 * (-27) * 16)) ÷ ((4 * 8 * 81) - 80) = (405 + 1728) ÷ (2592 - 80) = 0.84912 to five decimal places. The prior answerer seems to have made an error with his calculus, as it looks like he differentiated incorrectly by differentiating both factors of the term x⁴y⁴ at once rather than correctly using the product rule, d(uv) = u dv + v du.

The prior answerer made three mistakes in evaluating the correct expression that he got for the derivative. First, he inserted an extra factor of 3 into the term that gave me 1728, thus arriving at 5184; second, he inserted an extra factor of 9 into the term that gave me 2592; and fianlly, he substituted y = 3 instead of the correct y = 2 when evaluating that same term, thus causing it to come out as 78732 instead of 2592. It's possible that the mistakes were caused by his not using the handy equation editor that Wyzant's Ask an Expert provides; his expression for the derivative is hard to read.

I checked to some extent by plotting this equation in Mathematica 13, and the slope of the curve at x = - 3 does look close to 0.84912.

Answer 2

You would need to apply implicit differentiation to your problem to figure out the expression for dy/dx then plug in the point given to you x = -3 and y = 2

Problem: x⁵ - x⁴y⁴ + y⁵ = -1507

Derivative: 5x⁴ - [4x³y⁴ + (x⁴)(4y³dy/dx)] + 5y⁴(dy/dx) = 0

Explanation (by terms): (Power Rule) (Product Rule and attaching dy/dx when differentiating y term) (Power rule) (differentiating constants results in 0)

Algebraically isolate dy/dx

1) Move 5x⁴ term to the other side of the equal sign.

2) Combine the like (dy/dx) terms.

3) Isolate dy/dx

4) Plug in given point (-3,2) into dy/dx and obtain value

Result of dy/dx evaluated at (-3,2) should be 0.8491242038

Answer 3

f(x,y) = x^5 -x^4y^4 +y^5 = -1507

dy/dx = y'= 5x^4 - 4x^3y^4 -4x^4y^3y' +5y^4y'

the middle two terms come from using the product rule

1st term times derivative of 2nd term, plus 2nd term times derivative of the 1st term

1st term is -x^4, 2nd term is y^4

derivative of the 1st =-4x^3 times 2nd = -4x^3y^4

derivative of the 2nd = 4y^3y' times the 1st = -4x^4y^3y'

when x=-3 and y=2, substitute -3 and 2 into the above expression,

y' = 5(-3)^4 - 4(-3)^3(2^4) -4(-3)^4(2)^3(y') +5(2)^4(y') =0

another method, try graphing it with a graphing calculator

then look to see the slope at x=-3

you can at least get a general idea of the range of slope, at least whether it's - or +

nearer 1, 0 or +/- infinity

I'd say check the point on the graph (-3,2) but it's not on the curve, if the top calculations were correct

but you may need a 3 D graph, z=x^5 -x^4y^4 + y^5, hard to do by hand or visualize

if you can't find a graph, plot some points and connect them with a smooth curve, then look at x=2 to see the slope = dy/dx, but this becomes a partial differential equation problem

z' = dz/dx = 0 = 5x^4 -4x^3y^4 -4x^4y^3y' +5y^4y' where y'=dy/dx

replace x with -3, y with 2, and solve for y'

405 -4((-27(16) -4(81(8))y' +80y' = 0

y'(-2592+80) = -64(27)-405

y'(-2512) = -1728-405 = -2133

y'=2133/2512=.84912

dy/dx =.84912

or see what another tutor came up with