Evaluate dy/dt for 2xy-2x+2y^3=-10 with the conditions dx/dt=-4 x=2 y=-1

dy/dt=?

use implicit derivatives ....

2x dy/dt + 2y dx/dt + 6y^2 dy/dt = 0
now just plug in the given stuff.

I keep getting -4/5 which is incorrect.

2(2)dy/dt+(2)(-1)(-4)+6(-1)^2=0
10 dy/dt+8=0
10dy/dt=-8

dos soluciones para cada ecuacion

x-y=7

y=x-4

y=2-3x

Just curious. How do you get from

2(2)dy/dt+(2)(-1)(-4)+6(-1)^2=0

to

10 dy/dt+8=0

???

your 2(2)dy/dt+(2)(-1)(-4)+6(-1)^2=0

should have been
2(2)dy/dt + 2(-1)(-4) + 6(-1)^2 dy/dt = 0
4 dy/dt + 8 + 6dy/dt = 0
10dy/dt = -8
dy/dt = -4/5

I don't see any errors.

just notice I skipped the -2x term

so our derivative should be .

2(2)dy/dt + 2(-1)(-4) - 2dx/dt + 6(-1)^2 dy/dt = 0
2(2)dy/dt + 2(-1)(-4) - 2(-4) + 6(-1)^2 dy/dt = 0
4dy/dt + 8 + 8 + 6dy/dt = 0
dy/dt = -16/10
= -1.6

To evaluate dy/dt for the given equation 2xy - 2x + 2y^3 = -10, we need to use implicit differentiation.

Implicit differentiation involves differentiating both sides of the equation with respect to the variable (in this case, t) while considering all variables as implicitly functions of t.

Let's differentiate both sides of the equation with respect to t:

d/dt(2xy) - d/dt(2x) + d/dt(2y^3) = d/dt(-10)

Now, let's differentiate each term separately using the chain rule:

For the term 2xy, we have two variables x and y, both dependent on t. Applying the product rule:

d/dt(2xy) = 2x(dy/dt) + 2y(dx/dt)

For the term 2x, we only have x as a variable dependent on t. So:

d/dt(2x) = 2(dx/dt)

For the term 2y^3, we have y as a variable dependent on t:

d/dt(2y^3) = 6y^2(dy/dt)

Finally, the constant term -10 has no variables dependent on t, so its derivative is zero.

Now let's substitute the given conditions: dx/dt = -4, x = 2, and y = -1 into the equation:

2(2)(dy/dt) + 2(-1)(-4) + 6(-1)^2(dy/dt) = 0

4(dy/dt) + 8 + 6(dy/dt) = 0

10(dy/dt) + 8 = 0

10(dy/dt) = -8

dy/dt = -8/10

Simplifying the fraction, we get:

dy/dt = -4/5

Therefore, dy/dt for the given equation with the given conditions is -4/5.