Please help don't understand how to solve it.
3xy-6x+2y^2
dx/dt=-6
x=2
y=-3
thank you
Neither do I. You haven't asked a question.
If you want dy/dt at (2,-3) then you just need to do
3y + 3x dy/dt - 6 + 4y dy/dt
but you can't evaluate anything, because you have no equation.
Why not try posting the entire question, not just part of it?
Assume x and y are functions of t, Evaluate dy/dt,
3xy-6x+2y^2
dx/dt=-6
x=2
y=-3
dy/dt
Still no good, but let's assume you have
3xy-6x+2y^2 = 0
Then
3y dx/dt + 3x dy/dt - 6 dx/dt + 4y dy/dt = 0
(3x+4y) dy/dt = (6-3y) dx/dt
dy/dt = 3(2-y)(dx/dt)/(3x+4y)
dy/dt = 3(5)(-6)/(6-12) = 15
But that's so only if my original equation is close.
To solve this problem, we need to find the derivative of the expression 3xy - 6x + 2y^2 with respect to time (t). Given that dx/dt = -6, x = 2, and y = -3, we can substitute these values into the derivative expression to get the answer.
To find the derivative, we differentiate each term of the expression separately:
The derivative of the term 3xy with respect to t is given by:
d/dt (3xy) = 3y * (dx/dt)
Substituting in the value of dx/dt as -6:
d/dt (3xy) = 3y * (-6) = -18y
Then, the derivative of the term -6x with respect to t is:
d/dt (-6x) = -6(dx/dt) = -6(-6) = 36
Finally, the derivative of the term 2y^2 with respect to t is:
d/dt (2y^2) = 2 * (dy/dt) * y
Since y is a constant with respect to t, the derivative of y with respect to t is zero:
dy/dt = 0
Therefore, the derivative of the term 2y^2 becomes:
d/dt (2y^2) = 2 * 0 * y = 0
Now summing up all the derivatives, we have:
dx/dt = d/dt (3xy - 6x + 2y^2) = -18y + 36 + 0
Substituting the given values x = 2 and y = -3, we have:
dx/dt = -18(-3) + 36 + 0
= 54 + 36
= 90
So, the derivative of the expression 3xy - 6x + 2y^2 with respect to t, when x = 2 and y = -3, is 90.