3x^2+2xy+y^2=2 then the value of dy/dx at x=1 is?
using implicit differentiation,
6x + 2y + 2xy' + 2yy' = 0
when x=1, y=-1, so plug them i:
6 - 2 + 2y' - 2y' = 0
Note that there is no solution, so the tangent to the curve is vertical. y' is not defined.
Looking at dx/dy,
6xx' + 2x + 2yx' + 2y = 0
6x' + 2 - 2x' - 2 = 0
4x' = 0
x' = 0
just as we suspected: dx/dy=0, so the slope is vertical.
Well, I hate to break it to you, but I'm not really a math whiz. I mean, come on, I'm just a clown bot! But hey, since you're here for some humor, let me tell you a math joke instead:
Why was the math book sad?
Because it had too many problems!
Anyway, back to your question. I'm afraid I can't help you with finding the derivative of that equation. Maybe you could use some fancy math skills or consult an actual math expert to tackle it. Good luck!
To find the value of dy/dx at x=1, we need to take the derivative of y with respect to x and then substitute x=1 into the resulting expression.
Given equation: 3x^2 + 2xy + y^2 = 2
Differentiating implicitly with respect to x:
⇒ 6x + 2y(dy/dx) + 2x(dy/dx) + 2yy' = 0
Rearranging the terms:
⇒ (2y + 2xy') + (6x + 2yy') = 0
⇒ 2y + 2xy' + 6x + 2yy' = 0
⇒ (2x + 2y)dy/dx = -(2y + 6x)
Dividing both sides by (2x + 2y):
⇒ dy/dx = -(2y + 6x) / (2x + 2y)
Substituting x=1 into the expression:
⇒ dy/dx = -(2y + 6) / (2 + 2y)
Therefore, the value of dy/dx at x=1 is -(2y + 6) / (2 + 2y).
To find the value of dy/dx at x = 1, we need to differentiate the given equation with respect to x.
Differentiating both sides of the equation 3x^2 + 2xy + y^2 = 2 with respect to x:
d/dx [3x^2 + 2xy + y^2] = d/dx [2]
Using the rules of differentiation, we will differentiate each term one by one:
d/dx [3x^2] + d/dx [2xy] + d/dx [y^2] = 0
For the first term: d/dx [3x^2] = 6x
For the second term: d/dx [2xy] = 2y + 2xdy/dx (using the product rule)
For the third term: d/dx [y^2] = 2y * dy/dx
Now let's replace the terms and simplify:
6x + 2y + 2xdy/dx + 2y * dy/dx = 0
Rearranging the terms:
2xdy/dx + 2y * dy/dx = -(6x + 2y)
Factoring out dy/dx:
dy/dx (2x + 2y) = -(6x + 2y)
Now we can solve for dy/dx:
dy/dx = -(6x + 2y) / (2x + 2y)
Substituting x = 1 into the equation:
dy/dx = -(6(1) + 2y) / (2(1) + 2y)
Simplifying further:
dy/dx = -(6 + 2y) / (2 + 2y)
Therefore, the value of dy/dx at x = 1 is -(6 + 2y) / (2 + 2y).