1. If 3x^2+2xy+y2=2 then the value of dy/dx x = 1 is
A. -2 B. 0 C. 2 D. 4 E. not defined
how is it -1?
3x^2+2xy+y2=2
6x + 2y + 2xy' + 2yy' = 0
x=1 ==> y=-1
6(1) + 2(-1) + 2(1)y' + 2(-1)y' = 0
6 - 2 + 2y' - 2y' = 0
4 = 0
(e) undefined
The graph is an ellipse. At (1,-1) there is a vertical tangent
i hate you
Well, this question sure knows how to make things interesting! It's like a math equation doing a circus act! Let's see if we can unravel this puzzle together.
To find the value of dy/dx at x = 1, we need to take the derivative of the given equation with respect to x.
Taking the derivative of both sides with respect to x, we get:
6x + 2y(dy/dx) + 2xy' + 2y = 0
Now, let's plug in x = 1 into this expression:
6(1) + 2y(dy/dx) + 2(1)(y') + 2y = 0
Simplifying this equation, we get:
6 + 2y(dy/dx) + 2y' + 2y = 0
Since our goal is to find dy/dx at x = 1, we can plug in x = 1 into this expression as well:
6 + 2y(dy/dx) + 2y' + 2y = 0
Now, let's take a look at the answer choices and do some detective work:
A. -2
B. 0
C. 2
D. 4
E. not defined
Hmm, it seems like we need to do a bit more math to figure out the value of dy/dx at x = 1. Unfortunately, there's not enough information given in the question to solve for dy/dx at that particular value. So, the answer is E. not defined.
Well, life is full of mysteries, just like this math problem! Keep those brain cells limber and always be ready for a challenge.
To find the value of dy/dx at x = 1, we need to differentiate the given equation with respect to x.
The given equation is:
3x^2 + 2xy + y^2 = 2
To differentiate this equation with respect to x, we will use the chain rule.
Differentiating the first term, 3x^2, with respect to x gives us:
d(3x^2)/dx = 6x
For the second term, 2xy, we need to apply the product rule. The product rule states that if we have two functions u(x) and v(x), the derivative of their product is given by:
d(uv)/dx = u(dv/dx) + v(du/dx)
In this case, u(x) = 2x and v(x) = y(x).
Differentiating the second term, 2xy, with respect to x gives us:
d(2xy)/dx = 2x(dy/dx) + 2y
For the third term, y^2, we need to apply the chain rule, as y is a function of x.
Differentiating the third term, y^2, with respect to x gives us:
d(y^2)/dx = 2y(dy/dx)
Combining all of these derivatives, we have:
6x + 2x(dy/dx) + 2y = 0
Now, we need to find the value of dy/dx at x = 1. To do this, we substitute x = 1 into the equation we derived above:
6(1) + 2(1)(dy/dx) + 2y = 0
6 + 2(dy/dx) + 2y = 0
We are also given the equation:
3x^2 + 2xy + y^2 = 2
Substituting x = 1 into this equation, we get:
3(1)^2 + 2(1)y + y^2 = 2
3 + 2y + y^2 = 2
Rearranging this equation, we have:
y^2 + 2y - 1 = 0
Solving this quadratic equation, we find two possible values for y:
y = (-2 ± √(2^2 - 4(1)(-1)) / 2
y = (-2 ± √(4 + 4)) / 2
y = (-2 ± √8) / 2
y = (-2 ± 2√2) / 2
y = -1 ± √2
Since the question only asks for the value of dy/dx at x = 1, we can choose either value of y. Let's choose y = -1 + √2.
Substituting y = -1 + √2 into the equation 6 + 2(dy/dx) + 2y = 0, we have:
6 + 2(dy/dx) + 2(-1 + √2) = 0
Simplifying this equation:
2(dy/dx) = -6 + 2 - 2√2
2(dy/dx) = -4 - 2√2
(dy/dx) = (-4 - 2√2) / 2
(dy/dx) = -2 - √2
Therefore, the value of dy/dx at x = 1 is -2 - √2.
So, the correct answer is A. -2.