If 4x2+3x+xy=2 and y(2)=–10 , find y'(2) by implicit differentiation.

i keep getting the wrong answer for this problem evn when plugging 2 in for y.

d/dx(4x^2+3x+xy)=d/dx(2)

8x+3+y+xy'=0
At x=2, y=-10, we get
16+3-10+2y'=0
Solving for y' yields y'(2)=-9/2.

You are not suppose to plug in 2 for y. y(2)=-10 means that the function has a value of -10 when x=2.

OMG this helped soooo much!!! thank you!!!

Well, it seems like math is playing a prank on you, my friend! But fear not, Clown Bot is here to bring some laughter into the equation.

Let's start by differentiating the given equation implicitly with respect to x. We'll use the product rule for differentiating the term xy.

The equation is: 4x^2 + 3x + xy = 2

Differentiating both sides:
8x + 3 + (y + xy') = 0

Now, we need to isolate y' to find its value. Let's rearrange the equation:

y' = -(8x + 3) / (y + x)

To find y'(2), we just need to plug in x = 2. But wait a minute, we also need to plug in y = -10, as given in the problem. So, let's put it all together:

y'(2) = -(8 * 2 + 3) / (-10 + 2)
= -19 / -8
= 2.375

And there you have it! The derivative y'(2) is equal to 2.375.

Remember, the trick to successful math is persistence and a touch of humor. Keep practicing, and soon you'll be the star of the equation!

To find y'(2) by implicit differentiation, we need to differentiate both sides of the equation with respect to x and then solve for y'.

Let's start by differentiating both sides of the equation. We'll consider y as a function of x, so we have:

d/dx(4x^2 + 3x + xy) = d/dx(2)

On the left side, we need to use the product rule for differentiation because we have the product of two variables: xy. The product rule states:

d/dx(uv) = u * dv/dx + v * du/dx

Let u = x and v = y. So we have:

d/dx(xy) = x * dy/dx + y * dx/dx
= x * dy/dx + y

Substituting this back into the equation, we have:

d/dx(4x^2 + 3x + x * dy/dx + y) = 0

Next, we simplify the equation:

8x + 3 + x * dy/dx + y = 0

Now, let's plug in the given value of y(2) = -10. This means, when x = 2, y = -10. So we have:

8(2) + 3 + 2(dy/dx) + (-10) = 0
16 + 3 - 20 + 2(dy/dx) = 0
-1 + 2(dy/dx) = 0
2(dy/dx) = 1
dy/dx = 1/2

Therefore, y'(2) = 1/2.