Suppose 5x^2+2x+xy=2 and y(2)=−11. Find y′(2) by implicit differentiation.
5x^2 + 2x + xy = 2
10x + 2 + y + xy' = 0
Now just plug in your numbers and solve for y'.
What numbers am i plugging in?
Solving for y' gives me (-y-10x-2)/x.
Plugging in 2 for x doesn't give me the right answer, so I don't know where to go from there.
To find y'(2) using implicit differentiation, we need to differentiate both sides of the equation 5x^2 + 2x + xy = 2 with respect to x.
Step 1: Differentiate both sides of the equation with respect to x.
d/dx(5x^2 + 2x + xy) = d/dx(2)
Step 2: Differentiate each term separately using the product rule and chain rule when necessary.
Differentiating 5x^2: (d/dx)(5x^2) = 10x
Differentiating 2x: (d/dx)(2x) = 2
Differentiating the product term xy: (d/dx) (xy) = x(dy/dx) + y
Note: Since y is a function of x, we use the chain rule and multiply by dy/dx.
On the right side, the derivative of a constant is 0.
The differentiated equation becomes:
10x + 2 + x(dy/dx) + y = 0
Step 3: Rearrange the equation to solve for dy/dx, which is y'.
x(dy/dx) + y = -10x - 2
Step 4: Isolate dy/dx, which is y'.
x(dy/dx) = -10x - 2 - y
Step 5: Solve for dy/dx, which is y'.
dy/dx = (-10x - 2 - y) / x
Step 6: Substitute x = 2 and y(2) = -11 into the equation to find y'(2).
dy/dx = (-10(2) - 2 - (-11)) / 2
dy/dx = (-20 - 2 + 11) / 2
dy/dx = -11/2
Therefore, y'(2) is equal to -11/2.