find the equation of the tangent line at the given value of x on the curve
2xy-2x+y=-14, x=2
2xy-2x+y=-14, x=2
We want dy/dx
2xdy/dx + 2xdx/dx - 2dx/dx + dy/dx = 0
dx/dx = 1
(2x+1)dy/dx + 2x - 2 = 0
dy/dx = -2(x-1)/(2x+1)
now put in x = 2
To find the equation of the tangent line at a given value of x on a curve, we need to find the derivative of the curve, evaluate it at the given x-value, and use this derivative to form the equation of the tangent line.
Given the equation of the curve: 2xy - 2x + y = -14
First, we need to rewrite it in the form of y = f(x), which is the standard form of a curve equation. So let's rearrange to solve for y:
2xy + y = 2x - 14
y(2x + 1) = 2x - 14
y = (2x - 14) / (2x + 1)
Now, let's differentiate the equation with respect to x to find the derivative:
dy/dx = [(2x + 1)(2) - (2x - 14)(2)] / (2x + 1)^2
dy/dx = (4x + 2 - 4x + 28) / (2x + 1)^2
dy/dx = (30) / (2x + 1)^2
dy/dx = 30 / (2x + 1)^2
Next, we need to evaluate the derivative at the given x-value, x = 2:
dy/dx = 30 / (2(2) + 1)^2
dy/dx = 30 / (4 + 1)^2
dy/dx = 30 / (5)^2
dy/dx = 30 / 25
dy/dx = 6/5
Now, we have the slope of the tangent line at x = 2, which is 6/5. To find the equation of the tangent line, we also need a point that lies on the tangent line. We can use the given x-value and substitute it back into the original curve equation to find the corresponding y-value:
2xy - 2x + y = -14
2(2)y - 2(2) + y = -14
4y - 4 + y = -14
5y - 4 = -14
5y = -10
y = -2
So the point on the curve is (2, -2).
Now we can use the point-slope form of a line, with the point (2, -2) and slope 6/5, to get the equation of the tangent line:
y - y1 = m(x - x1)
y - (-2) = (6/5)(x - 2)
y + 2 = (6/5)(x - 2)
y + 2 = (6/5)x - (6/5)(2)
y + 2 = (6/5)x - 12/5
y = (6/5)x - 12/5 - 2
y = (6/5)x - 12/5 - 10/5
y = (6/5)x - 22/5
Therefore, the equation of the tangent line at x = 2 on the curve 2xy - 2x + y = -14 is y = (6/5)x - 22/5.