Final answer:
After using implicit differentiation on the equation x³ + y² - 2x = 3y + 2, we obtained y' = (3x² - 2) / (2y - 3). Substituting the point (1,4) into this derivative gives us the slope of the tangent line as 1/5, leading to the tangent line equation y = 1/5x + 16/5.
Step-by-step explanation:
To find y'x (the derivative of y with respect to x) for x³ + y² - 2x = 3y + 2, we first use the implicit differentiation technique. Taking the derivative of both sides of the equation with respect to x gives us:
- 3x² + 2yy' - 2 = 3y' + 0.
Solving for y', we can isolate y' on one side:
- 2yy' - 3y' = 3x² - 2,
- y'(2y - 3) = 3x² - 2,
- y' = (3x² - 2) / (2y - 3).
Next, we need to find the equation of the tangent line at the point (1,4). First, we'll substitute x = 1 and y = 4 into the derivative to find the slope (m) of the tangent:
- m = (3(1)² - 2) / (2(4) - 3) = 1 / 5.
With the slope and the point (1,4), we can use the point-slope formula to find the equation of the tangent line:
- y - y1 = m(x - x1),
- y - 4 = 1/5(x - 1).
So the equation of the tangent line at (1,4) is y = 1/5x + 16/5.