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Find y′x for x³+y²-2x=3y+2.Then find an equation of the tangent line at the point (1,4)

User STRML
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Final answer:

To find the derivative of y with respect to x for the equation, we use implicit differentiation and solve for dy/dx. At the point (1, 4), the slope of the tangent line is -1/5. The equation of the tangent line at this point is y - 4 = (-1/5)(x - 1).

Step-by-step explanation:

To find y′ (also known as the derivative of y with respect to x, or dy/dx) for the equation x³ + y² - 2x = 3y + 2, we need to differentiate both sides of the equation with respect to x. Because the equation involves both x and y and they are interdependent, we use implicit differentiation. This means we differentiate y terms with respect to y and multiply by dy/dx due to the chain rule.

To differentiate the given equation implicitly, we get:

3x² + 2y(dy/dx) - 2 = 3(dy/dx) + 0

Now, we solve for dy/dx (y′):

2y(dy/dx) - 3(dy/dx) = 2 - 3x²

dy/dx(2y - 3) = 2 - 3x²

dy/dx = (2 - 3x²) / (2y - 3)

To find the equation of the tangent line at the point (1, 4), we plug these values into the derivative:

dy/dx at (1, 4) = (2 - 3(1)²) / (2(4) - 3) = -1 / 5

The slope of the tangent line at (1, 4) is -1/5. Using the point-slope form y - y1 = m(x - x1), the equation of the tangent line is:

y - 4 = (-1/5)(x - 1)

This is the equation of the tangent line at the point (1, 4).

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