Use implicit differentiation to find the slope of the tangent line to the curve at the point (1,3)
-4x^2+4xy-2y^3=-46
m=____?
ty guys so much!
-8x + 4x dy/dx + 4y - 6y^2 dy/dx = 0
dy/dx(4x - 6y^2) = 8x - 4y
dy/dx = (4x - 2y)/(2x - 3y^2)
plug in x = 1 and y = 3 to find the slope
To find the slope of the tangent line to the curve at the point (1,3), we can use implicit differentiation.
Implicit differentiation involves differentiating both sides of the equation with respect to x and then solving for dy/dx (the derivative of y with respect to x).
Given equation: -4x^2 + 4xy - 2y^3 = -46
1. Differentiate both sides of the equation with respect to x:
d/dx (-4x^2 + 4xy - 2y^3) = d/dx (-46)
Using the power rule and product rule of differentiation:
-8x + 4y + 4x(dy/dx) - 6y^2(dy/dx) = 0
2. Rearrange the terms to solve for dy/dx:
4x(dy/dx) - 6y^2(dy/dx) = 8x - 4y
Factor out dy/dx:
(dy/dx)(4x - 6y^2) = 8x - 4y
3. Solve for dy/dx:
dy/dx = (8x - 4y) / (4x - 6y^2)
4. Substitute the values of x and y using the point (1, 3):
dy/dx = (8(1) - 4(3)) / (4(1) - 6(3)^2)
dy/dx = (8 - 12) / (4 - 54)
dy/dx = -4 / -50
dy/dx = 2/25
The slope of the tangent line to the curve at the point (1,3) is 2/25.