Find the slope of the tangent line to the curve x^2+y^3=5x^2y+5 at the point (1,−1).
Please help!
x^2+y^3=5x^2y+5
2x dx+3y^2 dy=10xy dx+5x^2 dy
dx(2x-10xy)=dy(5x^2-3y^2
dy/dx = slope = (5x^2-3y^2)/(2x-10xy) put in 1,-1 and solve
Thank you!! Do you mind checking my other question to see if I did it right?
To find the slope of the tangent line to the curve at the given point, we can use the derivative. The derivative represents the rate of change of a function at any given point.
1. Start by differentiating both sides of the equation with respect to x. This will give you the derivative of the function implicitly.
d/dx(x^2 + y^3) = d/dx(5x^2y + 5)
2. Differentiate each term separately using the rules of differentiation.
2x + 3y^2 * dy/dx = 10xy + 5x^2 * dy/dx
3. Rearrange the terms to solve for dy/dx, which is the derivative of y with respect to x.
3y^2 * dy/dx - 10xy = 10x^2 - 2x
4. Substitute the given coordinates of the point (1, -1) into the equation to obtain a value for dy/dx.
3(-1)^2 * dy/dx - 10(1)(-1) = 10(1)^2 - 2(1)
3 * dy/dx + 10 = 10 - 2
3 * dy/dx = -2
dy/dx = -2/3
The slope of the tangent line to the curve at the point (1, -1) is -2/3.