(y/(x+9y))= (x^2)-7
at the point (1, -6/55)
Use implicit differentiation to find the slope of the tangent line to the curve
cross-multiply first
y = x^3 - 7x + 9x^2y - 63y
y' = 3x^2 - 7 + 9x^2y' + 18xy - 63y' , where y' = dy/dx
64y' - 9x^2y' = 3x^2 + 18xy-7
y' = dy/dx = (3x^2 + 18xy - 7)/(64 - 9x^2)
I will leave it up to you to do the remaining arithmetic
To find the slope of the tangent line to the curve at the point (1, -6/55), we can use implicit differentiation.
Step 1: Differentiate both sides of the equation with respect to x.
Let's take the derivative of both sides of the equation: (y/(x+9y)) = x^2 - 7
Using the quotient rule, the left side becomes:
[(x+9y)(d/dx)(y) - y(d/dx)(x+9y)] / (x+9y)^2 = 2x
Simplifying it:
[(x+9y)(dy/dx) - y(1 + 9(dy/dx))] / (x+9y)^2 = 2x
Step 2: Solve for (dy/dx).
Let's rearrange the equation to solve for (dy/dx):
[(x+9y)(dy/dx) - y(1 + 9(dy/dx))] / (x+9y)^2 = 2x
[(x+9y)(dy/dx) - y - 9y(dy/dx)] / (x+9y)^2 = 2x
[(x+9y - 9y)(dy/dx)] / (x+9y)^2 = 2x + y
(dy/dx) / (x+9y)^2 = 2x + y
dy/dx = (2x + y) * (x+9y)^2
Step 3: Plug in the values at the point (1, -6/55).
Now we can plug in the values of x and y at the point (1, -6/55):
dy/dx = (2(1) + (-6/55))(1 + 9(-6/55))^2
dy/dx = (2 - 6/55)(1 - 54/55)^2
dy/dx = (104/55)(1/55)^2
dy/dx = (104/55)(1/3025)
dy/dx = 104/166375
Therefore, the slope of the tangent line to the curve at the point (1, -6/55) is 104/166375.