Use implicit differentiation to find the equation of the tangent line to the curve xy^3+2xy=9 at the point (31). The equation of this tangent line can be written in the form y=mx+b where m is
x(3y^2dy) + y^3(dx) +2xdy + 2ydx = 0
dy (3xy^2+2x) = -dx (y^3+2y)
dy/dx = -(y^3+2y)/(3xy^2+2x)
I assume you mean the point (3,1)
x = 3
y = 1
so
m = dy/dx = -(1+2)/(9+6) = - 3/15 = -1/5
put in point
1 = m * 3 + b
1 = -3/5 + b
b = 8/5
y = -x/5 + 8/5
5y = 8-x
To find the equation of the tangent line using implicit differentiation, follow these steps:
Step 1: Start by differentiating both sides of the given equation with respect to x. Treat y as a function of x and use the chain rule when differentiating y terms.
For the given equation: xy^3 + 2xy = 9
Differentiating both sides with respect to x:
d/dx [xy^3 + 2xy] = d/dx [9]
Apply the product rule and chain rule accordingly:
y^3 * (d/dx[x]) + x * d/dx[y^3] + 2 * y * (d/dx[x]) + 2 * x * (d/dx[y]) = 0
Simplifying this expression gives:
y^3 + 3xy^2 * (dy/dx) + 2y + 2x * (dy/dx) = 0
Step 2: Now, find dy/dx by isolating it in the equation obtained in step 1.
Group the terms with dy/dx together:
3xy^2 * (dy/dx) + 2x * (dy/dx) = -y^3 - 2y
Factor out dy/dx:
(dy/dx)(3xy^2 + 2x) = -y^3 - 2y
Isolate dy/dx:
(dy/dx) = (-y^3 - 2y) / (3xy^2 + 2x)
Step 3: Substitute the given point (31) into the expression obtained in step 2 to find the slope of the tangent line.
Substitute x = 3 and y = 1 into the equation:
(dy/dx) = (-(1)^3 - 2(1)) / (3(3)(1)^2 + 2(3))
(dy/dx) = (-1 - 2) / (9 + 6)
(dy/dx) = -3/15
(dy/dx) = -1/5
Therefore, the slope of the tangent line is -1/5.
Step 4: Use the point-slope form of the equation of a line to write the equation of the tangent line.
Using the point (3, 1) and the slope -1/5:
y - y1 = m(x - x1)
y - 1 = (-1/5)(x - 3)
Rearranging and simplifying to the y = mx + b form:
y = (-1/5)x + 14/5
So, the equation of the tangent line is y = (-1/5)x + 14/5, where m = -1/5.