Find dy/dx for: 2y^3 - 3xy = 4
Write an equasion for the line tangent to the curve at (1,2)
Find d^2y/dx^2 at (1,2)
As you can see, it is not easy to solve for y in terms of x only and then differentiate that equation. It is much easier to use the method called "implicit differentiation" in which you take the derivative of both sides of the equation, treating y as a function of x. This results in:
6y^2*dy/dx - 3x*dy/dx -3y = 0
dy/dx*(2y^2-x) = y
dy/dx = y/(2y^2-x)
At (1,2), dy/dx = 2/(8-1) = 2/7
Use that slope and the coordinates (1,2) that the line must pass through to get the equation of the tangent line.
For the second derivative, differentiate an equation containing dy/dx impliticly with respect to x.
dy/dx*(2y^2-x) = y
d^2y/dx^2*(2y^2-x)
+ dy/dx (4y*dy/dx -1) = dy/dx
Insert the value of dy/dx = 2/7 that you already know at the point (1,2), and solve for d^2y/dx^2 at the same point.
I did that and got 2... my friend said the answer is just 2/7. Is she right?
To find dy/dx for the given equation 2y^3 - 3xy = 4, we need to use implicit differentiation.
Step 1: Differentiate both sides of the equation with respect to x.
d/dx (2y^3 - 3xy) = d/dx(4)
Step 2: Apply the chain rule to differentiate the terms involving y.
6y^2 * (dy/dx) - [3x * (dy/dx) + y * (3)] = 0
Step 3: Simplify the equation and solve for dy/dx.
6y^2 * (dy/dx) - 3xy * (dy/dx) - 3y = 0
(6y^2 - 3xy) * (dy/dx) = 3y
dy/dx = (3y) / (6y^2 - 3xy)
Now, we can proceed to find the equation of the tangent line at the point (1,2).
Step 1: Substitute x = 1 and y = 2 into the expression we found for dy/dx.
dy/dx = (3*2) / (6*(2^2) - 3*1*2) = 6 / (24 - 6) = 6 / 18 = 1/3
So, the slope of the tangent line at (1,2) is 1/3.
Step 2: Use the point-slope form of a line to find the equation of the tangent line:
y - y₁ = m(x - x₁)
y - 2 = (1/3)(x - 1)
3(y - 2) = x - 1
3y - 6 = x - 1
x - 3y + 5 = 0
Therefore, the equation of the line tangent to the curve at (1,2) is x - 3y + 5 = 0.
To find d²y/dx² at (1,2), we first need to find the second derivative dy/dx.
Step 1: Differentiate the expression we found for dy/dx with respect to x.
d/dx [(3y) / (6y^2 - 3xy)] = [d/dx(3y) * (6y^2 - 3xy) - (3y) * d/dx(6y^2 - 3xy)] / (6y^2 - 3xy)^2
Step 2: Apply the product and chain rules to differentiate the terms. Also, substitute x = 1 and y = 2.
d²y/dx² = [(3 * (dy/dx) * (6y^2 - 3xy) - (3y) * (12y * (dy/dx) - 3x))] / (6y^2 - 3xy)^2
d²y/dx² = [(3 * (1/3) * (6(2^2) - 3(1)(2)) - (2) * (12(2) * (1/3) - 3(1)))] / (6(2^2) - 3(1)(2))^2
d²y/dx² = [2 * (21 - 20)] / (24 - 6)^2
d²y/dx² = 2 / 18^2
d²y/dx² = 2 / 324
d²y/dx² = 1 / 162
Therefore, d²y/dx² at (1,2) is 1/162.