Use implicit differentiation to find the slope of the tangent line to the curve at the point (4,1)
-1x^2 - 4xy + 3y^3 = -29
I differentiated both sides and solved for dy/dx and got 2x-4y/9y^2-4x. Then I plugged in X and Y and got -4/7
but when i enter my answer it's wrong. What is the right answer? Where did I go wrong?
I got (2x+4y)/(9y^2-4x) for dy/dx
I bet your error was in the derivative of
-4xy
which is -4x(dy/dx) - 4y
check your signs.
Thanks a lot, it was the signs that was wrong.
Differentiate both sides of the equation with respect to x, with y being a function of x.
-2x -4y -4x*dy/dx +9y^2*dy/dx = 0
dy/dx = [2x +4y]/[9y^2 -4x]
When x=4 and y = 1, I get
dy/dx = (8 + 4) /(9 - 16) = -12/7
Thanks :)
To find the slope of the tangent line to the curve at the point (4, 1), you correctly used implicit differentiation to find the derivative of the equation. However, there seems to be an error in your calculation for dy/dx.
Let's go through the steps again to find where the mistake occurred.
Starting with the equation: -x^2 - 4xy + 3y^3 = -29
1. Differentiate both sides of the equation with respect to x, treating y as a function of x:
(d/dx) [-x^2 - 4xy + 3y^3] = (d/dx) [-29]
The derivative of each term can be calculated as follows:
-2x - 4y(dy/dx) + 9y^2(dy/dx) = 0
2. Move all terms involving dy/dx to one side of the equation:
-4y(dy/dx) + 9y^2(dy/dx) = 2x
3. Factor out dy/dx:
(dy/dx)(-4y + 9y^2) = 2x
4. Solve for dy/dx by dividing both sides by (-4y + 9y^2):
dy/dx = 2x / (-4y + 9y^2)
Now, let's substitute the values x = 4 and y = 1 into the equation:
dy/dx = 2(4) / (-4(1) + 9(1)^2)
dy/dx = 8 / (-4 + 9)
dy/dx = 8 / 5
Therefore, the slope of the tangent line to the curve at the point (4, 1) is 8/5.
Make sure to double-check your calculations for the derivative and the substitution of values.