A) If x^2+y^3−xy^2=5, find dy/dx in terms of x and y.

dy/dx=________

B) Using your answer for dy/dx, fill in the following table of approximate y-values of points on the curve near x=1 y=2.

0.96 ______
0.98 ______
1.02 ______
1.04 ______


C) Finally, find the y-value for x=0.96 by substituting x=0.96 in the original equation and solving for y using a computer or calculator.
y(0.96)= ________

D) How large (in magnitude) is the difference between your estimate for y(0.96) using dy/dx and your solution with a computer or calculator?
___________

I am not going to do this for you. What are you stuck on, or don't understand?

In the first part... I got (-2x+y^2)/(3y^2+x2y) but apparently that's wrong which will make everything else wrong.... I just need to know how to do that part right...

I see what is wrong with my equation... now I have (-2x+y^2)/(3y^2-x2y) and for the second part I tried to plug 1 for x and all the other values for y and then in the third part I plugged 2 for y and 0.96 for x and I got them all wrong:

my values are:
0.96 = -1.2765
0.98= -1.1285
1.02= -0.8875
1.04= -0.7885

and for part C) 0.255
and For part D) -1.5315

A) To find dy/dx, we need to differentiate the given equation with respect to x.

Differentiating x^2 + y^3 - xy^2 = 5 with respect to x, we get:
2x + 3y^2(dy/dx) - y^2 - 2xy(dy/dx) = 0

Rearranging the terms, we have:
2x - y^2 + (3y^2 - 2xy)(dy/dx) = 0

To find dy/dx, isolate it by dividing through by (3y^2 - 2xy):
(dy/dx) = (y^2 - 2x)/(2x - 3y^2)

Therefore, dy/dx = (y^2 - 2x)/(2x - 3y^2).

B) To fill in the table of approximate y-values, we can use the given point (x=1, y=2) along with the expression we derived for dy/dx in terms of x and y.

We can use the equation dy/dx = (y^2 - 2x)/(2x - 3y^2) to find the slope at x=1 and y=2, which will allow us to approximate the values of y for the given x-values.

Using the point (x=1, y=2) in dy/dx = (y^2 - 2x)/(2x - 3y^2), we get:
dy/dx = (2^2 - 2(1))/(2(1) - 3(2)^2)
= (4 - 2)/(2 - 12)
= 2/-10
= -1/5

Now we can use this slope to approximate the y-values for the given x-values in the table:

For x = 0.96:
To find the approximate y-value, we use the point-slope form of a line equation:

y - y1 = m(x - x1)

Using (x1, y1) = (1, 2) and m = -1/5, we have:
y - 2 = (-1/5)(0.96 - 1)
y - 2 = (-1/5)(-0.04)
y - 2 = 0.008
y = 2.008

So the approximate y-value for x = 0.96 is 2.008.

C) To find the y-value for x = 0.96, we substitute x = 0.96 in the original equation and solve for y using a computer or calculator.

Substituting x = 0.96 in x^2 + y^3 - xy^2 = 5, we have:
(0.96)^2 + y^3 - 0.96y^2 = 5

Solving this equation using a computer or calculator, we find:
y ≈ 2.020

So y(0.96) ≈ 2.020.

D) To determine the magnitude of the difference between the estimate for y(0.96) using dy/dx and the solution obtained with a computer or calculator, we calculate the absolute difference between the two values.

The absolute difference is given by:
|y(0.96) using dy/dx - y(0.96) using calculator|

Taking |2.008 - 2.020|, we find that the magnitude of the difference is approximately 0.012.