Answer:
Part A)
![\displaystyle \frac{dy}{dx}=\frac{2xy}{y-x^2}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cfrac%7Bdy%7D%7Bdx%7D%3D%5Cfrac%7B2xy%7D%7By-x%5E2%7D)
Part B)
![y=x-2](https://tex.z-dn.net/?f=y%3Dx-2)
Part C)
![(0, \sqrt{3})\text{ and } (0, -\sqrt3)](https://tex.z-dn.net/?f=%280%2C%20%5Csqrt%7B3%7D%29%5Ctext%7B%20and%20%7D%20%280%2C%20-%5Csqrt3%29)
Part D)
![\displaystyle \frac{d^2y}{dx^2}_{(1, -1)}=-\frac{1}{2}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cfrac%7Bd%5E2y%7D%7Bdx%5E2%7D_%7B%281%2C%20-1%29%7D%3D-%5Cfrac%7B1%7D%7B2%7D)
Step-by-step explanation:
We have the equation:
![y^2-2x^2y=3](https://tex.z-dn.net/?f=y%5E2-2x%5E2y%3D3)
Part A)
We want to find the derivative of the equation. So, dy/dx.
Let’s take the derivative of both sides with respect to x. Therefore:
![\displaystyle \frac{d}{dx}\Big[y^2-2x^2y\Big]=\frac{d}{dx}[3]](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cfrac%7Bd%7D%7Bdx%7D%5CBig%5By%5E2-2x%5E2y%5CBig%5D%3D%5Cfrac%7Bd%7D%7Bdx%7D%5B3%5D)
Differentiate. We will need to implicitly different on the left. The second term will also require the product rule. Therefore:
![\displaystyle 2y\frac{dy}{dx}-4xy-2x^2\frac{dy}{dx}=0](https://tex.z-dn.net/?f=%5Cdisplaystyle%202y%5Cfrac%7Bdy%7D%7Bdx%7D-4xy-2x%5E2%5Cfrac%7Bdy%7D%7Bdx%7D%3D0)
Rearranging gives:
![\displaystyle \frac{dy}{dx}\Big(2y-2x^2\Big)=4xy](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cfrac%7Bdy%7D%7Bdx%7D%5CBig%282y-2x%5E2%5CBig%29%3D4xy)
Therefore:
![\displaystyle \frac{dy}{dx}=\frac{4xy}{2y-2x^2}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cfrac%7Bdy%7D%7Bdx%7D%3D%5Cfrac%7B4xy%7D%7B2y-2x%5E2%7D)
And, finally, simplifying:
![\displaystyle \frac{dy}{dx}=\frac{2xy}{y-x^2}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cfrac%7Bdy%7D%7Bdx%7D%3D%5Cfrac%7B2xy%7D%7By-x%5E2%7D)
Part B)
We want to write the equation for the line tangent to the curve at the point (1, -1).
So, we will require the slope of the tangent line at (1, -1). Substitute these values into our derivative. So:
![\displaystyle \frac{dy}{dx}_{(1, -1)}=\frac{2(1)(-1)}{(-1)-(1)^2}=1](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cfrac%7Bdy%7D%7Bdx%7D_%7B%281%2C%20-1%29%7D%3D%5Cfrac%7B2%281%29%28-1%29%7D%7B%28-1%29-%281%29%5E2%7D%3D1)
Now, we can use the point-slope form:
![y-y_1=m(x-x_1)](https://tex.z-dn.net/?f=y-y_1%3Dm%28x-x_1%29)
Substitute:
![y-(-1)=1(x-1)](https://tex.z-dn.net/?f=y-%28-1%29%3D1%28x-1%29)
Simplify:
![y+1=x-1](https://tex.z-dn.net/?f=y%2B1%3Dx-1)
So:
![y=x-2](https://tex.z-dn.net/?f=y%3Dx-2)
Part C)
If the line tangent to the curve is horizontal, this means that dy/dx=0. Hence:
![\displaystyle 0=\frac{2xy}{y-x^2}](https://tex.z-dn.net/?f=%5Cdisplaystyle%200%3D%5Cfrac%7B2xy%7D%7By-x%5E2%7D)
Multiplying both sides by the denominator gives:
![0=2xy](https://tex.z-dn.net/?f=0%3D2xy)
Assuming y is not 0, we can divide both sides by y. Hence:
![2x=0](https://tex.z-dn.net/?f=2x%3D0)
Then it follows that:
![x=0](https://tex.z-dn.net/?f=x%3D0)
Going back to our original equation, we have:
![y^2-2x^2y=3](https://tex.z-dn.net/?f=y%5E2-2x%5E2y%3D3)
Substituting 0 for x yields:
![y^2=3](https://tex.z-dn.net/?f=y%5E2%3D3)
So:
![y=\pm\sqrt{3}](https://tex.z-dn.net/?f=y%3D%5Cpm%5Csqrt%7B3%7D)
Therefore, two points where the derivative equals 0 is:
![(0, \sqrt{3})\text{ and } (0, -\sqrt3)](https://tex.z-dn.net/?f=%280%2C%20%5Csqrt%7B3%7D%29%5Ctext%7B%20and%20%7D%20%280%2C%20-%5Csqrt3%29)
However, we still have to test for y. Let’s go back. We have:
![0=2xy](https://tex.z-dn.net/?f=0%3D2xy)
Assuming x is not 0, we can divide both sides by x. So:
![0=2y](https://tex.z-dn.net/?f=0%3D2y)
Therefore:
![y=0](https://tex.z-dn.net/?f=y%3D0)
And going back to our original equation and substituting 0 for y yields:
![(0)^2-2x^2(0)=0\neq3](https://tex.z-dn.net/?f=%280%29%5E2-2x%5E2%280%29%3D0%5Cneq3)
Since this is not true, we can disregard this case.
So, our only points where the derivative equals 0 is at:
![(0, \sqrt{3})\text{ and } (0, -\sqrt3)](https://tex.z-dn.net/?f=%280%2C%20%5Csqrt%7B3%7D%29%5Ctext%7B%20and%20%7D%20%280%2C%20-%5Csqrt3%29)
Part D)
Our first derivative is:
![\displaystyle \frac{dy}{dx}=\frac{2xy}{y-x^2}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cfrac%7Bdy%7D%7Bdx%7D%3D%5Cfrac%7B2xy%7D%7By-x%5E2%7D)
Let’s take the derivative of both sides again. Hence:
![\displaystyle \frac{d^2y}{dx^2}=\frac{d}{dx}\Big[\frac{2xy}{y-x^2}\Big]](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cfrac%7Bd%5E2y%7D%7Bdx%5E2%7D%3D%5Cfrac%7Bd%7D%7Bdx%7D%5CBig%5B%5Cfrac%7B2xy%7D%7By-x%5E2%7D%5CBig%5D)
Utilize the quotient and product rules and differentiate:
![\displaystyle \frac{d^2y}{dx^2}=\frac{(2y+2x\frac{dy}{dx})(y-x^2)-2xy(\frac{dy}{dx}-2x)}{(y-x^2)^2}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cfrac%7Bd%5E2y%7D%7Bdx%5E2%7D%3D%5Cfrac%7B%282y%2B2x%5Cfrac%7Bdy%7D%7Bdx%7D%29%28y-x%5E2%29-2xy%28%5Cfrac%7Bdy%7D%7Bdx%7D-2x%29%7D%7B%28y-x%5E2%29%5E2%7D)
Let dy/dx=y’. Therefore:
![\displaystyle \frac{d^2y}{dx^2}=\frac{(2y+2xy^\prime)(y-x^2)-2xy(y^\prime-2x)}{(y-x^2)^2}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cfrac%7Bd%5E2y%7D%7Bdx%5E2%7D%3D%5Cfrac%7B%282y%2B2xy%5E%5Cprime%29%28y-x%5E2%29-2xy%28y%5E%5Cprime-2x%29%7D%7B%28y-x%5E2%29%5E2%7D)
For (1, -1), we already know that y’ is 1 at (1, -1). Therefore:
![\displaystyle \frac{d^2y}{dx^2}_{(1, -1)}=\frac{(2(-1)+2(1)(1))((-1)-(1)^2)-2(1)(-1)((1)-2(1))}{((-1)-(1)^2)^2}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cfrac%7Bd%5E2y%7D%7Bdx%5E2%7D_%7B%281%2C%20-1%29%7D%3D%5Cfrac%7B%282%28-1%29%2B2%281%29%281%29%29%28%28-1%29-%281%29%5E2%29-2%281%29%28-1%29%28%281%29-2%281%29%29%7D%7B%28%28-1%29-%281%29%5E2%29%5E2%7D)
Evaluate:
![\displaystyle \frac{d^2y}{dx^2}_{(1, -1)}=-\frac{1}{2}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cfrac%7Bd%5E2y%7D%7Bdx%5E2%7D_%7B%281%2C%20-1%29%7D%3D-%5Cfrac%7B1%7D%7B2%7D)