Consider the curve given by the equation y^2-2x^2y=3. a) Find dy/dx .
b) Write an equation for the line tangent to the curve at the point (1, –1).
c) Find the coordinates of all points on the curve at which the line tangent to the curve at that point is horizontal.
d) Evaluate d^2y/dx^2 at the point (1, –1)

Part A)

Part B)

Part C)

Part D)

Step-by-step explanation:

We have the equation:

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:

Differentiate. We will need to implicitly different on the left. The second term will also require the product rule. Therefore:

Rearranging gives:

Therefore:

And, finally, simplifying:

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:

Now, we can use the point-slope form:

Substitute:

Simplify:

So:

Part C)

If the line tangent to the curve is horizontal, this means that dy/dx=0. Hence:

Multiplying both sides by the denominator gives:

Assuming y is not 0, we can divide both sides by y. Hence:

Then it follows that:

Going back to our original equation, we have:

Substituting 0 for x yields:

So:

Therefore, two points where the derivative equals 0 is:

However, we still have to test for y. Let’s go back. We have:

Assuming x is not 0, we can divide both sides by x. So:

Therefore:

And going back to our original equation and substituting 0 for y yields:

Since this is not true, we can disregard this case.

So, our only points where the derivative equals 0 is at:

Part D)

Our first derivative is:

Let’s take the derivative of both sides again. Hence:

Utilize the quotient and product rules and differentiate:

Let dy/dx=y’. Therefore:

For (1, -1), we already know that y’ is 1 at (1, -1). Therefore:

Evaluate:

sup

step-by-step explanation:

answer:

c. the third graph