Answer:
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: