I need help in Applications of differentiation

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Well we are doing this free response paper and I was able to finish the first one correctly, I think, but I need help on the 2nd problem:

Consider the curve given by the equation y^3 + 3x^2(y) + 13=0

(a) Find dy/dx.

(b) Write an equation for the line tangent ot the curve at the point (2, -1).

(c) Find the minimum y-coordinate of any point on the curve. Justify your answer.

I worked on finding the derivative of the function but I got a bit confused because it said dy/dx. This means the same thing as y' right?
If so what I did was that I first used the quotient rule on y^3 and turned it into 3y^2
Then I used the product rule on 3x^2(y) and got 3x^2 +6xy
Did I do this right?

Please Respond and Thank you in advance

With Respect,
Carlos
 
You can use implicit differentiation. Notice the x and y in the function?.

After you differentiate, sub x=2 and y=-1 into dy/dx to find the slope at that point. You can then use y=mx+b to find your equation of the tangent line.

Give it a shot and write back with your results. We will help
 
See that is kind of where I stumble. I don't think I am finished with my derivative and I don't know whether to plug the points for x and y one at a time or both at the same time. My derivative thus far is:

3y^2 + 3x^2 + 6xy=0

Do I plugg in the numbers now or do I move my y's to one side and only plugg in one?

I am not sure and I would surley appreciate some help

Thank You
Carlos
 
Where is your "dy/dx" in your implicit derivation? For instance, the derivative of "y<sup>3</sup>" would be "3y<sup>2</sup>(dy/dx)", not just "3y<sup>2</sup>".

Thank you.

Eliz.
 
Oh yeah! Thanks but that only applys to the y right because its with respect to x?

If so, my work is as follows:
3y^2 dy/dx + 3x^2 + 6xy dy/dx=0

3y^2 dy/dx + 6xy dy/dx= -3x^2

3y^2 dy/dx + dy/dx= -x/3y

2dy/dx= -x/9y^3

dy/dx= -x/18y^3

I don't know if my work is right but then I am supposed to plugg in the given
(2, -1) and I got 1/9 as the slope. Is this correct?

Thank You
Carlos.
 
You forgot something........the product rule.

\(\displaystyle \L\\y^{3}+\underbrace{3x^{2}y}_{\searrow}+13=0\)
\(\displaystyle \L\\3y^{2}\frac{dy}{dx}+\underbrace{3x^{2}\frac{dy}{dx}+6xy}_{\text{product rule}}=0\)

\(\displaystyle \L\\\frac{dy}{dx}=\frac{-6xy}{3y^{2}+3x^{2}}\)

\(\displaystyle \L\\\frac{-3(2xy)}{3(x^{2}+y^{2})}\)

\(\displaystyle \H\\\frac{dy}{dx}=\frac{-2xy}{x^{2}+y^{2}}\)

Now, do that slope thing again
 
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