Derivitive Questions?

  • Thread starter sjaguar13
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There are three questions on my test which I don't know what I did wrong.

The first one is:
Differentiate the following f(x)=Cot^2(csc^2(3x-1))
I wrote down some + which are crossed out. Should it be:
2cot(csc^2(3x-1))(-csc^2(csc^2(3x-1)))(2)(csc(3x-1))(-csc(3x-1)cot(3x-1))(3) ?

The second is to find the the second derivitive. The first is y/(2y-x). If y' equals that and you differentiate it to get (2y-x)y"-y(2-x)/(2y-x)^2, what does the y' become? If it's zero, you can move everything but the y" over and you end up with y"=0. If it becomes y" also, how to get both y" on the same side? I just seem to move the junk over and end up with the same problem, a y" on each side.

The third question I got completely wrong. A cone shaped coffee filter of radius 5cm and height 10cm contains water which drips out a hole at the bottom at a constant rate of 2cm^3 per second. How fast is the water level falling when the height is 8cm? (v=1/3(pi)r^2h) I got 1 out of 10 points on it. I am confused as to what the 2cm^3 is. I know I need to figure out the fomula, substitute the rate I know along with the known variables and solve for the rate of the height changing. I know the height rate should be negative, but I can't figure out what the formual is supposed to be. When the height is 8, the radius is 4, but that didn't get me anywhere. I tried using V-rate of the water dripping = something. I tried solving the volume equation for h, but I don't know. How is it supposed to be done?
 

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sjaguar13 said:
The third question I got completely wrong. A cone shaped coffee filter of radius 5cm and height 10cm contains water which drips out a hole at the bottom at a constant rate of 2cm^3 per second. How fast is the water level falling when the height is 8cm? (v=1/3(pi)r^2h) I got 1 out of 10 points on it. I am confused as to what the 2cm^3 is. I know I need to figure out the fomula, substitute the rate I know along with the known variables and solve for the rate of the height changing. I know the height rate should be negative, but I can't figure out what the formual is supposed to be. When the height is 8, the radius is 4, but that didn't get me anywhere. I tried using V-rate of the water dripping = something. I tried solving the volume equation for h, but I don't know. How is it supposed to be done?
Set up a diagram here. You need to use related rates. Here is a hint:

[tex] \frac{dv}{dt} = \frac{dV}{dH} \cdot \frac{dH}{dt} [/tex]

You know [tex] \frac{dV}{dt} = \frac{2.0cm^3}{s} [/tex]

As for the first two questions, I have not had time to look at them. I'm sure somebody will come up with the solution.

Regards,

Nenad
 
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