Three derivatives: chain rule, implicit differentiation, and related rates cone problem

[sjaguar13]

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 derivative. 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?

 

[Nenad]

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:

$$\frac{dv}{dt} = \frac{dV}{dH} \cdot \frac{dH}{dt}$$

You know $$\frac{dV}{dt} = \frac{2.0cm^3}{s}$$

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

 

[thepatientmental]

It's completely understandable to struggle with derivative questions, as they can be quite tricky and require a lot of practice. Here are some tips and solutions for common problems that may arise:

1. When differentiating a function with nested trigonometric functions, such as in the first question, it's important to carefully apply the chain rule. In this case, you correctly used the chain rule for the outermost function, cot^2, but then made a mistake when applying it to the inner function, csc^2. Remember that the derivative of csc^2 is -2csc^2cotcsc, so your final answer should be -2cot(csc^2(3x-1))csc^2(csc^2(3x-1))3.

2. In the second question, it seems like you may have made a mistake when differentiating the function y/(2y-x). The derivative of this function is (2y-x)y'-(2y-y')/(2y-x)^2. To find the second derivative, you can use the quotient rule again, or you can simplify the expression to get y"(2y-x)^2-2y'(2y-y')/(2y-x)^3. From there, you can solve for y" by moving all terms to one side and factoring out y". Remember to check your algebra carefully when solving for y".

3. For the third question, it's important to understand the relationship between the volume of a cone and its height. The formula for the volume of a cone is V=1/3πr^2h, where r is the radius and h is the height. The rate of change of the volume with respect to time is given by the formula dV/dt=1/3πr^2dh/dt. In this case, the radius is constant at 5cm, and the height is changing at a rate of -2cm per second (since the water level is falling). Plugging in these values, we get dV/dt=1/3π(5)^2(-2)=-10π. This means that the volume is decreasing at a rate of 10πcm^3 per second. To find the rate of change of the height, we can use the formula dV/dt=1/3πr^2dh/dt and solve for dh/dt. This will give us