how do you find the derivative (dy/dt)??

Question

find the derivative:

y= (1/9)Cot(3x-1)



help!

i mean dy/dx

the back of my book has a different answer :o

instead it says 2csc²(3x-1)cot(3x-1)???

Answer 1

From list of standard derivatives:-
y = cot x
dy / dx = - cosec ² x

Now y = (1/9) cot (3x - 1)
let u = 3x - 1
du/dx = 3
y = (1/9) cot u
dy/du = (- 1/9) cosec ² u

dy/dx = (dy/du) (du/dx)
dy/dx = (-1/9) cosec² u (3)
dy/du = (-1/3) cosec ² (3x - 1)

Answer 2

aren't you sick of people just giving answers and not telling you how to do it so you can learn?

First off, (1/9) is a constant, temporarily ignore it and just leave it there. Next you see a cotangent of (3x-1). Because this is a function (cot) of another function (3x-1), the derivative involves a process called Chain Rule.
First you take the derivative of the cot function (or more generally, whatever is the outside function). In this step leave the inside alone, but when you're done with this step, you multiply by the derivative of the inside function (here 3x-1). Lastly you put everything together with that constant, and simplify any fractions or integers.

Answer 3

It's not dy/dt. It is dy/dx.
The derivative of cot(x) is -csc^2(x)
the derivative of cot(3x-1) is -3csc^2(3x-1)
You have a (1/9) to multiply.
(1/9)(-3)csc^2(3x-1)
=(-1/3)csc^2(3x-1)

Answer 4

y= (1/9)cot(3x-1)
=> dy/dx = - (1/9) * 3 * csc^2 (3x - 1)

dy/dt
= (dy/dx) * (dx/dt)
= - (1/3) * csc^2 (3x - 1) * (dx/dt)

You can evaluate dx/dt if you know x as a function of t and substitute in the above reult to get the final answer.

Answer 5

madhukar is correct, u need the value of dx/dt which is probably given to u in the problem, so you need to include that to get the answer in the back of the book

Answer 6

dy/dx=-3(1/9)cosec**2(3x-1)
=-(1/3)cosec**2(3x-1)

Answer 7

dy/dx = (1/9) *3*-csc^2x = -(1/3)csc^2 x

Answer 8

Big Hint: (d/dx) cot(u) = -csc²(u) * du/dx ☺

Doug

Answer 9

y' = -1/(3 sin^2(3x-1))

Answer 10

y'=(3/9)Tg^(-1)(3x-1)