Question

Find the derivative of y=cotx/(1+cotx)

Answer 1

\[y= \frac{cotx}{1+cotx}\]

Answer 2

Sure :D

Answer 3

Im just trying to show how far I gotten with the problem

Answer 4

\[\frac{(1+cotx)(-\csc^2x)-(cotx)(-\csc^2x)}{(1+\cot^2)}\]

Answer 5

I see one error, on the denominator, the squared should be in the outside of the parenthesis. (1 + cotx)^2

Answer 6

oh yea you're right my mistake

Answer 7

Otherwise, you did the quotient rule right and is just a matter of simplifying!

Answer 8

\[\frac{(1+cotx)(-\csc^2x)-(cotx)(-\csc^2x)}{(1+\cot)^2}\]

Answer 9

\(\color{black}{ \displaystyle y=\frac{\cot x}{1+\cot x}=\frac{\sin x}{\sin x+\cos x}}\).
(I would differentiate then.)

Answer 10

I'm just i guess intimidated on how to simplify it honestly. I don't know what cot*csc^2 is

Answer 11

Did you just use the ratios to simply the given equation?

Answer 12

\[\frac{(1+cotx)(-\csc^2x)-(cotx)(-\csc^2x)}{(1+\cot)^2}\]
I'm still stick on figuring out how to simplify this :( And I don't know how solomon got to the alternate form of the equation.

Answer 13

\[\large\rm \frac{(1+cot x)\color{orangered}{(-\csc^2x)}-(cot x)\color{orangered}{(-\csc^2x)}}{(1+\cot x)^2}\]We could try factoring this orange thing out of each term, ya?

Answer 14

Multiplying everything together seems like a messy alternative :P

Answer 15

:O

Answer 16

I didn't even see that omg thank you!

Answer 17

I'll continue the problem on paper then type the work up here.

Answer 18

k

Answer 19

Did I factor it out right?
\[\frac{(-\csc^2x)(1+cotx)-(cotx)}{(1+\cot^2)}\]

Answer 20

The -csc^2x should be multiplying `everything` in the numerator,
so you might want another set of brackets to make that clear,\[\large\rm \frac{(-\csc^2x)\left[(1+cotx)-(cotx)\right]}{(1+\cot^2)}\]

Answer 21

Ah ok

Answer 22

Oh woops we've got that denominator messed up again :p\[\large\rm \frac{(-\csc^2x)\left[(1+cotx)-(cotx)\right]}{(1+\cot x)^2}\]

Answer 23

Haha yea I just copy and pasted it my bad.

Answer 24

So would the cotx cancel out in the numerator?

Answer 25

\[\large\rm \frac{(-\csc^2x)\left[1\right]}{(1+\cot x)^2}\]Yes.

Answer 26

Alright :D So the denominator would equal 1 due to the identity right?

Answer 27

Sorry internet is being kind of slow right now...

Answer 28

Equals 1? Hmm I'm not sure what you mean..

Answer 29

Oh no i read it wrong...

Answer 30

Remember that our Pythagorean Identity looks like this,\[\large\rm 1+\cot^2x=\csc^2x\]Not like this,\[\large\rm (1+\cot x)^2\ne \csc^2x\]

Answer 31

If that's maybe what you were referring to.

Answer 32

From this point,
I actually don't see a nice path using any cosecant or cotangent identity.
I think it's time to convert to sines and cosines and continue on from there.

Answer 33

Ok do you usually convert to sin and cos at the end?

Answer 34

Honestly, it just depends what your teacher expects.

This is a fine and dandy answer:\[\large\rm \frac{-\csc^2x(1+\cot x)-\csc^2x(\cot x)}{(1+\cot x)^2}\]
This answer is also good:\[\large\rm \frac{-\csc^2x}{(1+\cot x)^2}\]Or in terms of sines and cosines will also give a nice looking answer.

Just depends.

Answer 35

Based on problems you've already done,
you should have a good idea of what your teacher expects for a good looking answer.

Hopefully :d hehe

Answer 36

Ok well in the book it has the answer converted to sin and cos actually.

Answer 37

Ok ok fair enough :d so we'll keep going then.

Answer 38

So for converting im having trouble visualizing it. This is what I have right now
\[\frac{ (1)(\sin) }{ (sinx^2+\cos +1) }\]

Answer 39

Forgot the negative sign

Answer 40

and forgot to square the denominator...

Answer 41

Hmm I was thinking of it like this,\[\large\rm \frac{\color{orangered}{-\csc^2x}}{(1+\color{royalblue}{\cot x})^2}\quad=\quad \frac{\color{orangered}{-\frac{1}{\sin^2x}}}{\left(1+\color{royalblue}{\frac{\cos x}{\sin x}}\right)^2}\]

Answer 42

Oh ok. Still the fraction ception is kinda messing me up still.

Answer 43

This might seem like a weird step,
you can lemme know if it's confusing...

In order to get rid of these fractions upon fractions,
we need to multiply the top and bottom by sin^2x.

Answer 44

Oh ok to cancel it out i see.

Answer 45

\[\large\rm \frac{\left(-\frac{1}{\sin^2x}\right)}{\left(1+\frac{\cos x}{\sin x}\right)^2}\left(\frac{\sin^2x}{\sin^2x}\right)\]

Answer 46

So your left with -1 in the numerator. I'm not sure about the denominator since its sinx*sin^2x

Answer 47

Ok bottom is a little weird.\[\large\rm \sin^2x\left(1+\frac{\cos x}{\sin x}\right)^2\]We have to drop the square on our sine in order to bring it into this other square.

Answer 48

Example of this type of idea:\[\large\rm 2^2(3)^2\quad=\quad (2\cdot3)^2\]I had to drop the square on that first 2 in order to bring it into the other group.
Hope that's not too confusing.

Answer 49

Oh ok.

Answer 50

\[\large\rm \sin^2x\left[\left(1+\frac{\cos x}{\sin x}\right)\right]^2\quad=\quad \left[\sin x\left(1+\frac{\cos x}{\sin x}\right)\right]^2\]

Answer 51

I see. So the denominator would be 1+cosx right?

Answer 52

Oh no (1+cos)^2

Answer 53

Woops, you didn't multiply the 1 by sinx.

Answer 54

Make sure you distribute the sinx to each term :)

Answer 55

Oh i see my bad. It would be (sinx + cosx)^2

Answer 56

Mmm good good, that looks better.

Answer 57

Oh I guess we could simplify further :\
Ughhh

Answer 58

Actually no. The answer we got so far is
\[\frac{ 1 }{ (sinx+cosx)^2 }\]

Answer 59

Right*?

Answer 60

I'm really sorry i meant it like that's the answer in the book.

Answer 61

\[\large\rm \frac{-1}{(sinx+cosx)^2}\]with a negative up top?
Yes :) looks good.

If that's where your book stopped, then that's fine.

I was just going to point out that we COULD distribute the square in the bottom,\[\large\rm (\sin x+\cos x)^2\quad=\quad \sin^2x+2\sin x \cos x+\cos^2x\]And then apply your Pythagorean Identity with the sin^2x+cos^2x,\[\large\rm =1+2\sin x \cos x\]

Answer 62

Oh i see lol. I didn't think about it like that actually

Answer 63

Boy that was a doozy :P
But we did it! Yayyyy! \c:/ Zeppy and Zappy!

Answer 64

Omg thank you so much for helping me through this problem. I literally spent like prolly 3 hours now on that problem.. I honestly really appreciate you for stick with me. Thank you.

Answer 65

np

Answer 66

If that 'multiplying top and bottom by sin^2x` was a confusing step,
you had another option.

You could instead look for a common denominator in your denominator,
in the square,
and the apply your weird "keep change flip" rule or however you learned it,
flipping the bottom fraction, rewriting it as multiplication.

And the sin^2x's would have cancelled out that way.

But multiplying through by LCM of denominators is a really useful tool to have in your goody bag, so I wasn't to make sure we at least touched on it.

Answer 67

Ah ok. To be honest I liked the way you first showed it. I was just having a super hard time visualizing the conversion on paper.

Answer 68

I think its a good time to take a break from math for a bit. My brain is kinda having a hard time functioning right now lol. But, I can't thank you enough for helping me with this problem. You truly are a kind person :D