Question

IS the derivative of y=x^2*Cot(x)-1/x^2 is equal to
2x*cot(x)-x^2*csc(x)-2/x^3?

Answer 1

i got this answer by my self, and my ti-89 got \[\frac{ 2x^4\sin(X)Cos(x)+2(\sin(x))^2-x^5 }{x^3(\sin(x))^2 }\]

Answer 2

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

correct?

Answer 3

yes sir!

Answer 4

Alrighty, cool. So let's see, I get:

2xcot(x) - x^2csc^2(x)+ 2/(x^3)

Should be something like that O.o

Answer 5

doing it by my self i got \[2xcot(x)-x^2cscx-\frac{ 2 }{ x^3 }\]

Answer 6

So same thing I get but wrong sign on the last part xD

Answer 7

yeah i got the same thing, but my ti is throwing that answer to me

Answer 8

why you got it possitve?

Answer 9

ooo it was already negative so negative times negative equal positive

Answer 10

Well, the ti LOVES to turn things into sin and cos, so that makes it confusing. And if I had to go through it the long way:

\[-\frac{ 1 }{ x^{2} }=-x^{-2} \implies -(-2)x^{-3} = \frac{ 2 }{ x^{3} }\]

Answer 11

cool thanks!!

Answer 12

Yeah, sure :3

Answer 13

so the answer the TI gave me is correct just in the long way?

Answer 14

Likely. In fact, it looks like it turned cot into cos/sin and did quotient rule, then combined everything into one common denominator, lol. It just made a mess of things.

Answer 15

damn

Answer 16

Yeah, thatll be the bad thing about checking derivatives with it x_x

Answer 17

hey what is the derivative of (sin(x))*(Tan(x)). cosx*sec^2x?

Answer 18

sin times tan?

cos(x)tan(x) + sin(x)sec^2(x)

Answer 19

derivative of (sin(x))*(Tan(x)). My answer is cosx*sec^2x

Answer 20

yeah sin times tan

Answer 21

Yeah, should be right what I put.

Answer 22

yup your right!

Answer 23

I sped through it, so gotta double check xD

Answer 24

i check it and i say is correct.

Answer 25

and like always my fluttering TI giving other pellet

Answer 26

can always use: http://www.wolframalpha.com/ to check derivative answers too

Answer 27

are you there psymon?

Answer 28

Yep.

Answer 29

i have a new one, \[G(x)= Csc(x)Cot(x)\]

Answer 30

Too bad thats not integral of, then its easy, lol.

Answer 31

and my answer is \[\cot(x)(-\csc(x)\cot(x))-\csc(x)(\csc^2x)\]

Answer 32

should i do something? like add the cot(x)'s and the csc(x)'s?

Answer 33

-cot(x)csc(x)cot(x) - csc(x)csc^2(x)
So yeah, looks the same, just combine what you can from there.

Answer 34

Yei, this isn't hard at all my TI and my professor makes it complicated. damn

Answer 35

Lol, yeah, if you are used to doing them, theyre really simple. But just like everything else when you get to calculus level, its more tedious and possible to mess up than it is hard.

Answer 36

i got other one \[y= \frac{ \cot x }{1+ \cot x }\]. let me get the answer

Answer 37

This is one id so cheat on, haha.

Answer 38

the answer is \[-\frac{ Csc^2 x }{ 1-\csc^2x }\]

Answer 39

Thats your answer, right?

Answer 40

i think is 1

Answer 41

yeah i got that

Answer 42

Possible our answers are the same, but I do that kind of derivative in a sneaky way.

Answer 43

how you do it then?

Answer 44

Well, doing it the normal way, I get:
\[-\frac{ \csc^{2}x }{ (1+cotx)^{2} }\]

Answer 45

explain me why numerator got possitive and to the second power, pleaasseee:D

Answer 46

Yeah. Well, if you do quotient rule, the denomintor is supposed to be squared normally. I assume you know the quotient rule, so Ill just write out what you have after the derivative of everything.
\[\frac{ -\csc^{2}x(1+cotx) + \cot(x)\csc^{2}x }{ (1+cotx)^{2} }\]

Answer 47

oooo i got it

Answer 48

Yep. But I can show you how I do it a tricky way if you want.

Answer 49

yeah that would be could! please

Answer 50

\[\frac{ cotx }{ 1+cotx }\]
Add 1 and subtract 1 at the same time in the numerator. This is a trick where you technically add 0, but it allows you to make more than one fraction. So I do this:

\[\frac{ cotx + 1 - 1 }{ 1+cotx }\]
Now you are allowed to make a new fraction for every term in the numerator of a fraction or any combination of terms in the numerator. So that means I can rewrite the above like this:

\[\frac{ cotx + 1 }{ 1+cotx }- \frac{ 1 }{ 1+cotx }\implies 1 - \frac{ 1 }{ 1+cotx }\]
So now if I take the derivative, the 1 goes away completely and the other fraction becomes
\[-(1+cotx)^{-1} \implies (1+cotx)^{-2}*-\csc^{2}x \implies \frac{ -\csc^{2}x }{ (1+cotx)^{2} }\]

Answer 51

heyyyy! good trick ! let me try it in other

Answer 52

Lol, alright then xD

Answer 53

this time is \[y=\frac{ \cos x}{ 1+\sin x }\]

Answer 54

damn i got stuck in
\[\frac{ \cos x +1 }{ 1 +\sin x } - \frac{1 }{ 1 +\sin x }\]

Answer 55

Yeah, that would work. Its just having cosx+1/1+sinx doesnt do you much good, lol. Probably not a situation where you want to do + 1 - 1

Answer 56

should i use quotient rule?

Answer 57

Quotient or product rule, you can do either one. But yeah, no real super trick for this one xD

Answer 58

hey i got \[-\frac{ \cos x \cos x }{(1+\sin x) ^2 }\]. is it correct?

Answer 59

I dont get that, no.

Answer 60

damn

Answer 61

what about\[\frac{ (- \cos x)^2+\sin (x)(\sin (x) +1) }{ (1+\sin (x))^2 }\]

Answer 62

I did product rule, so let me check that method.

Answer 63

Looks like youre missing a negative. Gah, hating doing it the quotient rule way xD

\[\frac{ -sinx(1+sinx) - \cos(x)\cos(x) }{ (1+sinx)^{2} }\]
\[\frac{ -sinx-\sin^{2}x-\cos^{2}x }{ (1+sinx)^{2} }\]
\[\frac{ -sinx-(\sin^{2}x+\cos^{2}x) }{ (1+sinx)^{2} }\]
\[\frac{ -sinx-1 }{ (1+sinx)^{2} }=\frac{ -(sinx+1) }{ (sinx+1)^{2} }\implies \frac{ -1 }{ (sinx+1) }\]

Answer 64

where the sin^2 x come frome, in the second fraction

Answer 65

I multiplied the -sinx through. -sinx(1+sinx) = -sinx-sin^2(x)

Answer 66

got it!

Answer 67

Lol, alright, cool xD

Answer 68

\[\frac{−sinx−(\sin2x+\cos2x) }{ (1+sinx)2 }\] so this suppose to be the answer?

Answer 69

or \[\frac{ −1−(\sin^2x+\cos^2x) }{ (1+sinx)2 }\]

Answer 70

You can keep going like I did. Can you follow what I did?

Answer 71

Yes i do,

Answer 72

Okay, cool. Yeah, so the answer turns out to be something pretty simplified it seems.

Answer 73

what an idiot of me, i wasnt doing the sum of sin^2x+ cos^x

Answer 74

Yeah, simplifies to 1, lol.

Answer 75

what about \[ Y= X^2 \sin x + 2x \cos x - 2 \sin x\]

Answer 76

i used the sum and difference rule, which it gave me the result of \[2x \cos x - 2 \sin x- cosx\]

Answer 77

ups\[2xcosx−2sinx+cosx\]

Answer 78

Thats a new problem, right?

Answer 79

the new problem is\[Y=X^2sinx+2xcosx−2sinx\]

Answer 80

2xsinx + x^2cosx + 2cosx -2xsinx - 2cosx
Everything cancels to x^2cosx.

Answer 81

i dont get it

Answer 82

i know the answer is x^2cosx but where did x^2cosx come from?

Answer 83

oooooooo i get it, you differentiate each value by its own?

Answer 84

Yeah. Kinda have to. I mean, there may be things you can do, but I go term by term. First term is product rule, 2nd term is another product rule, last one is straightforward.

Answer 85

straightfoward?

Answer 86

As in nothing tricky. It just is a simple derivative.

Answer 87

what about this one \[Y=x^2Cosx-2xSinx-2cosx\]

Answer 88

and my answer is\[-x^2sinx\]

Answer 89

Yeah, it even looks the same:
2xcosx - x^2sinx - 2sinx -2xcosx + 2sinx
simplifies to -x^2sinx

Answer 90

looks the same as mine

Answer 91

Yeah, guess so xD I get the same answer.

Answer 92

i need to remember this rules all of them:(

Answer 93

Yeah, remembering product rule and quotient rule and them being able to do them correctly. Takes practice x_x

Answer 94

i have my test in three weeks and i dont know very well Limit theme, some of them are easy just by substituting, the others are hard as hell

Answer 95

Well, these arent limits, lol. If youre worried about those why not bring some of those up?

Answer 96

because right know im doing some homework about trigonometric Function differentiation

Answer 97

i want to bring some up, but homework first then i study limits

Answer 98

Ah. Fair enough, lol.