Question

differentiate xylog(x+y)=1

Answer 1

Is this log base 10 or base e?

Answer 2

it it N.C.E.R.T. Question ?

Answer 3

no

Answer 4

@Psymon its not mentioned

Answer 5

I see. Well, we can try it with base e or base 10, which do you think is more likely?

Answer 6

base 10

Answer 7

Alright then. So log base 10 derivative is:
\[\frac{ 1 }{ (\ln10)u }*\frac{ du }{ dx}\]
So this is going to be a product rule of 3 terms
\[f'(x)g(x)h(x) + f(x)g'(x)h(x) + f(x)g(x)h'(x)\]
Does it say if we have to differentiate with respect to x or y by chance?

Answer 8

w.r.t x

Answer 9

Alright, just making sure :3 Although it looks like genius12 has got theproblem cooking up, lol. Let's see what hecomes up with xD

Answer 10

actually this is a proof
dy/dx=-y(x^2y+x+y)/x(xy^2+x+y)

Answer 11

Ah. Either way, I'm sure genius has got it xD No need for me to interrupt his work.

Answer 12

lol:D

Answer 13

no one got the answer???

Answer 14

o.o no-one posted an answer O.o Um, I could try....as long as Im not interrupting someone else.

Answer 15

pls try

Answer 16

how will we get dy if we differentiate w.r.t x?

Answer 17

Alrighty. So the first portion of the product rule would be this then:
\[(1)(y)\log(x+y)\]
Next we would have:
\[(x)(1)\frac{ dy }{ dx }\log(x+y)\] and the 3rd term:
\[(x)(y)\frac{ 1 }{ (\ln10)(x+y) }*(1+\frac{ dy }{ dx })\]

So now we need to separate our dy/dx terms from the other terms.

Answer 18

Since we have a dy/dx inside of a parenthesis in that 3rd term, I have to multiply that through:
\[\frac{ xy }{ (\ln10)(x+y) }+\frac{ xy }{ (\ln10)(x+y) }*\frac{ dy }{ dx }\]

Answer 19

We have 2 terms with dy/dx and 2 terms without it, so I'll separate those on opposite sides of the equal sign
\[\frac{ dy }{ dx }( xlog(x+y)+\frac{ xy }{ (\ln10)(x+y) }) = -(ylog(x+y) + \frac{ xy }{ (\ln10)(x+y) }\]
Now I need to divide both sides by what is with dy/dx

Answer 20

\[-(ylog(x+y) + \frac{ xy }{ (\ln10)(x+y) })*(\frac{ 1 }{ (xlog(x+y)) }+\frac{ (\ln10)(x+y) }{ xy })\]
Now let me simplify this on paper, haha.

Answer 21

Yeah, looking at your proof now, this must be log base e O.o

Answer 22

actually this is a proof
dy/dx=-y(x^2y+x+y)/x(xy^2+x+y)

Answer 23

Yeah, I noticed it now, which makes me think this would be log base e and not log base 10.

Answer 24

ssry my mistake

Answer 25

u were tying smthing wht happened

Answer 26

I guess I'm wondering where all the ln's went. I'm not sure how to get them to all disappear in your proof like that. Maybe I'm not sure how to do it then. I mean, I solve for (dy/dx), but it's nothing like that

\[\frac{ dy }{ dx } =\frac{ -y[(x+y)\ln(x+y)+x] }{ x[(x+y)\ln(x+y)+y] }\]


I must flat out be missing something, lol x_x

Answer 27

i know even i dont know where log vanished
i tried this sum 5 times still i didn't get

Answer 28

It won't disappear for log or ln. My calculator gave me a middle finger when I tried with that. Wish i could say I knew where that came from O.o

Answer 29

wht i should do now

Answer 30

this question is there in RD sharma but unsolved

Answer 31

did anyone try substituting x+y= u ?

Answer 32

Would that actually get rid of the log when you finish, though? I would think some of the terms would still have the log even if you did that substitution.

Answer 33

yep

Answer 34

i tried and i got rid of log

Answer 35

as of now i am here, simplification is remaining
1+ dy/dx = (x+y) d/dx (1/(xy))

Answer 36

I see. So would you have to make y = u-x as well then?

Answer 37

what i did is
x+y= u (just for convenience, we can do it without doing this too)
1+ dy/dx = du

yep, y=u-x

x(u-x) log u = 1

u= e^ (1/ [x u-x])
now differentiate both sides w.r.t x

Answer 38

***u= e^ (1/ [x (u-x)])

Answer 39

shalini, are you following so far ?

Answer 40

not exactly

Answer 41

ok, which step you have doubt in ?

Answer 42

first u complete

Answer 43

i understood

Answer 44

eh, complete ?
i want you to learn, by interaction....giving you entire solution at once would not help much....try to get each and every step

Answer 45

and i gave you a headd start. i would be glad if you can think of next steps...

Answer 46

i understood till where u hv done

Answer 47

so, can you differentiate w.r.t x that for me ??
u= e^ (1/ [x (u-x)])

du/dx =.. ?

Answer 48

sir,i m very bad in differentiation

Answer 49

i will try

Answer 50

what id d/dx of e^x ?

Answer 51

its e^x

Answer 52

good, now lets see whether you know chain rule,
what is d/dx of e^x^2 ?

Answer 53

2xe^x^2

Answer 54

oh wow, you know chain rule :)
so you can surely and easily differentiate this.

d/dx (e^anything) = e^anything d/dx (anything)
u= e^ (1/ [x (u-x)]) du/dx =.. ?

Answer 55

just one next step, not entire thing..

Answer 56

w8

Answer 57

Is this supposed to come out as complicated as I'm getting it? I mean, I've done the derivative and isolated du/dx, but it's looking a bit insane O.o

Answer 58

thats why i asked to give me just one next step....you need not derivate the remaining part or isolate "du/dx"

Answer 59

Well, I didn't want to post steps for him, I was just trying to do it on my own, sorry.

Answer 60

i think
du/dx= e^1/x(u-x)*-[u-2x]/(x(u-x))^2

Answer 61

don't be sorry, we are learning....you can message me the next step, and i will tell you what u can do further

Answer 62

ehh....see this became complicated...but ok,
what i was expecting is this,

du/dx = e^ (1/ [x (u-x)]) d/dx [e^ (1/ [x (u-x)])]

but we have e^ (1/ [x (u-x)]) = u !!!
so,
du/dx = u d/dx (e^ (1/ [x (u-x)]))
got this ?

Answer 63

***du/dx =u d/dx(1/ [x(u-x)])

Answer 64

now we don't need u, we can bring back our 'y'

1+dy/dx = (x+y) d/dx (1/xy)

now this isn't very complicated

Answer 65

this was important step i guess, so any of you have any doubts, please get it cleared

Answer 66

Still staring at it, haha.

Answer 67

no continue

Answer 68

Alright, I guess I see what ya did.

Answer 69

let me write all steps together, i made some typos earlier

u= e^ (1/ [x (u-x)])

du/dx = e^ (1/ [x (u-x)]) d/dx [(1/ [x (u-x)])]

but u= e^ (1/ [x (u-x)])
so, du/dx = u d/dx[1/x(u-x)]
bringing back 'y' , {u=x+y so, du/dx = 1+dy/dx and u-x=y}

1+dy/dx = (x+y) d/dx (1/xy)

Answer 70

shalini, i guess you know both chain and product rule, so can you try this for me, separately ??

d/dx (1/xy)

Answer 71

No need to differentiate the (x+y)?

Answer 72

nopes...because of chain rule we had to diff. 1/(xy)

Answer 73

So the (x+y) will just be there to multiply with the result of d/dx(1/xy).

Answer 74

correct

Answer 75

-xdy/dx+y/x^2y^2

Answer 76

did you mean
\(\huge -\dfrac{xy'+y}{x^2y^2}\)
??
where y' = dy/dx , just for convenience

Answer 77

yes
y its not correct

Answer 78

so now you have
\(\huge 1+y'=\huge -(x+y)(\dfrac{xy'+y}{x^2y^2})\)
try to isolate y', this work is purely algebraic

Answer 79

oh and i forgot to ask, any doubts till that step ?

Answer 80

Alright, so since we got to that point I'll now ask my question, lol. So I would have:
\[-(\frac{ 1 }{ x ^{2}y } + \frac{ xy' }{ (xy)^{2} })\]
Now because the x is being multiplied by y', am I not allowed to cancel it out with one of the x's in the denominator or can I?

Answer 81

you can, surely,
xy' / x^2y^2 = y'/ xy^2

Answer 82

Alright, making sure :3

Answer 83

no doubt

Answer 84

good, so i think distributing x+y over that big bracket, like psymon did , will be a good step to start simplifying...

Answer 85

remember our aim is to isolate y'

Answer 86

then the next good step, to get rid of the denominator, will be to multiply both sides by x^2 y^2

Answer 87

i didn't get Psymon's step

Answer 88

then try it by yourself ?

Answer 89

arre why r u getting angry

Answer 90

angry ? lol, not at all...if i don't put smileys doesn't mean i am getting angry
i am just encouraging you to first try it by yourself, if you don't get it, we are here to help you :)

Answer 91

That turned out pretty awesome x_x And my step was instinctive to just try and start to isolate y'. After the quotient rule, I split the result into two separate fractions. That was all i did : )

Answer 92

from where psymon got
−(1x2y+xy′(xy)2)

Answer 93

how will you simplify (x+y)*z ?

Answer 94

xz+yz

Answer 95

I'll show ya what I did.
\[-(\frac{ xy'+y }{ x ^{2}y ^{2} })\]
So this was the result of our quotient rule. For every term in the numerator, I can make a new fraction, using that term over the whole denominator. So I split it into:
\[-(\frac{ xy' }{ x ^{2}y ^{2} }+\frac{ y }{ x ^{2}y ^{2} })\]
Even thoughthis is what I did, don't let it confuse you I just insticitvely broke it apart, it is probably more efficient to do what hartnn said and just multiply out x^2y^2 from the beginning xD

Answer 96

k

Answer 97

so, (x+y) (xy′+y/x^2y^2) will be just
=x (xy′+y/(x^2y^2)) + y (xy′+y/(x^2y^2))
right ?
yep,all have different ways to simply things, i will surely multiply x^2y^2 right now to get rid of denominator

Answer 98

I agree, just first was my first thought, doesn't mean it was most efficient :P