Question

find dy/dx of (X^y.Y^x=1)

Answer 1

\[x^y y^x = 1\]

Answer 2

yes how to solve it.. please tell...

Answer 3

take log on both sides
\[\log(x^y) + \log(y^x) = 0\]
\[xlog(y) = -ylog(x)\]

Answer 4

but we have to find dy/dx NOT Log...

Answer 5

lol wait

Answer 6

You know implicit differentiation ?

Answer 7

No i dont know...

Answer 8

Differentiate this last equation w.r.t x
\[\frac{dx}{dx}\log(y) + \frac{x dy}{y dx} = -\frac{dy}{dx} - \frac{y}{x}\]

Answer 9

Solve for dy/dx from this equation !

Answer 10

@Ghayyas.mubashir

Answer 11

i cant..
tell me some eazy method..

Answer 12

Here y can't be totally expressed in terms of x... so implicit differentiation is the only way :)

Answer 13

i Actually dont know implicit..

Answer 14

@Meepi @amistre64

Answer 15

A quick and handy way to do this sort of differentiation... (although, the method described above is the basis for it)...
define a function
f(x,y)=((x^y)(y^x))-1
if you know partial derivatives then \[dy/dx=-(\delta(f)/\delta(x))/(\delta(f)/\delta(y))\]
where \[\delta\] represents partial derivative of function with respect to the variable.

Answer 16

@Ghayyas.mubashir - the method described by @vikrantg4 is the easiest and most basic of all the methods with which the problem can be solved

Answer 17

i dont understand his method

Answer 18

You know chain rule and product rules ?

Answer 19

@Ghayyas.mubashir - whose method ?

Answer 20

no i dont know..

Answer 21

he's referring to my method @xavier123

Answer 22

Ghayyas, if you dont know implicit differentiation yet, then you are really not prepared for this question

Answer 23

Here is an into to implicit differenctiation @Ghayyas.mubashir
If you an equation expressed "implicitly" and not explicitly, i.e. not given in the y =... form but rather in a form like x^2 + y^2 = 1 - 2xy or something like that, then the way differentiate is by taking the derivative term by term with respect to x.

If a function was given in y = .. form, how would you differentiate? You would obviously differentiate y with respect to x, so the derivative of a function like y = x^3 would be:\[y=x^3 \rightarrow \frac{ dy }{ dx }=3x^2\] Notice how the "dy/dx" notation really means that you are taking the chance in y with respect to x? You do the exact same with implicit differentiation. You look at the equation, something like the one I gave earlier, x^2 + y^2 = 1 - 2xy, and you treat y as a function of x. So let's differentiate this:\[x^2+y^2 = 1-2xy \rightarrow 2x+2y \frac{ dy }{ dx }=-2y-2x \frac{ dy }{ dx }\] See how each term containing "x" was differentiated normally but each term containing "y" was differentiate using chain rule. Since we are treating y as a function of x, the x are differentiated the way you would differentiate them in an equation like y = x^2, but y, is a function of x, so using chain rule, we have to do 2y and then multiply it by dy/dx to show that y is a function changing with respect to x. And then you just rearrange and factor the equation to solve for dy/dx. And once again, your dy/dx will also be expressed implicitly, i.e. it will have both x and y in it.

Answer 24

@genius12 - nice explanation :)