Question

How do I get the d/dx of
x^(x) = y^(y)

Answer 1

you mean dy/dx ?

Answer 2

If possible, please show me each step, and yes dy/dx

Answer 3

Okay, you know that you will have to have a y' (chain rule) for y, since you are deriving dy/dx ?

Answer 4

So, lets do a logarithmic differentiation. Lets take the natural log of both sides, okay?

Answer 5

Yes

Answer 6

x^x = y^y
ln(x^x) = ln(y^y)
xln(x)=ln(y)

Answer 7

can you tell me what the left side is going to be (product rule) ?

Answer 8

the derivative of ln(x)=1/x

Answer 9

correct

Answer 10

x(lnx)

Answer 11

give me a sec

Answer 12

sure.

Answer 13

post the steps here please, don't use wolfram or anything like this... please....

Answer 14

okay

Answer 15

d/dx (F times G) = F' G + G' F

Answer 16

think there might be a slight typo above
\[x\ln(x)=y\ln(y)\] maybe better

Answer 17

Oops,
xln(x)=yln(y)

Not, xln(x)=ln(y)

Answer 18

My bad, it is
xln(x)=yln(y)

since I am taking the exponents out from ln(x^x)=ln(y^y)

Yes, sorry sate.

Answer 19

And the reason derivative of ln(x)=1/x is,

y=ln(x)
x=e^y
deriving...
1=y' e^y
y'= 1/e^y
and we know the e^y=x, so
y'=1/x

Answer 20

\[[x (\frac{ 1 }{ x })+ (1)(\ln(x))]\]

Answer 21

That's the left side

Answer 22

yes

Answer 23

the right is going to be similar, but for each derivative of y, you will have to multiply times y prime.

Answer 24

yes

Answer 25

go for it:)

Answer 26

xln(x)=yln(y)
x(1/x)+1*ln(x)=y*(1/y)*y'+1*ln(y)*y'

Answer 27

See the similarity and the difference?

Answer 28

\[[(y)(\frac{ 1 }{ y }* \frac{ dy }{ dx })] + [ \frac{ dy }{ dx } * lny]\]

Answer 29

you are close, but....

Answer 30

You are deriving like this,

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

Answer 31

without the second equal sign on the last line.

Answer 32

and for each derivative (dy/dx) of y, you will not only derive the function as you did by the x, but also MULTIPLY times y prime, since you are deriving dy/dx.

Makes sense?
What are you going to get after you derive it?

Answer 33

I'm really confused. I thought that:
The second equation we can break up yln(y) where according to the product rule, f= y
and f'= dy/dx. g= ln(y) and g'= 1/y * dy/dx

So using the product rule of f*g'+f'*g we would get what I put..?

Answer 34

I meant the RHS

Answer 35

yes the product rule on both sides, the y-side and the x-side, but the y-derivative differs in that, that you have to multiply each y' derivative times y'.

Answer 36

I'll show you.

Answer 37

\[x \frac{dy}{dx}[~\ln(x)~]+\ln(x)\frac{dy}{dx}[~x]=y \frac{dy}{dx}[~\ln(y)~]+\ln(y)\frac{dy}{dx}[~y]\]\[x(1/x)+\ln(x)*(1)=y(1/y)*y'+\ln(y)*(1)*y'\]

Answer 38

this becomes,
\[1+\ln(x)=1*y'+\ln(y)*y'\]

Answer 39

Then,
\[1+\ln(x)=y'[1+\ln(y)]\]\[y'=\frac{1+\ln x}{1 + \ln y}\]

Answer 40

ohhh

Answer 41

I made a mistake

Answer 42

I guess you can leave it like this... but I am not sure what your teacher requires you to do with the derivative once you get it.

Answer 43

Where did you make a mistake?

Answer 44

I have studied this about a week or 2 ago:)

You will get to a point where you can master these. Or at least do them decently like me.

Answer 45

I gtg, bye and yw

Answer 46

Thank you!