HELP!

I used a digital pad with a pen to write... Very hard to do because you look on the screen and write on the pad... My handwriting is neater than that.
 
Luis G said:
It is incomplete.

You need to express the differentiation in terms of "x", not x and y ;)

y' = something expressed in terms of x
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Luis G said:
My bet is that SL will do it :D
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OK, you've awakened the math man in me which hasn't been exercised nearly enough. I am trying to remember these formulas from scratch (my calculus book is missing at the moment), so if ay of them are wrong, let me know. Let me see...

y ln x = x ln y

(y ln x)' = (x ln y)'

Using product rule: (ab)' = a'b + b'a

Using (ln x)' = 1/x

y'ln(x) + (y/x) = ln(y) + (x/y)

Solve for y'

y' = ( ln(y) + (x/y) - (y/x) ) / (ln(x))

That's about as simplified as it can be. :D It's impossible to get a pure substitution for y.
 
Stop Laughing said:
y' = ( ln(y) + (x/y) - (y/x) ) / (ln(x))

That's about as simplified as it can be. :D It's impossible to get a pure substitution for y.
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You have the same error as Jerrek, remember than you do not express y' in terms of "x and y".

y' = something expressed in x's only.
 
Luis G said:
You have the same error as Jerrek, remember than you do not express y' in terms of "x and y".

y' = something expressed in x's only.
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Luis get a grip. There is no error in mine. I had my math professor with a doctorate in mathematics look it over and he said it was just fine. You can't isolate all the y's.
 
ash r said:
i'm just so very completely bad at math. i cry when i get confused about it. :(
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Thats ok ;) Math ain't for everyone. Everyone has fields they excel in. And some they don't do well in.
 
Jerrek said:
Luis G said:
You have the same error as Jerrek, remember than you do not express y' in terms of "x and y".

y' = something expressed in x's only.
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There is no error in mine.
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I said your was incomplete, the error is to leave it incomplete.

And if your professor is uncapable of getting rid of the y's, then your professor might need a new doctorate, this time in algebra.

Btw, in the last pic i attached i have an error, this is the right value for "p", and p= 1/lny, so i got rid of all the "y", and now i have y' and x only. Following the process i found that y' = 1 . Except for y' = 4(q - q^2)
 
Jerrek said:
Luis get a grip. There is no error in mine. I had my math professor with a doctorate in mathematics look it over and he said it was just fine. You can't isolate all the y's.
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Jerrek, tell your teacher he's an asshole and whenever he likes i can teach him how to differentiate.

Why is he an asshole?, because he said yours was just fine. And it is not.
 
Just one minute looking at it? hmm, now i wonder if he saw the "y" in the right side. If he saw it, and he said it was ok, ask him how would he obtain the speed (first differentiation) of the particle whose movement is described by the equation x^y = y^x
 
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