Find dy/dx, which way should I use?

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Homework Statement

.[/B]
Find dy/dx
Y^x = X^y

Homework Equations


F'( Ln(x)) = 1/x
Lnx^y = yLnx

The Attempt at a Solution


Ln(Y^x) = Ln(X^y)
X • Lny = Y • Lnx , should I differentiate now or :
Y = X • Lny/Lnx
Y = X • Logxy , then differentiate ?
 
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It does not matter when you differentiate as long as the equality you are differentiating is equivalent to the original one.
 
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Mamdoh Abughalion said:

Homework Statement

.[/B]
Find dy/dx
Y^x = X^y

Homework Equations


F'( Ln(x)) = 1/x
Lnx^y = yLnx

The Attempt at a Solution


Ln(Y^x) = Ln(X^y)
X • Lny = Y • Lnx , should I derive now or :
Y = X • Lny/Lnx
Y = X • Logxy , then derive?
To find a derivative, you differentiate. The word derive means something else.
 
Orodruin said:
It does not matter when you differentiate as long as the equality you are differentiating is equivalent to the original one.
I know that , but my instructer told me that the second way is too much " dangerous " and may lead to wrong solution .
 
Also, please don't write ##F'(\ln x) = \frac 1 x##. There is no ##F##. What you mean, and should write, is ##(\ln x)' = \frac 1 x## when you want to give a derivative formula.
 
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Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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