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November 14th, 2016, 09:18 AM   #1
ZMD
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Topological Space III

Anyone?
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November 14th, 2016, 03:03 PM   #2
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$\displaystyle y=ln(x)$
$\displaystyle x=e^y$
$\displaystyle \frac{dx}{dy}=e^y=x$
$\displaystyle \frac{dy}{dx}=\frac{1}{x}$
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November 17th, 2016, 05:36 PM   #3
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I disagree with the previous post (though the answer is fine). The expression $\frac{dy}{dx}$ is not a fraction and it can't be "flipped" like this. Nevertheless, the formula can be justified as follows:

For any $x > 0$ we can write $e^{\ln (x)} = x$ since these are inverse functions. Differentiating both sides with the chain rule on the left we get $e^{\ln(x)}\cdot \frac{d}{dx} \ln(x) = 1$ and by dividing we obtain $ \frac{d}{dx} \ln(x) = \frac{1}{e^{\ln x}} = \frac{1}{x}$.
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November 18th, 2016, 02:52 PM   #4
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Quote:
Originally Posted by SDK View Post
I disagree with the previous post (though the answer is fine). The expression $\frac{dy}{dx}$ is not a fraction and it can't be "flipped" like this. Nevertheless, the formula can be justified as follows:

For any $x > 0$ we can write $e^{\ln (x)} = x$ since these are inverse functions. Differentiating both sides with the chain rule on the left we get $e^{\ln(x)}\cdot \frac{d}{dx} \ln(x) = 1$ and by dividing we obtain $ \frac{d}{dx} \ln(x) = \frac{1}{e^{\ln x}} = \frac{1}{x}$.
I am not claiming it is a fraction.. However $\displaystyle \frac{dy}{dx}\times \frac{dx}{dy} = 1$ as long as both can be defined.
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November 22nd, 2016, 08:45 PM   #5
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Originally Posted by mathman View Post
I am not claiming it is a fraction.. However $\displaystyle \frac{dy}{dx}\times \frac{dx}{dy} = 1$ as long as both can be defined.
I don't assume you think this is a fraction. However, students who see this notation while initially learning calculus don't realize this. I just wanted to lend another point of view which gives the same result without usingg notation which appears like a fraction.
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