November 14th, 2016, 09:18 AM  #1 
Member Joined: Nov 2016 From: Kansas Posts: 48 Thanks: 0  Topological Space III
Anyone?

November 14th, 2016, 03:03 PM  #2 
Global Moderator Joined: May 2007 Posts: 6,216 Thanks: 493 
$\displaystyle y=ln(x)$ $\displaystyle x=e^y$ $\displaystyle \frac{dx}{dy}=e^y=x$ $\displaystyle \frac{dy}{dx}=\frac{1}{x}$ 
November 17th, 2016, 05:36 PM  #3 
Senior Member Joined: Sep 2016 From: USA Posts: 114 Thanks: 44 Math Focus: Dynamical systems, analytic function theory, numerics 
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}$. 
November 18th, 2016, 02:52 PM  #4  
Global Moderator Joined: May 2007 Posts: 6,216 Thanks: 493  Quote:
 
November 22nd, 2016, 08:45 PM  #5 
Senior Member Joined: Sep 2016 From: USA Posts: 114 Thanks: 44 Math Focus: Dynamical systems, analytic function theory, numerics  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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