Logarithm differentiation + chain rule

  • Thread starter JamesGoh
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  • #1
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Main Question or Discussion Point

For this function

[itex]y=\sqrt{2ln(x)+1}[/itex]

if I use the chain rule properly, should I be getting this answer?

[itex]\frac{dy}{dx}=\frac{2}{x} \times \frac{1}{2} \times \frac{1}{\sqrt{2ln(x)+1}}[/itex]

My aim of doing this is to verify that

[itex]\frac{dy}{dx}=\frac{1}{xy}[/itex]
 

Answers and Replies

  • #2
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That's right and equal to what you want!
 
  • #3
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That's right and equal to what you want!

Exactly what Shaun said.
If you equate that to what information you need.

BINGO! You will get your answer
 
  • #4
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As a simpler alternative to using the chain rule, you can do this:

##y = \sqrt{2\ln(x) + 1} ##
##\Rightarrow y^2 = 2\ln(x) + 1##
Differentiating implicitly,
##2y\frac{dy}{dx} = \frac{2}{x}##
##\Rightarrow \frac{dy}{dx} = \frac{1}{xy}##
 
  • #5
45
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Yes, that is correct. Just to show the math:

d/dx[√2ln(x)+1]
=1/2√2ln(x)+1 * 2/x
=2/x * 1/2 * 1/√2ln(x)+1
 
  • #6
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You are missing several sets of parentheses.
Yes, that is correct. Just to show the math:

d/dx[√2ln(x)+1]
Most would interpret the part in brackets above as √2 * ln(x)+1, rather than √(2ln(x)+1).
=1/2√2ln(x)+1 * 2/x
Most would interpret the above as
1/2 * √2ln(x) + 2/x, which I don't think is what you intended.
=2/x * 1/2 * 1/√2ln(x)+1
 
  • #7
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You're right, perhaps I should have put 2ln(x) + 1 in parentheses, but I think he gets what I mean haha. Nonetheless, I appreciate the correction. Thank you!
 
  • #8
As a simpler alternative to using the chain rule, you can do this:

##y = \sqrt{2\ln(x) + 1} ##
##\Rightarrow y^2 = 2\ln(x) + 1##
Differentiating implicitly,
##2y\frac{dy}{dx} = \frac{2}{x}##
##\Rightarrow \frac{dy}{dx} = \frac{1}{xy}##
Hm, how does this take in to account the fact that squaring the equation will give multiple values?

##y^2=2\ln(x)+1## is a set of two equations, namely $$y= \sqrt{2\ln(x) + 1}$$ and $$y= -\sqrt{2\ln(x) + 1}$$
Will the derivative also be multiple valued? It doesn't seem like it since ##\frac{1}{xy}## is a singled valued funtion. What am I missing?
 
  • #9
33,152
4,836
As a simpler alternative to using the chain rule, you can do this:

##y = \sqrt{2\ln(x) + 1} ##
##\Rightarrow y^2 = 2\ln(x) + 1##
Differentiating implicitly,
##2y\frac{dy}{dx} = \frac{2}{x}##
##\Rightarrow \frac{dy}{dx} = \frac{1}{xy}##
paradoxymoron said:
Hm, how does this take in to account the fact that squaring the equation will give multiple values?
Actually it doesn't in this case. That's something that I considered but didn't mention before. In the first equation, the square root term is necessarily nonnegative, which means that y must also be nonnegative. So squaring both sides doesn't introduce extraneous solutions that weren't present in the original equation.
paradoxymoron said:
##y^2=2\ln(x)+1## is a set of two equations, namely $$y= \sqrt{2\ln(x) + 1}$$ and $$y= -\sqrt{2\ln(x) + 1}$$
Will the derivative also be multiple valued?
I don't understand what you're asking. The derivative will not be multiple valued.
paradoxymoron said:
It doesn't seem like it since ##\frac{1}{xy}## is a singled valued funtion. What am I missing?
 
  • #10
I don't understand what you're asking. The derivative will not be multiple valued.
What I'm trying to ask is that since we found the derivative is ##\frac{dy}{dx}=\frac{1}{xy}##, we have two options for ##y## now, the positive or negative root, since we squared the term to make the differentiation simpler. Why do we necessarily choose the positive one? I know I'm just being technical, sorry about my arrogance, but it's a pet-peeve of mine.
 
  • #11
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What I'm trying to ask is that since we found the derivative is ##\frac{dy}{dx}=\frac{1}{xy}##, we have two options for ##y## now, the positive or negative root, since we squared the term to make the differentiation simpler.
No we don't in this case, which I explained in post #9. Since y is the square root of some expression, y is necessarily nonnegative. Take a look again at what I wrote, and see if that answers your questions.


Why do we necessarily choose the positive one? I know I'm just being technical, sorry about my arrogance, but it's a pet-peeve of mine.
 

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