log differentiation

Sep 2008
631
2
Differentiate with respect to 'x'. \(\displaystyle log_{4}(x^{2}) \)

I know that \(\displaystyle \frac{d}{dx}(log_a{x}) = \frac{1}{lna} \times \frac{1}{x} \)

so, based on this formula, I got the answer: \(\displaystyle \frac{1}{ln4} \times \frac{1}{x^{2}} \)

But this is not the correct answer, can someone please show me how to correctly differentiate this function?

Thank you!
 

Chris L T521

MHF Hall of Fame
May 2008
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2,046
Chicago, IL
Differentiate with respect to 'x'. \(\displaystyle log_{4}(x^{2}) \)

I know that \(\displaystyle \frac{d}{dx}(log_a{x}) = \frac{1}{lna} \times \frac{1}{x} \)

so, based on this formula, I got the answer: \(\displaystyle \frac{1}{ln4} \times \frac{1}{x^{2}} \)

But this is not the correct answer, can someone please show me how to correctly differentiate this function?

Thank you!
Hint: \(\displaystyle \log_4(x^2)=2\log_4x\).

Can you continue? :)

EDIT: If you continue doing it your way, you have to apply chain rule to the \(\displaystyle x^2\) term.
 
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chiph588@

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Sep 2008
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Differentiate with respect to 'x'. \(\displaystyle log_{4}(x^{2}) \)

I know that \(\displaystyle \frac{d}{dx}(log_a{x}) = \frac{1}{lna} \times \frac{1}{x} \)

so, based on this formula, I got the answer: \(\displaystyle \frac{1}{ln4} \times \frac{1}{x^{2}} \)

But this is not the correct answer, can someone please show me how to correctly differentiate this function?

Thank you!
You need to apply the chain rule: \(\displaystyle \frac{d}{dx}\left(f(g(x))\right) = f'(g(x))\cdot g'(x) \)

In this case \(\displaystyle f(x) = \log_4(x) \) and \(\displaystyle g(x)=x^2 \).

So \(\displaystyle \left(\log_4(x^2)\right)' = \frac{1}{\ln(4)}\frac{1}{x^2}\cdot 2x = \frac{2}{x\cdot \ln(4)} \)
 
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Sep 2008
631
2
Hint: \(\displaystyle \log_4(x^2)=2\log_4x\).

Can you continue? :)

EDIT: If you continue doing it your way, you have to apply chain rule to the \(\displaystyle x^2\) term.
I think so....

\(\displaystyle 2 \times \frac{1}{ln4} \times \frac{1}{x} \)

\(\displaystyle \frac{2}{ln4} \times \frac{1}{x} \)

\(\displaystyle \frac{1}{ln2} \times \frac{1}{x} \)


Are you 'allowed' to factor out the '2' like this? Even though there is ln 'attached' to 4?

Is this method correct?
 

Chris L T521

MHF Hall of Fame
May 2008
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2,046
Chicago, IL
I think so....

\(\displaystyle 2 \times \frac{1}{ln4} \times \frac{1}{x} \)

\(\displaystyle \frac{2}{ln4} \times \frac{1}{x} \)

\(\displaystyle \frac{1}{ln2} \times \frac{1}{x} \)

Are you 'allowed' to factor out the '2' like this? Even though there is ln 'attached' to 4?
You can't do that. :/

The answer should remain as \(\displaystyle \frac{2}{x\ln 4}\).
 
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chiph588@

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You can't do that. :/

The answer should remain as \(\displaystyle \frac{2}{x\ln 4}\).
Actually I think you can...

\(\displaystyle \ln(4) = \ln(2^2) = 2\ln(2) \)
 
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Sep 2008
631
2
But is my method correct? As I simply used the formula in my book, I did not use the chain rule.
 

chiph588@

MHF Hall of Honor
Sep 2008
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429
Champaign, Illinois
But is my method correct? As I simply used the formula in my book, I did not use the chain rule.
If you used \(\displaystyle \log_4(x^2)=2\log_4(x) \), then the chain rule is not needed.
 
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Apr 2010
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Canada
If you used \(\displaystyle \log_4(x^2)=2\log_4(x) \), then the chain rule is not needed.
As a matter of learning, i think it's important to identify that the chain rule is still used! But it's the fact that the derivative of x is equal to 1, which allows us to neglect directly thinking about the chain rule.

This is an obvious result but I think for those having trouble, it might be important!