log differentiation

**Tweety** · May 25th 2010, 10:07 AM

Differentiate with respect to 'x'. $log_{4}(x^{2})$

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

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

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

Thank you!

**Chris L T521** · May 25th 2010, 10:09 AM

Originally Posted by Tweety

Differentiate with respect to 'x'. $log_{4}(x^{2})$

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

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

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

Thank you!

Hint: $\log_4(x^2)=2\log_4x$ .

Can you continue?

EDIT: If you continue doing it your way, you have to apply chain rule to the $x^2$ term.

**chiph588@** · May 25th 2010, 10:10 AM

Originally Posted by Tweety

Differentiate with respect to 'x'. $log_{4}(x^{2})$

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

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

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: $\frac{d}{dx}\left(f(g(x))\right) = f'(g(x))\cdot g'(x)$

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

So $\left(\log_4(x^2)\right)' = \frac{1}{\ln(4)}\frac{1}{x^2}\cdot 2x = \frac{2}{x\cdot \ln(4)}$

**Tweety** · May 25th 2010, 10:14 AM

Originally Posted by Chris L T521

Hint: $\log_4(x^2)=2\log_4x$ .

Can you continue?

EDIT: If you continue doing it your way, you have to apply chain rule to the $x^2$ term.

I think so....

$2 \times \frac{1}{ln4} \times \frac{1}{x}$

$\frac{2}{ln4} \times \frac{1}{x}$

$\frac{1}{ln2} \times \frac{1}{x}$

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** · May 25th 2010, 10:15 AM

Originally Posted by Tweety

I think so....

$2 \times \frac{1}{ln4} \times \frac{1}{x}$

$\frac{2}{ln4} \times \frac{1}{x}$

$\frac{1}{ln2} \times \frac{1}{x}$

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 $\frac{2}{x\ln 4}$ .

**chiph588@** · May 25th 2010, 10:16 AM

Originally Posted by Chris L T521

You can't do that. :/

The answer should remain as $\frac{2}{x\ln 4}$ .

Actually I think you can...

$\ln(4) = \ln(2^2) = 2\ln(2)$

**Tweety** · May 25th 2010, 10:18 AM

But is my method correct? As I simply used the formula in my book, I did not use the chain rule.

**Chris L T521** · May 25th 2010, 10:19 AM

Originally Posted by chiph588@

Actually I think you can...

$\ln(4) = \ln(2^2) = 2\ln(2)$

Oh wow...why didn't I see that? XD

But yes...that's right... XD

**chiph588@** · May 25th 2010, 10:21 AM

Originally Posted by Tweety

But is my method correct? As I simply used the formula in my book, I did not use the chain rule.

If you used $\log_4(x^2)=2\log_4(x)$ , then the chain rule is not needed.

**AllanCuz** · May 25th 2010, 11:53 AM

Originally Posted by chiph588@

If you used $\log_4(x^2)=2\log_4(x)$ , 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!

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