1. ## log differentiation

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!

2. Originally Posted by Tweety
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.

3. Originally Posted by Tweety
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 $\displaystyle \left(\log_4(x^2)\right)' = \frac{1}{\ln(4)}\frac{1}{x^2}\cdot 2x = \frac{2}{x\cdot \ln(4)}$

4. Originally Posted by Chris L T521
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?

5. Originally Posted by Tweety
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 $\displaystyle \frac{2}{x\ln 4}$.

6. Originally Posted by Chris L T521
You can't do that. :/

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

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

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

8. Originally Posted by chiph588@
Actually I think you can...

$\displaystyle \ln(4) = \ln(2^2) = 2\ln(2)$
Oh wow...why didn't I see that? XD

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

9. 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 $\displaystyle \log_4(x^2)=2\log_4(x)$, then the chain rule is not needed.

10. Originally Posted by chiph588@
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!