# log differentiation

#### 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!

#### Chris L T521

MHF Hall of Fame
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.

Tweety

#### chiph588@

MHF Hall of Honor
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 $$\displaystyle \left(\log_4(x^2)\right)' = \frac{1}{\ln(4)}\frac{1}{x^2}\cdot 2x = \frac{2}{x\cdot \ln(4)}$$

Tweety

#### Tweety

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

Tweety

#### chiph588@

MHF Hall of Honor
You can't do that. :/

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

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

Tweety

#### Tweety

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

#### Chris L T521

MHF Hall of Fame
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

#### chiph588@

MHF Hall of Honor
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.

Tweety

#### AllanCuz

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!