1. ## Derivative

Hello.

Could someone explain me the following in the subject of derivative.

y = ln x ; y'= 1/x

Following that rule why doesn't y =ln x^5 equal y' = 1/x^5 but instead y' = 5/x
If someone could explain me this I would be very greatful.

PS! I am not an english speaking person so my english is limited.

2. Originally Posted by Kurumax
Following that rule why doesn't y =ln x^5 equal y' = 1/x^5 but instead y' = 5/x
As a consequence of the Chain´s Rule:

$\displaystyle f(x)=\ln u(x)\Rightarrow f'(x)=\dfrac{1}{u(x)}\cdot u'(x)$

Fernando Revilla

3. Thank you!

4. I have a new question. How can i solve this derivative?

y = ln 1/x^2

I've been messing with this for some time now and the solution i got was y' = -1/x^2 but i do not believe it's right.

5. The question you are asking here is quite similar to your first question..

Look at FernandoRevilla's suggestion in post #2

$\displaystyle f(x)=\ln (u(x))$ where $\displaystyle u(x)=\dfrac{1}{x^2}$ and then find the derivative... show your work f you are stuck...

6. $\displaystyle y=ln \dfrac{1}{x^2} = -ln x^2$

$\displaystyle y' = -\dfrac{1}{x^2} \cdot \dfrac{2x}{1} = -\dfrac{2}{x}$

Is this correct?

7. Originally Posted by Kurumax

$\displaystyle y=ln \dfrac{1}{x^2} = -ln x^2$

$\displaystyle y' = -\dfrac{1}{x^2} \cdot \dfrac{2x}{1} = -\dfrac{2}{x}$

Is this correct?

EDIT: Yes .. it is correct..

8. Thank you for confirming the answer.

9. Originally Posted by Kurumax
$\displaystyle y=ln \dfrac{1}{x^2} = -ln x^2$
$\displaystyle y' = -\dfrac{1}{x^2} \cdot \dfrac{2x}{1} = -\dfrac{2}{x}$
An even easier alternative: $\displaystyle \displaystyle y = \ln{\left(\frac{1}{x^2}\right)} = \ln{(x^{-2})} = -2\ln{x}$.
So $\displaystyle \displaystyle \frac{dy}{dx} = -\frac{2}{x}$