Derivative

**Kurumax** · January 23rd 2011, 02:08 AM

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.

**FernandoRevilla** · January 23rd 2011, 02:14 AM

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:

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

Fernando Revilla

**Kurumax** · January 23rd 2011, 02:21 AM

Thank you!

**Kurumax** · January 23rd 2011, 09:56 AM

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.

**harish21** · January 23rd 2011, 10:01 AM

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

Look at FernandoRevilla's suggestion in post #2

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

**Kurumax** · January 23rd 2011, 10:13 AM

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

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

Is this correct?

**harish21** · January 23rd 2011, 10:17 AM

Originally Posted by Kurumax

Sorry about this thread being in the wrong section. My english math knowladge is not so good.

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

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

Is this correct?

EDIT: Yes .. it is correct..

**Kurumax** · January 23rd 2011, 10:36 AM

Thank you for confirming the answer.

**Prove It** · January 23rd 2011, 03:47 PM

Originally Posted by Kurumax

Sorry about this thread being in the wrong section. My english math knowladge is not so good.

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

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

Is this correct?

Correct.

An even easier alternative: $\displaystyle y = \ln{\left(\frac{1}{x^2}\right)} = \ln{(x^{-2})} = -2\ln{x}$ .

So $\displaystyle \frac{dy}{dx} = -\frac{2}{x}$

**Kurumax** · January 26th 2011, 05:14 AM

Thanks again everyone. Passed my test thanks to this help.

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