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If the derivative of lnx = 1/x why does the derivative of ln6 or ln3 or ln2844566 all equal 0?

- chiefboo
**Posts:**8**Joined:**Fri Mar 22, 2013 3:46 pm

chiefboo wrote:If the derivative of lnx = 1/x why does the derivative of ln6 or ln3 or ln2844566 all equal 0?

Because x is a variable, while 3 and 2844566 are both constants.

- nona.m.nona
**Posts:**249**Joined:**Sun Dec 14, 2008 11:07 pm

Ok that makes sense... what about:

ln(3x^(2)-1) = 1/(3x^(2)-1)

Why aren't the constants 0 in this case?

ln(3x^(2)-1) = 1/(3x^(2)-1)

Why aren't the constants 0 in this case?

- chiefboo
**Posts:**8**Joined:**Fri Mar 22, 2013 3:46 pm

chiefboo wrote:Ok that makes sense... what about:

ln(3x^(2)-1) = 1/(3x^(2)-1)

This equation is not true. Are you perhaps leaving out the "d()/dx" from the left-hand side (as well as the rest of the application of the Chain Rule)?

chiefboo wrote:Why aren't the constants 0 in this case?

Which constants? Had the constants all be set to zero, the logarithm would have been undefined.

Or are you perhaps making reference to the derivatives of constants when applying the Product Rule, etc?

- nona.m.nona
**Posts:**249**Joined:**Sun Dec 14, 2008 11:07 pm

http://www.wolframalpha.com/input/?i=de ... 82%29-1%29

Yes I am leaving out a bunch of steps because ultimately what happens in the problem is that the expression 3x^(2)-1 is moved into the denominator as per the rule for taking the derivative of natural log of x.

I don't understand why the derivative of the natural love of 3x would equal 1/x but the derivative of 3x^(2)-1 is 1/3x^(2)-1

Yes I am leaving out a bunch of steps because ultimately what happens in the problem is that the expression 3x^(2)-1 is moved into the denominator as per the rule for taking the derivative of natural log of x.

I don't understand why the derivative of the natural love of 3x would equal 1/x but the derivative of 3x^(2)-1 is 1/3x^(2)-1

- chiefboo
**Posts:**8**Joined:**Fri Mar 22, 2013 3:46 pm

chiefboo wrote:I don't understand why the derivative of the natural [log] of 3x would equal 1/x...

Apply the Chain Rule. Check the steps and verify the cancellation.

chiefboo wrote:...but the derivative of [the natural log of] 3x^(2)-1 is 1/3x^(2)-1

You might want to verify the actual derivative. The result displayed at your link does not match what you've posted.

To learn the "why", clearly write out each stage of the computation in each derivative. If you are unable to match the correct expressions, kindly please reply showing your work, so that errors may be located and explicated. Thank you.

- nona.m.nona
**Posts:**249**Joined:**Sun Dec 14, 2008 11:07 pm

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