Can someone help me with this.

thanks

Differentiate with respect to x

log(x + (1/x^2))

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# Differentiate log

Started by AM4R, Dec 29 2006 12:08 PM

5 replies to this topic

### #1

Posted 29 December 2006 - 12:08 PM

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### #2

Posted 29 December 2006 - 03:27 PM

Right... if you consider the general case for differentiating y=lnx, any log becomes straightforward.

Consider y = lnx (remember this is )

Then from the basic laws of logs, .... (i)

Differentiating both sides w.r.t x results in

Thus but from (i)

Thus

Try using this to differentiate your expression, keeping in mind "normal" log as in your case usually implies

Post back for a few more hints if necessary. Good luck!

Consider y = lnx (remember this is )

Then from the basic laws of logs, .... (i)

Differentiating both sides w.r.t x results in

Thus but from (i)

Thus

Try using this to differentiate your expression, keeping in mind "normal" log as in your case usually implies

Post back for a few more hints if necessary. Good luck!

### #3

Posted 31 December 2006 - 03:56 PM

Thanks for the help dfx.

Im still a little confuzzled.

What would e be equal to here ? 10?

Also would dy/dx = (x + (1/x^2))

Thanks

Im still a little confuzzled.

What would e be equal to here ? 10?

Also would dy/dx = (x + (1/x^2))

Thanks

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### #4

Posted 31 December 2006 - 10:28 PM

Also would dy/dx = (x + (1/x^2))

No, sorry I can't see where you're coming from?

Back to the question:

..... (i)

Differentiating both sides:

(remember the general case for differentiating = )

Thus rearrange for dy/dx and substitute in for ... (from step (i)) and that should give you what you want.

### #5

Posted 01 January 2007 - 11:33 AM

This seems a bit complicated to me! I think the question is probably asking for the derivative of

(This is different for different subjects, but in maths "log" usually means "log

This problem is a bit easier then, you just have to differentiate , which involves one application of the chain rule.

If you want to do this explicitly, then you could set , then the expression we want to differentiate is . And so (by the chain rule)

I hope this helps

(This is different for different subjects, but in maths "log" usually means "log

_{e}".)This problem is a bit easier then, you just have to differentiate , which involves one application of the chain rule.

If you want to do this explicitly, then you could set , then the expression we want to differentiate is . And so (by the chain rule)

I hope this helps

### #6

Posted 01 January 2007 - 03:11 PM

ah right

I understand now, all i did was use the chain rule

Thanks alot

I understand now, all i did was use the chain rule

Thanks alot

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