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Differentiate log

Started by AM4R, Dec 29 2006 12:08 PM

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#1 AM4R

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Posted 29 December 2006 - 12:08 PM

Can someone help me with this.

thanks

Differentiate with respect to x

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

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#2 dfx

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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 y=log_ex

)

Then from the basic laws of logs, e^y = x

.... (i)

Differentiating both sides w.r.t x results in $\frac {dy}{dx}(e^y) = 1$

Thus $\frac {dy}{dx} = \frac{1}{e^y}$ but from (i) e^y = x

Thus $\frac {dy}{dx} = \frac {1}{x}$

Try using this to differentiate your expression, keeping in mind "normal" log as in your case usually implies $log_{10}$

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

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#3 AM4R

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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

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#4 dfx

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Posted 31 December 2006 - 10:28 PM

QUOTE(AM4R @ Dec 31 2006, 03:56 PM)

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

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

Back to the question:

$y = log_{10}(x + (1/x^2))$

10^y = x + (1/x^2)

..... (i)

Differentiating both sides:

$10^yln10 \frac {dy}{dx} = 1 -2x^{-3}$

(remember the general case for differentiating a^x

=

)

Thus rearrange for dy/dx and substitute in x + (1/x^2)

for

... (from step (i)) and that should give you what you want.

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#5 Steve

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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

$\begin{align*}\ln\left(x + \frac{1}{x^2}\right)\end{align*}$

(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 $\ln(x + x^{-2})$ , which involves one application of the chain rule.

If you want to do this explicitly, then you could set $u=x + x^{-2}$ , then the expression we want to differentiate is $\ln(u)$ . And so (by the chain rule)

$\begin{align*}\frac{d}{dx}(\ln(u))&=\frac{1}{u}\times \frac{du}{dx}\end{align*}$

I hope this helps

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#6 AM4R

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Posted 01 January 2007 - 03:11 PM

ah right

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

Thanks alot

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