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Differentiate log - HSN forum

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Differentiate log


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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!

#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 smile.gif
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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) View Post

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 =  a^xlna )

Thus rearrange for dy/dx and substitute in  x + (1/x^2) for  10^y ... (from step (i)) and that should give you what you want.

#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 "loge".)

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 smile.gif
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#6 AM4R

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Posted 01 January 2007 - 03:11 PM

ah right biggrin.gif

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

Thanks alot
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