Help me with the derivate of this equation?

  • Thread starter roboredo
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  • #1
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Can sameone help me with the derivate of this equation?

y=a log10 (x) + B

Thank you
Marta
 

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  • #2
dextercioby
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What do you mean by "derivate this equation" ?

Daniel.
 
  • #3
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I am supposed to fit some data into this model (y= a log10 (x) + B, where a and b are constants), then I should calculate the slope of the isotherm in order to obtain an index
 
  • #4
dextercioby
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1.You posted this problem in the wrong forum. The homework one is just above.
2.The slope is given by the derivative

[tex] y(x)=a\lg x+b \Rightarrow \frac{dy(x)}{dx}=...? [/tex]

Daniel.
 
  • #5
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I am sorry, I have just realized it...which forum should I go to...general maths or homework?
But.. do you actually know the derivate of this equation?...
 
  • #6
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derivate

I am supposed to fit some data into this model
y= a log10 (x) + B, where a and b are constants
then I should calculate the slope of the isotherm in order to obtain an index

can someone help me?

Thank you , marta
 
  • #7
dextercioby
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What's the connection between the logarithm base 10 and the logarithm base "e" ?

Daniel.
 
  • #8
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I am no mathematician nor student, i just need help to solve this for work purposes and I have no maths books around..the only tool I have is internet.
 
  • #9
dextercioby
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Well, this is all you need

[tex] \lg x= \frac{\ln x}{\ln 10} [/tex]

and now use the derivative of the natural logarithm.

Daniel.
 
  • #10
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i am still struggling!
 
  • #11
dextercioby
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Well

[tex] \frac{d}{dx}\left(a\frac{\ln x}{\ln 10}\right)=\frac{a}{\ln 10}\frac{d \ln x}{dx} [/tex]

Daniel.
 
  • #12
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I still haven't figured it out...
but thank you any way!
 
  • #13
ZapperZ
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In case you haven't noticed, I've mearged both of your threads into this one. So at some point, the "flow" of the thread may not make any sense.

:)

Zz.
 
  • #14
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The answer is (for x>0)

[tex]\frac{a}{ln10} \frac{1}{x}=(\frac{a}{2.302585...} ) \frac{1}{x}[/tex]
 
Last edited:
  • #15
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Thank You!
 

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