derivates

[roboredo]

Can sameone help me with the derivate of this equation?

y=a log10 (x) + B

Thank you
Marta

[dextercioby]

What do you mean by "derivate this equation" ?

Daniel.

[roboredo]

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

[dextercioby]

1.You posted this problem in the wrong forum. The homework one is just above.
2.The slope is given by the derivative

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

Daniel.

[roboredo]

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?...

[roboredo]

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

[dextercioby]

What's the connection between the logarithm base 10 and the logarithm base "e" ?

Daniel.

[roboredo]

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.

[dextercioby]

Well, this is all you need

\lg x= \frac{\ln x}{\ln 10}

and now use the derivative of the natural logarithm.

Daniel.

[roboredo]

i am still struggling!

[dextercioby]

Well

\frac{d}{dx}\left(a\frac{\ln x}{\ln 10}\right)=\frac{a}{\ln 10}\frac{d \ln x}{dx}

Daniel.

[roboredo]

I still haven't figured it out...
but thank you any way!

[ZapperZ]

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.

[da_willem]

The answer is (for x>0)

\frac{a}{ln10} \frac{1}{x}=(\frac{a}{2.302585...} ) \frac{1}{x}

[roboredo]

Thank You!