# Help me with the derivate of this equation?

roboredo
Can sameone help me with the derivate of this equation?

y=a log10 (x) + B

Thank you
Marta

Homework Helper
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

Homework Helper
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
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

Homework Helper
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.

Homework Helper
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!

Homework Helper
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!

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