Question

How the differentiation happens of the expression ln t=ln r + ln h ?

when we differentiate the expression `ln t=ln r + ln h` it becomes `(deltat)/t=(deltar)/r+(deltah)/h` in this what we are differentiating respect to i understand `y= log x` which differentiated with respect to x it becomes `dy/dx=1/x` so how above differentiation works

Answer 1

Check the explanation below.

Explanation:

Well, I'm not sure this is accurate but this is as far as I could do to help from your question.

First, if `y=logx`
`dy/dx=1/(xln10)`

And if `y=lnx`
`dy/dx=1/x`

so the first equation's differentiation was correct but the second one was wrong.
as `lnt=lnr+lnh`
when you differentiate it with respect to another variable let's say it's z then you get
`1/t*(dt)/dz=1/r*(dr)/dz+1/h*(dh)/dz`

and by multiplying by `dz` you get:

`(dt)/t=(dr)/r+(dh)/h`