What is the ''difference quotient'' of d/dx[ln (x+3)] ?

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SUMMARY

The discussion focuses on calculating the difference quotient for the function d/dx[ln(x+3)]. The key equation derived is the limit as h approaches 0 of (ln(x+h+3) - ln(x+3))/h, which simplifies to 1/h * ln((x+h+3)/(x+3)). Participants emphasize the importance of defining the natural logarithm function clearly and suggest rewriting the logarithmic expression as (x+3+h)/(x+3) = 1 + (h/(x+3)) for further simplification.

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Michael Santos
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Homework Statement


What is the step by step difference quotient instructions of d/dx [ln (x+3)]?

Homework Equations

The Attempt at a Solution


I tried to solve but i got as far as (lim h--> 0 ..of.. 1/h * ln (x+h+3/x+3)[/B]
 
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How are you defining the natural log function?
 
vela said:
How are you defining the natural log function?
Define it with the difference quotient only

d/dx (f (x+h) - f (x))/h

d/dx[ln (x+3)] =

lim h --> 0 of (ln (x+h+3) - ln (x+3))/h

lim h --> 0 of( 1/h * ln ((x+h+3)/(x+3))

?
 
Last edited:
Michael Santos said:

Homework Statement


What is the step by step difference quotient instructions of d/dx [ln (x+3)]?

Homework Equations



The Attempt at a Solution


I tried to solve but i got as far as (lim h--> 0 ..of.. 1/h * ln (x+h+3/x+3)[/B]

Michael Santos said:
Define it with the difference quotient only

d/dx (f (x+h) - f (x))/h

d/dx[ln (x+3)] =

lim h --> 0 of (ln (x+h+3) - ln (x+3))/h

lim h --> 0 of( 1/h * ln ((x+h+3)/(x+3))

?
This thread appears to be related to two other threads you started today.
with this being the main thread.
These should all appear in one single thread.

The question you are attempting to answer is not altogether clear. Please state it word for word as it was given to you.

It appears that you are working with the function ##\ f(x) = ln(x+3) \ ## and want to find the difference quotient, perhaps with the goal of using that to find the derivative of ##\ f(x) \ ##. But details are quite uncertain.
 
Michael Santos said:
Define it with the difference quotient only
I'm asking what your definition of the log function is.

Anyway, after you combine the two logarithms, try rewriting the inside as
$$\frac{x+3+h}{x+3} = 1+\frac{h}{x+3}.$$
 

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