# Nth derivative of ln(x+c)

1. Dec 15, 2008

### islandboy401

Is there a formula or a method to derive the nth derivative of ln (x+c)...(where c is any number...such as 1, 2, 3, etc)?

I know that there is a formula for the nth derivative of ln x, but I was wondering if this formula can be extended to one we can use when we are dealing with the nth derivative of something slightly more complicated.

2. Dec 15, 2008

### slider142

Use the chain rule or the definition of the derivative to show that f(n)(x + c) is the same as f(n)(u) if you let u = x + c.

Last edited: Dec 15, 2008
3. Dec 15, 2008

### NoMoreExams

You should try to derive a formula for the derivative of y = ln(g(x)). The way I would do that is using implicit differentiation, namely:

$$y = ln(g(x))$$

$$e^{y} = g(x)$$

$$\frac{dy}{dx} e^{y} = \frac{dg}{dx}$$

$$\frac{dy}{dx} = \frac{g'(x)}{g(x)}$$

Hope this helps.

4. Dec 15, 2008

### lurflurf

yes
[log(x+C)]'=1/(x+C)
[log(x+C)]''=-1/(x+C)^2
(D^n)[log(x+C)]=(n-1)!(-1)^(n+1)/(x+C)^n
for n=1,2,3,4,...

is can be done for other simple functions like sin(x), exp(x), and x^a. The trouble with more complicated functions is complicated sums arise.