Can the Nth Derivative Formula for ln(x) Be Extended to ln(x+c)?

[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.

 

[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.

 

[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.

 

[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.

The derivative of the natural logarithm function, ln(x), with respect to x is given by:

$$\frac{d}{dx} \ln(x) = \frac{1}{x}$$

So, the derivative of ln(x) is 1/x. This is a fundamental result in calculus and is often used when finding derivatives of logarithmic functions.