Find f^(n) of (x) of: f(x)=ln(2x)

[MarkSA]

Hello,

I have a problem:

1) Find f^(n) of (x) of: f(x)=ln(2x)
f(x) = y

I think they're asking for a formula for the nth derivative. I got:
y' = 1/x
y'' = -1/(x^2)
y''' = 2/(x^3)
y'''' = -6/(x^4)
y''''' = 24/(x^5)

So there is a pattern.. each time it's negated, x's power is increased by 1, but i'm not sure what the numerators are doing. 1 to 1 to 2 to 6 to 24?
Also i'm not sure how to write a formula for this for the answer being negated every other time... any ideas?

 

[PAULK]

Hello,

I have a problem:

1) Find f^(n) of (x) of: f(x)=ln(2x)
f(x) = y

I think they're asking for a formula for the nth derivative. I got:
y' = 1/x
y'' = -1/(x^2)
y''' = 2/(x^3)
y'''' = -6/(x^4)
y''''' = 24/(x^5)

So there is a pattern.. each time it's negated, x's power is increased by 1, but i'm not sure what the numerators are doing. 1 to 1 to 2 to 6 to 24?
Also i'm not sure how to write a formula for this for the answer being negated every other time... any ideas?

The numerators look like factorials, So

y(fifth deriv) = 4!/x^5

The denominators are obvious powers. And if you want signs to alternate, just throw in (-1)^n. Maybe this does it:

y(nth deriv) = (n-1)!(-1)^n/x^n

 

[PAULK]

Oops. I think it should be (-1)^(n+1).