if f^n (a), denotes the n-th derivative of f at a. Then how does one find f^49 (3) for the function: .....?

Question

f(x) = ln(x^4 - 2x^2 +1) - ln(x^2 - 2x +1)

I assume there is some sort of trick/shortcut that has to do with series or a pattern or something, can anyone help me?

Answer 1

Yes, there is a shortcut: simplify the function first...

f(x) = ln (x^4 - 2x^2 + 1) - ln (x^2 - 2x + 1)
= ln ((x^2 - 1)^2) - ln ((x-1)^2)
= 2 ln (x^2 - 1) - 2 ln (x-1)
= 2 ln ((x-1)(x+1)) - 2 ln (x-1)
= 2(ln (x-1) + ln (x+1)) - 2 ln(x-1)
= 2 ln (x+1), for x not equal to 1.

f'(x) = 2/(x+1)
f^2(x) = -2 (x+1)^(-2)
f^3(x) = 4 (x+1)^(-3)
f^4(x) = -12 (x+1)^(-4)
etc.

You should be able to convince yourself that f^n(x) = (-1)^n . 2. (n-1)! . (x+1)^(-n). So f^49 (3) = (-1)^49 . 2. 48! . 4^(-49) = -2.48! / 4^49 = -48! / 2^97.

Answer 2

The first thing to do is to see what happens if you simplify the function:
f(x) = ln(x^4 - 2x^2 +1) - ln(x^2 - 2x +1) =
ln[((x^2 - 1)^2)/(x - 1)^2] =
ln[((x + 1)(x - 1))^2)/(x - 1)^2] =
ln((x + 1)^2) =
f(x) = 2ln(x + 1)

For now, let u = x + 1
f(u) = 2lnu
f'(u) = 2/u
f''(u) = 2(-1)u^-2
f'''(u) = 2(-1)(-2)u^-3
f4(u) = 2(-1)(-2)(-3)u^-4
fn(u) = 2((-1)^(n-1))(n-1)!u^-n
therefore
f49(x) = 2*48!/(x + 1)^49

f49(3) = 2*48!/4^49 ≈ 78.3428*10^31

Answer 3

First simplify the equation.

f(x) = ln(x^4 - 2x^2 +1) - ln(x^2 - 2x +1)
f(x) = ln(x^2 - 1)^2 - ln(x -1)^2
f(x) = ln[(x - 1)(x + 1)]^2 - ln(x -1)^2
f(x) = ln(x - 1)^2 +ln(x + 1)^2 - ln(x -1)^2
f(x) = ln(x + 1)^2 = 2ln(x + 1)

f'(x) = 2/(x+1) = 2(x+1)^(-1)
f''(x) = -2(x+1)^(-2)
f'''(x) = 4(x+1)^(-3)
...
f^(n) [x] = 2*(-1)^(n+1)*(n-1)!*(x+1)^(-n)

So
f^(49) [3] = 2*[(-1)^50]*(48!)*4^(-49)
f^(49) [3] = 2*(48!)/(4^49)

Answer 4

Based on my rigorous calculations, f^n(x) = (-1)^(n+1) . 2. (n-1)! . (x+1)^(-n). Hence, f^49 (3) = (-1)^50 . 2. 48! . 4^(-49) = 2.48! / 4^49 = 48! / 2^97.

Answer 5

I would start by simplifying the expression. Try factoring the two expressions in the ln()s. Then remember that ln(a) - ln(b) = ln(a/b)...