An expression for the nth derivative

[BW202]

Hi, I've been working on this problem that has me stumped:

f(x)=1/(1+2x)

"Find an expression for the nth derivative of f."

The way I've attempted to solve it is by finding the first few derivatives:

y'=-2/(1+2x)^2
y''=8/(1+2x)^3
y'''=-48/(1+2x)^4
y''''=384/(1+2x)^5

The expression I have so far for the nth derivative is:

(d^n)/(dx^n)[1/(1+2x)]=(?/(1+2x)^(n+1))(-1)^n

The only part I'm missing is the ? which should be a function relating the degree of the derivative to the absolute value of its numerator. Despite the absolute values of the numerators of the first four derivatives (2,8,48,384) having a recognizable relationship, I can't figure out how to express this relationship in terms of n.

Considering this is the first problem of its kind I've ever done, I'm afraid I'm overly complicating things, so if there's an easier way to solve this please let me know. Thanks in advance!

 

[stapel]

Despite the absolute values of the numerators of the first four derivatives (2,8,48,384) having a recognizable relationship, I can't figure out how to express this relationship in terms of n.

What is the relationship that you're seeing?

 

[BW202]

What is the relationship that you're seeing?

The numerator of the first derivative is the numerator of the function times 2. Numerator of the second derivative is the numerator of the first derivative times 4. Numerator of the third derivative is the numerator of the second derivative times 6. Numerator of the fourth derivative is the numerator of the third derivative times 8 and so on.

 

[stapel]

The numerator of the first derivative is the numerator of the function times 2. Numerator of the second derivative is the numerator of the first derivative times 4. Numerator of the third derivative is the numerator of the second derivative times 6. Numerator of the fourth derivative is the numerator of the third derivative times 8 and so on.

So you've got the seed value of the first derivative, being 2. The next value is 2*2. The next value is 2*2*6. The next value is 2*2*6*8. Can we relate this to n, being the counter for which derivative we're at? Dunno. Let's see what we can get by trying stuff.

It's often helpful to look at factorizations:

n = 1: 2
n = 2: 2*2*2
n = 3: 2*2*2*2*3
n = 4: 2*2*2*2*3*2*2*2

Do powers help?

n = 1: 2^1
n = 2: 2^3
n = 3: (2^4)*3
n = 4: (2^7)*3

Okay; that doesn't look great. Can we get the value for n at each stage into the expression somehow?

n = 1: 1*2
n = 2: 2*2*2
n = 3: 3*2*2*2*2
n = 4: 4*2*2*2*2*2*2*3

Maybe we're on to something now....

n = 1: n*2 = 2n
n = 2: n*n*n = n*n^2 = n*(2^n)
n = 3: n*2^4 = 2n*(2^n)
n = 4: n*2^n*3 = 3n(2^n)

Ooo... Maybe there's something with n, n-1, and 2^n that'll work...?

 

[BW202]

So you've got the seed value of the first derivative, being 2. The next value is 2*2. The next value is 2*2*6. The next value is 2*2*6*8. Can we relate this to n, being the counter for which derivative we're at? Dunno. Let's see what we can get by trying stuff.

It's often helpful to look at factorizations:

n = 1: 2
n = 2: 2*2*2
n = 3: 2*2*2*2*3
n = 4: 2*2*2*2*3*2*2*2

Do powers help?

n = 1: 2^1
n = 2: 2^3
n = 3: (2^4)*3
n = 4: (2^7)*3

Okay; that doesn't look great. Can we get the value for n at each stage into the expression somehow?

n = 1: 1*2
n = 2: 2*2*2
n = 3: 3*2*2*2*2
n = 4: 4*2*2*2*2*2*2*3

Maybe we're on to something now....

n = 1: n*2 = 2n
n = 2: n*n*n = n*n^2 = n*(2^n)
n = 3: n*2^4 = 2n*(2^n)
n = 4: n*2^n*3 = 3n(2^n)

Ooo... Maybe there's something with n, n-1, and 2^n that'll work...?

It looks like you're implying the function should be (n-1)(n)(2^n) but that wouldn't satisfy the need for f(4)=384 or f(1)=2

 

[stapel]

It looks like you're implying the function should be (n-1)(n)(2^n) but that wouldn't satisfy the need for f(4)=384 or f(1)=2

Okay. Now you fiddle with the numbers and see what you can come up with. That's kinda the point of the exercise.

 

[BW202]

Okay. Now you fiddle with the numbers and see what you can come up with. That's kinda the point of the exercise.

Got it!

(((n!)(2^n))/(1+2x)^(n+1))(-1)^n

Thank you for your help.

 

[Ishuda]

Just to add my 2 cents - nothing new but another way to look at it maybe:

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

Number of 2's = order of the derivative
Last number in (-1) (-2) (-3) ... series = order of the derivative
Sign = (-1)^{order of the derivative}
Exponent of 1+2x = negative of (order of derivative plus 1)

f⁽ⁿ⁾ = (-1)ⁿ n! 2ⁿ (1+2x)^-(n+1)

Quick check on f⁽ⁿ⁺¹⁾, so good to go.