derivatives of 0.5ln((1+x)/(1-x))?

Question

hello
Is the first derivative of this 0.5((1+x)(1-x))^-1

second derivative ((1+x)/(1-x)^-2) * (-1/(1-x)^2)

third derivative is too complicated to write down.
Could you tell me if this is the Maclaurin series for (tanh)^-1 (x) or am i completely wrong!
thanks!

Answer 1

You have the derivatives wrong - I worked them below

y = 0.5ln((1+x)/(1-x)) = 0.5ln[(1+x)(1-x)‾¹]

Using the Chain Rule
[ if f(x) = h(g(x)), then f'(x) = h'(g(x))g'(x) ]

And d(lnx)/dx = 1/x

y' = 0.5*(1/[(1+x)(1-x)‾¹])*d([(1+x)(1-x)‾¹])/dx

y' = 0.5(1-x)(1+x)‾¹d([(1+x)(1-x)‾¹])/dx

Using Product Rule

d([(1+x)(1-x)‾¹])/dx = x(1-x)‾¹ + (1+x)(1-x)‾²(-1)(-1) =
x(1-x)‾¹ + (1+x)(1-x)‾²

So

y' = 0.5(1-x)(1+x)‾¹[x(1-x)‾¹ + (1+x)(1-x)‾²]

y' = 0.5x(1+x)‾¹ + 0.5(1-x)‾¹ = 0.5[x(1+x)‾¹ + (1-x)‾¹]

y' = 0.5[(x(1-x) + (1+x))/(1+x)(1-x)]

y' = 0.5[(x-x²+1+x)/(1-x²)]

y' = -0.5(x²-2x-1)(1-x²)‾¹

Again using Product Rule

y'' = -0.5[(2x-2)(1-x²)‾¹ + (x²-2x-1)(1-x²)‾²(-2x)(-1)]

y'' = -0.5[2(x-1)(1-x²)‾¹ +2x(x²-2x-1)(1-x²)‾²]

y'' = -(1-x²)‾²[(x-1)(1-x²) + x(x²-2x-1)]

y'' = -(1-x²)‾²[x-x³-1+x²+x³-2x²-x]

y'' = -(1-x²)‾²(-x²-1)

y'' = (1+ x²)(1-x²)‾²

And then

y''' = 2x(1-x²)‾² + (1+ x²)(1-x²)‾³.(-2x)(-2)

y''' = (1-x²)‾³[2x(1-x²) + 4x²(1+ x²)]

y''' = 2x(1-x²)‾³[(1-x²) + 2x(1+ x²)]

y''' = 2x(1-x²)‾³[1-x² + 2x+ 2x³]

y''' = 2x(1-x²)‾³[2x³-x² + 2x+ 1]

y''' = 2x(2x³-x² + 2x+ 1)(1-x²)‾³

So

y = 0.5ln((1+x)/(1-x)) = 0.5ln[(1+x)(1-x)‾¹]

y' = -0.5(x²-2x-1)(1-x²)‾¹

y'' = (1+ x²)(1-x²)‾²

y''' = 2x(2x³-x² + 2x+ 1)(1-x²)‾³

The Taylor/Maclaurin series of (tanh)‾¹(x) (which is arctanh(x))

is

arctanh(x) = Sum[(2n+1)‾¹.x^(2n+1)]

And as MacLaurin series is defined as

Sum((f^n(a)/n!)x^n), where f^n is the nth derivative of f(n)

So for arctanh

f^1 = x^3/3 - which is not the same as y' as son on for other derivatives

Answer 2

Unfortunately, there's a sign mistake in the second term of Puggy's first expression for the first derivative. Here's what it should be, instead.

Let the function be y(x). First separate out the two ln terms implied by the "/" symbol. Then:

dy/dx = 0.5 d/dx[ ln(1 + x) - ln (1-x)] = 0.5 [1/(1 + x) + 1/(1 - x)]. ......(1)

Why? : because the "-" outside the "ln" multiplies a "-" from differentiating the (1 - x) expression. Carrying on, eqn. (1) gives:

dy/dx = 0.5 [2/(1 - x^2)] = 1/(1 - x^2). ......(2)

Then:

d^2 y/dx^2 = +2x/(1 - x^2)^2. ......(3)

How far were you expected to go? It looks like the derivatives will get quite messy from now on. For example:

d^3 y/dx^3 = 2 (1 + 3 x^2) / (1 - x^2)^3. ......(4)

(Perhaps one could hope to find the general expression by obtaining a recurrence relation, but I'm personally not very sanguine about it.)

And NO, I don't this is connected with the Maclaurin series for
(tanh)^-1 (x).

Good luck! Live long and prosper.

Answer 3

Take this as x * (1+x)^-1, then differentiate it as the product of two terms. So it's the derivative of the first times the second, plus the derivative of the second times the first. [1*(1+x)^-1] + [x * -1(1+x)^-2] = [1/(1+x)] - [x / (1+x)^2] = [(1+x)/(1+x)^2] - [x / (1+x)^2] = (1+x - x) / (1+x)^2 = 1 / (1+x)^2

Answer 4

y = 0.5 ln ( (1 + x) / (1 - x) )

Let's solve this using log properties .. BEFORE we take the derivative.

y = 0.5 [ ln(1 + x) - ln(1 - x) ]
y = 0.5ln(1 + x) - 0.5ln(1 - x)

y' = 0.5 (1/(1 + x)) - 0.5 (1/(1 - x)(-1))
y' = (0.5)/(1 + x) + (0.5)/(1 - x)

We can simplify this,

y' = (0.5) [1/(1 + x) + 1/(1 - x)]
y' = (0.5) [(1 - x) + (1 + x)]/(1 - x)(1 + x)
y' = (0.5) [2 / (1 - x^2)]
y' = 1/(1 - x^2)

Let's now solve for the second derivative, using the quotient rule.

y'' = [(0)(1 - x^2) - (1)(-2x)]/[1 - x^2]^2
y'' = 2x/[1 - x^2]^2

EDIT: The commenter who said there was a sign mistake in my answering your question was correct. It's been corrected; thanks Dr Spock!

Answer 5

First let's put it into a simpler form to differentiate.

0.5ln((1+x)/(1-x)) = (1/2)[ln(1 + x) - ln(1 - x)]

d{(1/2)[ln(1 + x) - ln(1 - x)]}/dx = (1/2){1/(1 + x) - (-1)/(1 - x)}
= (1/2){1/(1 + x) + 1/(1 - x)}
= (1/2){[(1 - x) + (1 + x)]/[(1 + x)(1 - x)}
= (1/2){2/(1 - x²)} = 1/(1 - x²)

The second derivative is:

d{1/(1 - x²)}/dx = (-1)(-2x)(1)/(1 - x²)²
= 2x/(1 - x²)²

Answer 6

http://mathworld.wolfram.com/MaclaurinSeries.html

Scroll down the page a little....not sure if this will answer your question