d/dx (tanh^-1 x)
1 / 1 – x^2
We can start by using the derivative of inverse functions formula. Let y = tanh^-1(x), then we have:
x = tanh(y)
Taking the derivative of both sides with respect to x, we get:
1 = sech^2(y) * dy/dx
dy/dx = 1 / sech^2(y)
Next, we need to express sech^2(y) in terms of x. Recall that sech(y) = 1/cosh(y), and cosh(y) = (e^y + e^(-y))/2. Therefore:
sech^2(y) = (1/cosh(y))^2 = 1/[ (e^y + e^(-y))/2 ]^2
sech^2(y) = 4/[ (e^y + e^(-y))^2 ]
Using x = tanh(y), we know that:
tanh(y) = (e^y – e^(-y)) / (e^y + e^(-y))
x = (e^y – e^(-y)) / (e^y + e^(-y))
x(e^y + e^(-y)) = e^y – e^(-y)
xe^y + x e^(-y) = e^y – e^(-y)
(e^y – xe^y) = e^(-y) * (1 + x)
(e^(2y) – xe^y e^y) = (1 + x)^2
(e^(2y) – x) = (1 + x)^2 / (1 – x)^2
Using this with the derivative formula from earlier, we have:
dy/dx = 1 / [4/(e^(2y) + 2 + 2x – x^2)^2]
Substituting for e^(2y), we get:
dy/dx = 1 / [4/(x^2 – 2x + 3)^2]
Thus:
d/dx (tanh^-1 x) = dy/dx = (x^2 – 2x + 3)^2 / 4
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