y =ln((x^2+1)/(x^2-1))^1/2
Find the first derivative for this ln square division.
y =ln((x^2+1)/(x^2-1))^1/2
y = 1/2 ln((x^2+1)/(x^2-1))
y = 1/2(ln(x^2+1) - ln(x^2-1))
y' = 1/2 (2x/(x^2+1) - 2x/(x^2-1))
y' = x/(x^2+1) - x/(x^2-1)
y' = -2x/(x^4-1)
Now, if you're a glutton for punishment, just apply the chain rule:
y =ln((x^2+1)/(x^2-1))^1/2
y' = (1/2)/[(x^2+1)/(x^2-1)]^(-1/2) * [(2x)(x^2-1) - 2x(x^2+1)]/(x^2-1)^2
I'm not!
Hmm. I seem to have misread the problem. You have the square root of the log?
In that case,
y' = -2x/[(x^4-1)*ln((x^2+1)/(x^2-1))^1/2]
go to wolframalpha and show steps
To find the first derivative of the given function, we will need to apply the chain rule.
The function can be rewritten as:
y = ln(((x^2+1)/(x^2-1))^1/2)
To differentiate this expression, let's break it down step by step:
Step 1: Rewrite the function using the laws of logarithms.
y = (1/2) * ln((x^2+1)/(x^2-1))
Step 2: Apply the chain rule. The chain rule states that if we have a composite function, f(g(x)), then the derivative of f(g(x)) with respect to x is given by f'(g(x)) * g'(x).
In this case, our composite function is f(g(x)), where f(u) = ln(u) and g(x) = (x^2+1)/(x^2-1).
First, let's find the derivative of g(x):
g'(x) = ((x^2-1)(2x) - (x^2+1)(2x))/(x^2-1)^2
= (2x(x^2-1) - 2x(x^2+1))/(x^2-1)^2
= (-4x)/(x^2-1)^2
Now, let's find the derivative of f(u):
f'(u) = 1/u
Step 3: Combine the derivatives using the chain rule.
dy/dx = f'(g(x)) * g'(x)
= (1/((x^2+1)/(x^2-1))) * (-4x)/(x^2-1)^2
= -(4x)/(x^2-1)(x^2-1)
So, the first derivative of y = ln(((x^2+1)/(x^2-1))^1/2) is -(4x)/(x^2-1)(x^2-1).