How do you find the first and second derivative of #y=ln [ x / (x^2 - 1) ]#?

Question

William Abdalla · Accepted Answer

#f'(x)=(1+x^2)/(x(1-x^2))#

#f''(x)=(x^4+4x^2-1)/(x^2(1-x^2)^2)# Since #f(x)=ln(g(x))->f'(x)=1/g(x)*g'(x)# and #g(x)=(n(x))/(d(x))->g'(x)=(n'(x)d(x)-n(x)d'(x))/(d(x))^2#, the first derivative is: #f'(x)=cancel((x^2-1))/x*(1(x^2-1)-x(2x))/(x^2-1)^cancel2# #=(x^2-1-2x^2)/(x(x^2-1))# #=(-x^2-1)/(x(x^2-1))# #=(1+x^2)/(x(1-x^2))# #f'(x)=(1+x^2)/(x-x^3)# The second derivative is: #f''(x)=(2x^2(1-x^2)-(1+x^2)(1-3x^2))/(x^2(1-x^2)^2)# #=(2x^2-2x^4-1-x^2+3x^2+3x^4)/(x^2(1-x^2)^2# #=(x^4+4x^2-1)/(x^2(1-x^2)^2)#

Evelyn Agard · Answer

undefined To find the first and second derivative of $ y = \ln\left(\frac{x}{x^2 - 1}\right) $:

First, find the first derivative $ y' $:
$$ y' = \frac{d}{dx} \ln\left(\frac{x}{x^2 - 1}\right) $$

Apply the chain rule and the derivative of ln:
$$ y' = \frac{1}{\frac{x}{x^2 - 1}} \cdot \frac{d}{dx} \left(\frac{x}{x^2 - 1}\right) $$
$$ y' = \frac{x^2 - 1}{x} \cdot \frac{d}{dx} \left(\frac{x}{x^2 - 1}\right) $$
$$ y' = \frac{x^2 - 1}{x} \cdot \left(\frac{(x^2 - 1) - x(2x)}{(x^2 - 1)^2}\right) $$
$$ y' = \frac{x^2 - 1}{x} \cdot \left(\frac{x^2 - 1 - 2x^2}{(x^2 - 1)^2}\right) $$
$$ y' = \frac{x^2 - 1}{x} \cdot \left(\frac{-x^2 - 1}{(x^2 - 1)^2}\right) $$
$$ y' = -\frac{x^2 - 1}{x(x^2 - 1)} $$

Now, find the second derivative $ y'' $:
$$ y'' = \frac{d}{dx} \left(-\frac{x^2 - 1}{x(x^2 - 1)}\right) $$
$$ y'' = \frac{(-x^2 - 1)(x(x^2 - 1)) - (-x^2 - 1)(x^2 - 1)}{(x(x^2 - 1))^2} $$
$$ y'' = \frac{-(x^3 - x + x^3 - x)}{(x^2(x^2 - 1))^2} $$
$$ y'' = \frac{-2x^3 + 2x}{x^4(x^2 - 1)} $$
$$ y'' = \frac{-2x(x^2 - 1)}{x^4(x^2 - 1)} $$
$$ y'' = -\frac{2}{x^3} $$

So, the first derivative is $ y' = -\frac{x^2 - 1}{x(x^2 - 1)} $ and the second derivative is $ y'' = -\frac{2}{x^3} $.