How do you find the second derivative of # ln(x/(x^2+1))# ?

Question

Scarlett Allison · Accepted Answer

#y=ln(x/(x^2+1))#

#y=ln(x)-ln(x^2+1)#

*See logarithmic rules .

Now, if #g(x)=ln(h(x))#, #g'(x)=(h'(x))/(h(x))#, which means that...

#(dy)/(dx)=x^-1-(2x)/(x^2+1)#

So far so good... Now from here, what we get is...

#(d^2y)/(dx^2)=-x^(-2)-((x^2+1)*2-(2x)^2)/((x^2+1)^2)#

Thanks to the quotient rule.

If we simplify the above we get...

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

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

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

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

Evelyn Agard · Answer

undefined To find the second derivative of $ \ln\left(\frac{x}{x^2+1}\right) $, follow these steps:

1. Find the first derivative of the function.
2. Find the second derivative of the resulting function.

Starting with the given function $ f(x) = \ln\left(\frac{x}{x^2+1}\right) $:

1. Find the first derivative, $ f'(x) $:

$$ f'(x) = \frac{d}{dx} \ln\left(\frac{x}{x^2+1}\right) $$

Using the chain rule and the derivative of natural logarithm:

$$ f'(x) = \frac{1}{\frac{x}{x^2+1}} \cdot \frac{d}{dx} \left(\frac{x}{x^2+1}\right) $$

$$ f'(x) = \frac{x^2 + 1}{x} \cdot \frac{(x^2+1) - x(2x)}{(x^2+1)^2} $$

$$ f'(x) = \frac{x^2 + 1}{x} \cdot \frac{x^2 + 1 - 2x^2}{(x^2+1)^2} $$

$$ f'(x) = \frac{x^2 + 1}{x} \cdot \frac{1 - x^2}{(x^2+1)^2} $$

$$ f'(x) = \frac{(x^2 + 1)(1 - x^2)}{x(x^2+1)^2} $$

$$ f'(x) = \frac{1 - x^4}{x(x^2+1)^2} $$

2. Find the second derivative, $ f''(x) $:

$$ f''(x) = \frac{d}{dx} \left(\frac{1 - x^4}{x(x^2+1)^2}\right) $$

$$ f''(x) = \frac{(x(x^2+1)^2)(-4x) - (1 - x^4)(x^2+1)^2 + (1 - x^4)(2x(x^2+1)(2x))}{(x(x^2+1)^2)^2} $$

$$ f''(x) = \frac{-4x(x^2+1)^2 - (1 - x^4)(x^2+1)^2 + 4x^3(1 - x^4)}{(x(x^2+1)^2)^2} $$

$$ f''(x) = \frac{-4x(x^2+1)^2 - (x^2+1)^2 + x^2(1 - x^4)}{(x(x^2+1)^2)^2} $$

$$ f''(x) = \frac{-4x(x^2+1)^2 - (x^4 + 2x^2 + 1) + x^2 - x^6}{(x(x^2+1)^2)^2} $$

$$ f''(x) = \frac{-4x^5 - 4x(x^2+1)^2 - x^4 - 2x^2 - 1 + x^2 - x^6}{(x(x^2+1)^2)^2} $$

$$ f''(x) = \frac{-5x^6 - 4x^5 - 3x^4 - 6x^2 - 1}{(x(x^2+1)^2)^2} $$

So, the second derivative of $ \ln\left(\frac{x}{x^2+1}\right) $ is:

$$ f''(x) = \frac{-5x^6 - 4x^5 - 3x^4 - 6x^2 - 1}{(x(x^2+1)^2)^2} $$