Derivatives and Antiderivatives. Calculate the first and second derivatives of

Question

Simplify your answers, then use your results to solve the IVP:

Answer 1

The absolute sign can be omitted for x<-1 and x>1 and in the region -1<x<1 f(x) = xln((1-x)/(1+x))

The initial value problem happens inside the -1 to 1 interval, so we can focus on that. You have a similar approach for the outer domain and actually get the same derivative (with an absolute sign in the log.

For f'(x) = ln(|(1-x)/(1+x)|) + 2x/(x²-1)

f''(x) = -4/(x²-1)²

y"=- -1/4 f"(x)

y' = -1/4f'(x) + C₁ f'(x) approaches 0 as x approaches 0, so C is -1

y = -1/4f(x) - x + C₂ f(x) goes to 0 as x approaches 0 , so C₂ = 3

y = -1/4 f(x) -x + 3