How do you differentiate #ln[sqrt(2+x^2)/(2-x^2)]#?
1 Answer
Mar 1, 2017
Explanation:
We will use the laws of logarithms to rewrite.
#y = ln[sqrt(2 + x^2)/(2 - x^2)]#
#y = lnsqrt(2 + x^2) - ln(2 - x^2)#
#y = ln(2 + x^2)^(1/2) - ln(2 - x^2)#
#y = 1/2ln(2 + x^2) - ln(2 -x^2)#
Now, by the chain rule, we have:
#y' = (2x)/(2(2 + x^2)) - (-2x)/(2 - x^2)#
#y' = x/(2 + x^2) + (2x)/(2 - x^2)#
#y' = (x(2 - x^2) + 2x(2 + x^2))/((2 + x^2)(2 - x^2))#
#y' = (2x - x^3 + 4x + 2x^3)/((2 + x^2)(2 - x^2))#
#y' = (x^3 + 6x)/((2 + x^2)(2 - x^2))#
#y' = (x(x^2 + 6))/((2 + x^2)(2 - x^2))#
Hopefully this helps!
