Starting with:
d / dx (-ln(x - (x^2+1)^(1/2)))ddx(−ln(x−(x2+1)12))
use the chain rule to get:
d / dx (-ln(x - (x^2+1)^(1/2)))ddx(−ln(x−(x2+1)12))
= -1/(x-(x^2+1)^(1/2)) *d/dx (x - (x^2+1)^(1/2))=−1x−(x2+1)12⋅ddx(x−(x2+1)12)
use the chain rule once again on the remaining derivative:
= -1/(x-(x^2+1)^(1/2)) * (1 - 1/2(x^2+1)^(-1/2)(2x))=−1x−(x2+1)12⋅(1−12(x2+1)−12(2x))
Simplify:
= (x/(x^2+1)^(1/2) -1 )/(x-(x^2+1)^(1/2))=x(x2+1)12−1x−(x2+1)12
Note that 1=(x^2+1)^(1/2) / (x^2+1)^(1/2)1=(x2+1)12(x2+1)12, then substitute this for 1:
= (x/(x^2+1)^(1/2) - (x^2+1)^(1/2) / (x^2+1)^(1/2) )/(x-(x^2+1)^(1/2))=x(x2+1)12−(x2+1)12(x2+1)12x−(x2+1)12
= ((x - (x^2+1)^(1/2)) / (x^2+1)^(1/2) )/(x-(x^2+1)^(1/2))=x−(x2+1)12(x2+1)12x−(x2+1)12
= ((x - (x^2+1)^(1/2)) / (x^2+1)^(1/2) ) divide (x-(x^2+1)^(1/2))=⎛⎜⎝x−(x2+1)12(x2+1)12⎞⎟⎠÷(x−(x2+1)12)
= ((x - (x^2+1)^(1/2)) / (x^2+1)^(1/2) ) * 1/(x-(x^2+1)^(1/2))=⎛⎜⎝x−(x2+1)12(x2+1)12⎞⎟⎠⋅1x−(x2+1)12
= (x - (x^2+1)^(1/2)) / (x-(x^2+1)^(1/2) ) * 1/((x^2+1)^(1/2))=x−(x2+1)12x−(x2+1)12⋅1(x2+1)12
= 1 * 1/((x^2+1)^(1/2)) =1⋅1(x2+1)12
= 1/((x^2+1)^(1/2)) = 1/sqrt((x^2+1)).=1(x2+1)12=1√(x2+1).
Finally:
d / dx (-ln(x - (x^2+1)^(1/2))) = 1/sqrt((x^2+1)).ddx(−ln(x−(x2+1)12))=1√(x2+1).
If you have any questions about the use of the chain rule or any other part of this solution, then please ask.
Rory.
