You can simply use the chain rule:
d/dx ln(x+sqrt(1+x^2)) = (1/(x+sqrt(1+x^2)) )d/dx(x+sqrt(1+x^2))ddxln(x+√1+x2)=(1x+√1+x2)ddx(x+√1+x2)
d/dx ln(x+sqrt(1+x^2)) = (1/(x+sqrt(1+x^2)) ) (1+x/sqrt(1+x^2))ddxln(x+√1+x2)=(1x+√1+x2)(1+x√1+x2)
d/dx ln(x+sqrt(1+x^2)) = (1/(x+sqrt(1+x^2)) ) (x+sqrt(1+x^2))/sqrt(1+x^2)ddxln(x+√1+x2)=(1x+√1+x2)x+√1+x2√1+x2
d/dx ln(x+sqrt(1+x^2)) = 1/sqrt(1+x^2)ddxln(x+√1+x2)=1√1+x2
or you can note that:
f(x) = ln(x+sqrt(1+x^2))f(x)=ln(x+√1+x2)
is the logarithmic form of the inverse hyperbolic sine function so that:
y= ln(x+sqrt(1+x^2)) <=> sinhy = xy=ln(x+√1+x2)⇔sinhy=x
differentiating the second expression implicitly:
d/dx (sinhy) = 1ddx(sinhy)=1
coshy dy/dx = 1coshydydx=1
dy/dx = 1/coshydydx=1coshy
and from the identity:
cosh^2y-sinh^2y = 1cosh2y−sinh2y=1
we then get:
coshy = sqrt(1+sinh^2y) = sqrt(1+x^2)coshy=√1+sinh2y=√1+x2
and:
dy/dx = 1/sqrt(1+x^2)dydx=1√1+x2
