Sätt f(x) = ln( x + (1+x^2)^(1/2)).
Sätt sedan
u(v) = ln(v),
w(z) = z^(1/2),
z(x) = 1+x^2,
v(x) = x + w(z(x))
så att
f(x) = u(v(x))
Vi har
du/dv = 1/v,
dw/dz = (1/2) z^(-1/2) = 1/(2 z^(1/2)),
dz/dx = 2x,
dw/dx = (1/2)(1+x^2)^(-1/2) * 2x = x/(1+x^2)^(1/2)
dv/dx = 1 + dw/dz * dz/dx = 1 + 1/(2 z^(1/2)) * 2x = 1 + x/z^(1/2)
Detta ger
df/dx = du/dv * dv/dx = 1/v * (1 + x/z^(1/2)) = 1/(x + w(z(x))) * (1 + x/z^(1/2))
= 1/(x + (1+x^2)^(1/2)) * (1 + x/(1+x^2)^(1/2))
= (1 + x/(1+x^2)^(1/2)) / (x + (1+x^2)^(1/2))
Förlängning med (1+x^2)^(1/2) får bort de inre divisionerna:
df/dx = ((1+x^2)^(1/2) + x) / ( (1+x^2)^(1/2) (x + (1+x^2)^(1/2)) )
Här kan vi förkorta bort (1+x^2)^(1/2) + x, varefter vi endast får kvar
df/dx = 1/(1+x^2)^(1/2)