Pochodna Analiza: ile wynosi pierwsza i druga pochodna z

	lnx
f(x)=
	√x

Janek191: ln x f(x) = −−−− ; x > 0 √x zatem ( ln x )' * √x − ln x * ( √x ) ' f '(x) = −−−−−−−−−−−−−−−−−−−−−−−−− = ( √x )² (1/x) * √x − ln x * ( 1 / 2 √x ) = −−−−−−−−−−−−−−−−−−−−−−− = x x⁻¹ * x^1/2 − ln x * 0,5 x ^{− 1/2} = −−−−−−−−−−−−−−−−−−−−−−−−−−−−− = x = x^{− 3/2} − ln x *0,5 x^{− 3/2} =========================== oraz f " (x) = ( − 3/2) x ^{− 5/2} − [ ( 1/x) *0,5 x^−3/2 + ln x *( −3/4) x ^−5/2 ] = = ( − 3/2) x ^{− 5/2} − 0,5 x ^{− 5/2} + ln x * ( 3/4) x ^{− 5/2} = = [ ( −6/4) − (3/4) + (3/4) ln x ] * x ^{− 5/2} = = [ 0,75 ln x − 2,25 ] * x ^{− 5/2} ==========================

Analiza: Dzięki