
what is the derivative of square root x ? (what is the derivative of root x ?)
f(x)=x⇒f′(x)=2x1
proof of derivative of square root x (proof of derivative of root x)
Way 1
f′(x)=h→0limhf(x+h)−f(x)
(x)′=h→0limhx+h−x
(x)′=h→0limhx+h−x.x+h+xx+h+x
(x)′=h→0limh.(x+h+x)(x+h−x).(x+h+x)
a2−b2=(a+b).(a−b)
(x)′=h→0limh.(x+h+x)(x+h)2−(x)2
(x)′=h→0limh.(x+h+x)(x+h)2−x2
(x)′=h→0limh.(x+h+x)x+h−x
(x)′=h→0limh.(x+h+x)h
(x)′=h→0limx+h+x1
(x)′=x+0+x1
(x)′=x+x1
(x)′=2.x1
Way 2
f(x)=x
[f(x)]2=(x)2
[f(x)]2=x2
[f(x)]2=x
{[f(x)]2}′=(x)′
f(x)=[g(x)]n⇒f′(x)=n.[g(x)]n−1.g′(x)
2.[f(x)]2−1.f′(x)=1
2.f(x).f′(x)=1
f′(x)=2.f(x)1
f′(x)=2.x1
Way 3
x=x21
lnx=lnx21
lnx=21.lnx
(lnx)′=(21.lnx)′
(lnx)′=21.(lnx)′
f(x)=lng(x)⇒f′(x)=g(x)g′(x)
x(x)′=21.x(x)′
x(x)′=21.x1
x(x)′=2.x1.1
x(x)′=2.x1
(x)′=x.2.x1
(x)′=2.xx.1
(x)′=2.xx
(x)′=2.x2x
(x)′=2.x.xx
(x)′=2.x.xx
(x)′=2.x1