http://www.ck12.org Chapter 2. Derivatives
f′(x) =limh→ 01 h[ x+h
x+h+ 1 −x
x+ 1]
=limh→ (^01) h
[(x+h)(x+ 1 )−x(x+h+ 1 )
(x+h+ 1 )(x+ 1 )
]
=limh→ (^01) h
[x (^2) +x+hx+h−x (^2) −xh−x
(x+h+ 1 )(x+ 1 )
]
=limh→ 01 h[ h
(x+h+ 1 )(x+ 1 )]
=limh→ 0 (x+h+^11 )(x+ 1 )=(x+^11 ) 2.Example 2:
Find the derivative off(x) =√xand the equation of the tangent line atx 0 =1.
Solution:
Using the definition of the derivative,
f′(x) =hlim→ 0 f(x+hh)−f(x)=hlim→ 0√x+h−√x
h
=hlim→ 0√x+h−√x
h√x+h+√x
√x+h+√x=hlim→ 01 h√xx++hh−+x√x=hlim→ 0 √x+^1 h+√x= 2 √^1 x.Thus the slope of the tangent line atx 0 =1 is
f′( 1 ) = 2 √^11 =^12.Forx 0 =1, we can findy 0 by simply substituting intof(x).
f(x 0 )≡y 0
f( 1 ) =√
1 = 1
y 0 = 1.Thus the equation of the tangent line is