1. Differentiability implies continuity
Theorem
If a function f is differentiable at a then f is continuous at a.
(So the theorem says if lim
xa
fx−fa
x−a
Proof
lim
fx
xa
= lim
fx − fa + fa
xa
fx − fa + lim
fa
= lim
xa
xa
fx−fa
= lim
x−a x − a + fa
xa
fx−fa
= lim
⋅ lim
x − a +fa
x−a
xa
xa
′
= f a ⋅ 0 +fa
= fa
= f ′ a (a finite number) then lim
fx = fa )
xa
Addition or subtraction of the same term
Limit law of addition
Multiplication and division by the same term
Limit law of multiplication
f is differentiable
So lim
fx = fa and therefore f is continuous in x = a.
xa
Note: If f is continuous in a then f is not necessarily differentiable in a.
Counter example: fx = |x| is continuous in 0 but not differentiable in 0.
2. (i) A positive derivative indicates that the function is increasing.
Definition of an increasing function:
A function f is increasing on an interval I if fx 2  > fx 1  for any x 1 , x 2 ∈ I such that x 2 > x 1.
Theorem: If f ′ > 0 on an interval I then f is increasing on the interval.
Proof: Pick any two points x 1, x 2 ∈ I such that x 2 > x 1. .
Consider the function f on the interval x 1, x 2 .
1. f is continuous on x 1, x 2 
2. f is differentiable on x 1, x 2 
(f is differentiable on I)
(f is differentiable on I
So, according to the Mean Value Theorem a c ∈ x 1, x 2  exists such that
′
fx 2  − fx 1 
f c =
x2 − x1
But f ′c > 0 and x 2 − x 1 > 0, therefore fx 2  − fx 1  > 0.
It follows that fx 2  > fx 1  and therefore f is increasing on I.
2 (ii) A negative derivative indicates that the function is decreasing.
Definition of an decreasing function:
A function f is decreasing on an interval I if fx 2  < fx 1  for any x 1 , x 2 ∈ I such that x 2 > x 1.
Theorem: If f ′ < 0 on an interval I then f is decreasing on the interval.
The proof of this theorem is similar to that given above.
3. Proof of the product rule for differentiation
d
dx
fx+hgx+h−fxgx
h
h→0
fx+hgx+h−fxgx+h+fxgx+h−fxgx
Subtracting
h
fx+h−fxgx+h+fxgx+h−gx
h
fx+h−fxgx+h
fxgx+h−gx
+lim
h
h
h→0
fx+h−fx
gx+h−gx
lim gx + h +lim fx lim
h
h
h→0
h→0
h→0
′
fx. gx =lim
=lim
h→0
=lim
h→0
=lim
h→0
=lim
h→0
and adding the term fxgx + h
= f ′ xgx + fxg x
4. Derivative of y = sin x
sinx+h−sin x
d
sin x =lim
dx
h
= lim
h→0
=lim
h→0
h→0
sin x cos h+cos x sin h−sin x
h
sin xcos h−1+cos x sin h
h
= sin x lim
= cos x
h→0
cos h−1
h
+ cos x lim
h→0
sin h
h
because lim
h→0
cos h−1
=0:
h
h→0
cos h−1
cos h−1 cos h+1
lim
=lim
. cos h+1
h
h
h→0
h→0
cos 2 h−1
=lim hcos h+1
h→0
−sin 2 h
=lim hcos
h+1
h→0
sin h
=lim h . lim sin h. lim cos−1h+1
h→0
h→0
h→0
cos h−1
h
= 0 and lim
h→0
We show why lim
= 1 × 0 × − 12  = 0
5. Other trig derivatives
d
sin x
tan x = dxd  cos
x 
dx
cos x cos x+sin x sin x
=
cos x 2
= cos12 x
= sec 2 x
d
dx
cos ecx
= dxd sin1 x
x
= −cos
sin 2 x
x
= − cos
. 1
sin x sin x
= − cos ecx. cot x
Derivatives of cot x and sec x follow similarly.
sin h
h
=1