Proofs
Derivative Proofs

Constant Rule (3.2)
The derivative of a constant function is 0. That is, if c is a real number, then
d
c  0 .
dx
Proof:
Let f(x) = c. Then by the definition of the derivative :
d
f ( x  x )  f ( x )
cc
0
 lim
 lim
 lim 0  0 .
 c   lim
x

0

x

0

x

0
dx
x
x
x x0
Quod erat demonstrandum.

Power Rule (3.3)
If n is a rational number not equal to 0 (constant case above), then the function
d
 x n   nx n 1 . For f to be differentiable at x=0, it must
f ( x )  x n is differentiable and
dx
be a number ∋ x n 1 is defined on an interval containing 0.
Proof #1 (by mathematical induction):
Case 1: n=1
d
x  x  x
x
 lim
 lim 1  1  1x 0 .
Let f ( x )  x1 , then
 x   lim
x 0
x 0 x
x 0
dx
x
d
 x n   nx n 1 , then
Case 2: Assume the rule works for n, and prove for n+1, i.e, that is
dx
d
 x n 1   ( n  1) x n .
dx
Let f ( x )  x n , and let g ( x )  x n1  x  x n  x  f ( x ) , then
d n 1
g ( x  x )  g ( x )
( x  x ) f ( x  x )  x  f ( x )
 x   lim
 lim


x

0

x

0
dx
x
x
( x  x )( x  x ) n  x  x n
x ( x  x ) n  x ( x  x ) n  x  x n
lim
 lim

x 0
x 0
x
x
x ( x  x )n  x n   x ( x  x ) n
  ( x  x ) n  x n  x ( x  x ) n 
lim 
 lim  x 


x 0
x  0
x

x

x




 f ( x  x )  f ( x ) 
x lim 
 lim ( x  x ) n  xf ( x )  x n 

x 0
x

 x  0
x  nx n 1   x n  nx n  x n  (n  1) x n
Quod erat demonstrandum.
Proof #2 (by using the binomial theorem):
Case 1: n ∈ : To complete this proof, we will apply the Binomial Theorem:
n
n
( x  a )n     x n i a i , i.e. for example, n=4:
i 0  i 
4
 4
4
 4
4
 4
4
( x  x )4     x 4i x i    x 4 x 0    x 3x1    x 2 x 2    x1x 3    x 0x 4 
i 0  i 
0
1 
 2
3
 4
n
n!
x 4  4 x 3x  6 x 2 x 2  4 xx 3  x 4 , where    nCi 
, thus for
i !( n  i )!
i 
 4
n
4!
24
n!
n(n  1)(n  2)...3  2  1
 2   2!2!  4  6 , and in general, for  1   1!(n  1)!  1  (n  1)(n  2)...3  2  1  n ,
 
 
n
n!
n(n  1)(n  2)(n  3)...3  2  1 n(n  1)
, and
 2   2!(n  2)!  2  1  ( n  2)( n  3)...3  2  1 
2
 
n 
n!
n(n  1)(n  2)...3  2  1
 n  1  (n  1)!1!  (n  1)(n  2)...3  2  1  1  n


Thus, employing this theorem, let f ( x )  x n , then apply the theorem and simplify.
1
x n  nx n 1x  x n 2 x 2  ...  x n  x n
d n
( x  x )n  x n
2
 x   lim
 lim

x  0
dx   x0
x
x
1


1
x  nx n 1  x n 2 x  ...  x n 1 
nx n 1x  x n 2 x 2  ...  x n
2


2
lim
 lim
x 0
x 0
x
x
1
1
lim nx n 1  x n 2 x  ...  x n 1  nx n 1  x n 2 (0)  ...  (0)  nx n 1
x 0
2
2
Case 2: n ∈ . This case requires application of the chain rule, together with the
natural number case above, since all rational numbers can be written as a combination
of roots and powers. Note that both these cases actually exclude the case of n=0 since
that case is handled by the constant case previously addresses.
Case 3: n ∈ R . This case requires advanced calculus to prove and is beyond the scope
of this course, but is essentially a limiting process involving the fact that all real numbers
can be approximated as closely as we like to a rational number, to which we can then
d 
 x    x 1 as we would expect.
apply case 2. So,
dx  
Quod erat demonstrandum.

Constant Multiple (3.4)
If f is a differentiable function and k ∈ R , then kf is also differentiable and
d
 kf ( x )  kf ( x )
dx
Proof:
Let g ( x )  kf ( x ) , then
k  f ( x  x )  f ( x )
d
g ( x  x )  g ( x )
kf ( x  x )  kf ( x )
 lim
 lim
 g ( x)  lim
x

0

x

0

x

0
dx
x
x
x
f ( x  x )  f ( x )
 kf ( x ) .
Since f is differentiable, k lim
x 0
x
f ( x) 1
 f ( x ) , so that the same theorem applies.
Note that
k
k
Quod erat demonstrandum.

Sum & Difference Theorem (3.5)
The sum (or difference) of two differentiable functions f and g is itself differentiable.
Moreover, the derivative of f+g (or f – g) is the sum (or difference) of f and g.
d
 f ( x )  g ( x )  f ( x )  g ( x ) .
dx
Proof:
The proof of the sum and difference rule follows directly from properties of limits.
Let h( x )  f ( x )  g ( x ) . Then
d
h( x  x )  h ( x )
f ( x  x )  g ( x  x )  f ( x )  g ( x )
 lim

 h( x )  lim
x 0
x  0
dx
x
x
f ( x  x )  f ( x )
g ( x  x )  g ( x )
lim
 lim
 f ( x )  g ( x )
x 0
x  0
x
x
(The difference version is left to the reader.)
Quod erat demonstrandum.

Sine and Cosine Derivatives (3.6)
d
d
sin( x )  cos( x)
cos( x)    sin( x)
dx
dx
Proof:
Both these proof will require the use of trigonometric identities for the sum/difference
of angles.
sin(   )  sin  cos   cos  sin  and cos(   )  cos  cos  sin  sin 
Case 1: sin(x)
d
sin( x  x )  sin( x )
sin( x ) cos( x )  cos( x )sin( x )  sin( x )
 lim
sin( x )  lim
x 0
x 0
dx
x
x
Collect the two terms with sin(x) and then factor out sin(x) , and collect the ∆x terms.
sin( x ) cos( x )  1  cos( x )sin( x )
sin( x ) cos( x )  sin( x )  cos( x )sin( x )
 lim
 lim
x  0
x  0
x
x
 cox ( x )  1 
 sin( x ) 
 lim sin( x ) 

lim
cos(
x
)
 x0
 x  
x  0
x

 cox ( x )  1 
 sin( x ) 
sin( x ) lim 
 cos( x ) lim 

x  0
x  0
x


 x 
cos( x )  1
 0 and
The values of these two limits can be determined numerically: lim
x 0
x
sin( x )
lim
 1 , thus we have sin( x ) 0  cos( x) 1  cos( x)
x 0
x
Case 2: cos(x)
d
cos( x  x )  cos( x )
cos( x ) cos( x )  sin( x )sin( x )  cos( x )
 lim
cos( x )  lim
x 0
x 0
dx
x
x
Collect the two terms with cos(x) and then factor out cos(x) , and collect the ∆x terms.
cos( x ) cos( x )  1  sin( x )sin( x )
cos( x ) cos( x )  cos( x )  sin( x )sin( x )
 lim
 lim
x 0
x 0
x
x
 cox ( x )  1 
 sin( x ) 
 lim cos( x ) 
 lim sin( x ) 


x  0
x

 x  0
 x 
thus we have
 cox ( x )  1 
 sin( x ) 
cos( x ) lim 
  sin( x ) lim
x  0
x 0 
x

 x 
cos( x) 0  sin( x) 1   sin( x)
Quod erat demonstrandum.

Exponential e (3.7)
d x
 e   e x
dx
Proof:
 ex  1
d x
e x x  e x
e x e x  e x
x
x
 e   lim
e
 lim
. Factor out the : e lim 
 . We can
x 0
x 0
x 0
dx
x
x
 x 
 e x  1
 ex  1
x
establish the value of lim 
,
thus

1
e
lim
 e x (1)  e x .

x 0
x 0  x 
x




Quod erat demonstrandum.

Product Rule (3.8)
The product of two differentiable functions f and g is itself differentiable. Moreover, the
derivative of fg is the first function times the derivative of the second plus the second
function times the derivative of the first.
d
 f ( x ) g ( x )  f ( x ) g ( x )  f ( x ) g ( x )
dx
Proof:
This proof requires a somewhat opaque step involving adding and subtracting the same
term at one point. Where that term comes from may appear more motivated if you
follow the proof from the end backwards. I’ll point the step out when we get to it.
d
f ( x  x ) g ( x  x )  f ( x ) g ( x )
add and subtract the term
 f ( x ) g ( x )  lim
x 0
dx
x
f ( x  x ) g ( x ) to the numerator to get
f ( x  x ) g ( x  x )  f ( x  x ) g ( x )  f ( x  x ) g ( x )  f ( x ) g ( x )
lim
. Separate the
x  0
x
first two terms and the second two terms, then factor out f ( x  x ) from the first two
terms, and g(x) from the second two. Because we are adding and subtracting the same
value, we have not changed the value of the expression.
 f ( x  x ) g ( x  x )  f ( x  x ) g ( x ) f ( x  x ) g ( x )  f ( x ) g ( x ) 
lim 

 
x 0
x
x

 f ( x  x ) g ( x  x )  f ( x  x ) g ( x ) 
 f ( x  x ) g ( x )  f ( x ) g ( x ) 
lim 
 lim 


x 0
x
x

 x0 
 g ( x  x )  g ( x ) 
 f ( x  x )  f ( x ) 
lim f ( x  x ) 
 lim g ( x ) 

 
x 0
x
x

 x0

 g ( x  x )  g ( x ) 
 f ( x  x )  f ( x ) 
lim f ( x  x ) lim 
 g ( x ) lim 

 
x 0
x 0

x

0
x
x



Because f and g are both differentiable, the second factor of each term are the
derivatives of the respective functions. The first factor of the first term goes to f(x) as
∆x goes to 0. lim f ( x  x) g ( x)  g ( x) f ( x)  f ( x) g ( x)  g ( x) f ( x) , and of course,
x 0
we can reorder by the commutative property.
Quod erat demonstrandum.

Quotient Rule (3.9)
The quotient f/g of two differentiable functions f and g is itself differentiable at all values
of x for which g(x)≠0. Moreover the derivative of f/g is given by the denominator times
the derivative of the numerator minus the numerator times the derivative of the
denominator all divided by the denominator squared.
d  f  f ( x ) g ( x )  g ( x ) f ( x )

dx  g 
 g ( x)2
Proof:
This proof is done very similarly to the proof of the product rule, as it uses the adding
and subtraction of an apparently unmotivated term. As before, you may find it easier to
see where the term comes from by looking at the proof in the reverse order.
f ( x  x ) f ( x )

f ( x)
d  f ( x) 
h( x  x )  h( x )
g ( x  x ) g ( x )
Let h( x ) 
, then
.
 lim
 lim
x 0
g ( x)
dx  g ( x )  x0
x
x
Simplify the complex fractions by multiplying by the common denominator
g ( x ) g ( x  x )
f ( x  x ) g ( x )  f ( x ) g ( x  x )
to get lim
. Here we add and subtract
x 0
g ( x ) g ( x  x )x
g ( x ) g ( x  x )
f(x)g(x) from the numerator.
f ( x  x ) g ( x )  f ( x ) g ( x )  f ( x ) g ( x )  f ( x ) g ( x  x )
lim
. We rearrange the fraction
x 0
g ( x ) g ( x  x )x
to get just ∆x by itself.
f ( x  x ) g ( x )  f ( x ) g ( x )  f ( x ) g ( x )  f ( x ) g ( x  x )
1
lim

. Separate
x 0
x
g ( x ) g ( x  x )
the first two terms of the big fraction and factor out g(x), and factor out -f(x) from the
two remaining terms.
1

 f ( x  x )  f ( x ) 
  g ( x )  g ( x  x )  
lim g ( x ) 
 f ( x) 





x 0 
x
x



  g ( x ) g ( x  x )

.
1

 f ( x  x )  f ( x ) 
  g ( x )  g ( x  x )  
 g ( x ) lim
  f ( x ) lim
   lim
x 0 
x 0 
x
x



 x0 g ( x ) g ( x  x )
Because we know that f(x) and g(x) are differentiable, the two terms in square brackets
are the respective derivatives. In the last factor, we see that g(x+∆x) goes to g(x) as ∆x
1
f ( x ) g ( x )  g ( x ) f ( x )

goes to zero.  g ( x ) f ( x )  f ( x ) g ( x )  
.
2
 g ( x )
 g ( x)2
Quod erat demonstrandum.

More Trig Functions Derivatives (3.10)
d
d
 tan( x )  sec2 ( x )
cot( x)    csc 2 ( x)
dx
dx
d
csc( x )   csc( x) cot( x)
dx
d
sec( x)   sec( x) tan( x)
dx
Proof:
All of the following proofs employ the quotient rule and basic trigonometric identities.
Case 1: tan(x)
d
d  sin( x )  cos( x ) cos( x )    sin( x )sin( x )  cos2 ( x )  sin 2 ( x )



 tan( x )  
dx
dx  cos( x ) 
cos2 ( x )
cos 2 ( x )
1
 sec 2 ( x )
2
cos ( x )
Case 2: cot(x)
2
2
d
d  cos( x )   sin( x )sin( x )  cos( x ) cos( x )  cos ( x )  sin ( x ) 



cot( x)  
dx
dx  sin( x ) 
sin 2 ( x )
sin 2 ( x )
1
 csc2 ( x )
2
sin ( x )
Case 3: sec(x)
d
d  1  (0) cos( x )    sin( x )  (1) sin( x )
sin( x )
1




sec( x)  
2
2

dx
dx  cos( x ) 
cos ( x )
cos ( x ) cos( x ) cos( x )
 tan( x )sec( x )
Case 4: csc(x)
d
d  1  (0)sin( x )  cos( x )(1)  cos( x )
cos( x )
1




csc( x )  
2
2

dx
dx  sin( x ) 
sin ( x )
sin ( x )
sin( x ) sin( x )
  cot( x ) csc( x )
Quod erat demonstrandum.

Chain Rule (3.11)
If y = f(u) is a differentiable function of u and u = g(x) is a differentiable function of x,
dy dy du


then y = f(g(x)) is a differentiable function of x and
or equivalently
dx du dx
d
 f ( g ( x ))  f ( g ( x )) g ( x ) .
dx
Proof:
While this proof can be done with advanced calculus using the definition of the
derivative we have used in previous proofs, that is somewhat beyond the scope of this
course. A more obvious proof is obtainable from one of the alternate forms of the
f ( x )  f (c)
derivative: lim
.
xc
xc
Let h( x )  f ( g ( x )) . Then
d
h( x )  h(c)
f ( g ( x ))  f ( g (c ))
 lim
. Multiply
 h( x )  lim
x c
x c
dx
xc
xc
g ( x )  g (c)
and then collect one of these terms with the numerator and one with
g ( x )  g (c)
f ( g ( x ))  f ( g (c)) g ( x )  g (c)

the denominator. lim
. This second term is clearly the
x c
g ( x )  g (c)
xc
derivative of g(x). The first term is the equivalent for the derivative form of f(u), thus
f ( g ( x )) g ( x ) .
by
Quod erat demonstrandum.
[A good mnemonic for this rule is “the derivative of the outside times the derivative of
the inside”.]

General Power Rule (3.12)
If y  [u( x )]n , where u is a differentiable function of x and n is a real number, then
dy
d n
n 1 du
u   nu n 1u .
 n u( x )
or equivalently
dx
dx
dx
Proof:
Here, we apply the chain rule
together to get
dy dy du dy n
du
 u   nu n 1 , and

 .
 u( x ) . Put them
dx du dx du
dx
dy dy du
n 1


 n  u ( x )  u .
dx du dx
Quod erat demonstrandum.

Natural Logarithm Derivative (3.13)
Let u be a differentiable function of x.
d
1 du u 
ln(u)     , u  0 .
dx
u dx u
d
1
ln( x )  , x  0 and
dx
x
Proof:
The second part of the theorem is a direct application of the chain rule, once the first
part is true, so we will prove on the first part. This proof is a form of implicit
differentiation.
d y
d
 e    x  . Apply the chain rule on the right.
Let y = ln(x), then x  e y 
dx
dx
d
dy
dy 1
dy 1
 e y   e y
 1 . Divide by e y to get
 y . Replace e y  x to get
 .
dx
dx
dx e
dx x
Quod erat demonstrandum.

Derivatives of Bases Other than e (3.15)
Let a be a positive real number (a≠1) and let u be a differentiable function of x.
d
d u
d
1
d
u
 a x   ln(a )a x
a   ln(a )a uu
loga ( x) 
loga (u) 
dx
dx
dx
ln(a ) x dx
ln(a )u
Proof:
As with the logarithm proof, we will consider only the cases with functions of x. The
composite cases are direct applications of the chain rule based on the more general
case.
Case 1: a x
d
d ln( a ) x
d
 a x  
 e
  eln( a ) x  ln(a ) x   eln( a ) x ln(a )  ln(a )a x
dx
dx
dx
Case 2: loga ( x )
d ln( a ) y
d
e
   x   1 . Thus,
dx
dx
dy
1
 y
 a y ln(a )  1 . Divide by a y ln( a ) .
,
dx a ln(a )
1

.
ln(a ) x
Let y  log a ( x )  a y  x  eln( a ) y  x 
d
dy
ln(a ) y   eln( a ) y ln(a )
dx
dx
dy
and then replace a y  x to get
dx
eln( a ) y 
Quod erat demonstrandum.

Derivative of an Inverse Function (3.17)
Let f be a function that is differentiable on an interval I. If f has an inverse function g,
then g is differentiable at any x for which f ( g ( x ))  0 . Moreover,
1
g ( x ) 
, f ( g ( x ))  0 .
f ( g ( x ))
Proof:
Let g ( x )  f 1 ( x ) . Then by properties of inverse functions, we have f ( g ( x ))  x .
d
d
 f ( g ( x ))   x  . By the chain rule on the left, we have f ( g ( x )) g ( x )  1 . Divide
dx
dx
1
1
  f 1  ( x ) 
by f ( g ( x )) and we are done. g ( x ) 
.
f ( g ( x ))
f ( x )
Quod erat demonstrandum.

Derivatives of the Inverse Trigonometric Functions (3.18)
d
1
d
1
d
1
arccos( x ) 
arctan( x ) 
arcsin( x) 


dx
1  x2
1  x 2 dx
1  x 2 dx
d
1
arc cot( x) 
dx
1  x2
d
1
arcsec( x ) 
dx
| x | x2  1
d
1
arc csc( x ) 
dx
| x | x2  1
Proof:
Each of these proofs can be done entirely algebraically, or geometrically. I’ll show the
first one algebraically and geometrically, but the rest will be done only geometrically. In
all problems, x may be either positive or negative, depending on the quadrant of the
angle y. The only exception will be in the arcsecant and arccosecant functions, where x
represents the length of the hypotenuse of the triangle, which is why |x| appears in the
formulas instead. I will do these proofs assuming that x is positive. It should also be
noted here that arcsin( x)  sin 1 ( x) . Different books use different notations but they
mean the same thing.
Case 1: arcsin(x)
Algebraic version:
Let y  arcsin( x ) or by definition, sin( y )  x . Then by the chain rule we have
d
dy d
sin( y )  cos y   x   1 . Divide by cos(y).
dx
dx dx
dy
1
1
1
1




. The rest of the algebra here is more
dx cos( y )
cos2 y
1  sin 2 y
1  x2
apparent working backwards that it appears to be motivated working forwards.
Geometric version:
x
Opposite
sin( y )  
When we build our triangle using this information, we can
1 Hypotenuse
find the final side by using the Pythagorean Theorem.
Thus,
dy
1
Hypotenuse
1

 sec( y ) 

dx cos( y )
Adjacent
1  x2
x
Case 2: arccos(x)
Let y  arccos( x ) or by definition, cos( y )  x . Then by the chain rule we have
d
dy d
cos( y )   sin( y )   x   1 . Divide by -sin(y).
dx
dx dx
Thus,
dy
1
Hypotenuse
1

  csc( y ) 

dx sin( y )
Opposite
1  x2
1
y
1  x2
1
1  x2
y
x
Case 3: arctan(x)
Let y  arctan( x ) or by definition, tan( y )  x . Then by the chain rule we have
d
dy d
 tan( y )  sec2 ( y )   x   1 . Divide by sec2 ( y ) .
dx
dx dx
x
Thus,
dy
1
Adjacent 2
2


cos
(
y
)


dx sec2 ( y )
Hypotenuse2
12

1 x
2


1
1  x2
1  x2
y
Case 4: arccot(x)
1
Let y  arc cot( x ) or by definition, cot( y )  x . Then by the chain rule we have
d
dy d
cot( y )   csc2 ( y )   x   1 . Divide by  csc2 ( y ) .
dx
dx dx
x
Thus,
dy
1
Opposite2
2



sin
(
y
)


dx csc2 ( y )
Hypotenuse2

12
1  x2


1
1  x2
1  x2
y
1
Case 5: arcsec(x)
Let y  arcsec( x ) or by definition, sec( y )  x . Then by the chain rule we have
d
dy d
sec( y )  sec( y ) tan( y )   x   1 . Divide by sec( y ) tan( y ) . Thus,
dx
dx dx
dy
1
Adjacent Adjacent

 cos( y ) cot( y ) 


dx sec( y ) tan( y )
Hypotenuse Opposite
1
1
1


x x2  1 | x | x2  1
x = |x|
1
y
x2  1
Case 6: arccsc(x)
Let y  arc csc( x ) or by definition, csc( y )  x . Then by the chain rule we have
d
dy d
csc( y )   csc( y ) cot( y )   x   1 . Divide by csc( y ) cot( y ) . Thus,
dx
dx dx
dy
1
Opposite Opposite

  sin( y ) tan( y )  


dx csc( y ) cot( y )
Hypotenuse Adjacent
1
1
1
 

2
x x  1 | x | x2  1
Quod erat demonstrandum.
x = |x|
x2  1
y
1