Course: Accelerated Engineering Calculus I Instructor: Michael Medvinsky
10.2
Derivatives of Elementary functions (3.1):
Ex 1.
Ex 2.
Ex 3.
Ex 4.
=
( )
( )
=
=
;
;
( )
( )
'
= h lim
→
0
'
= h lim
→
0
( ) = x 2 ;
( )
'
= lim h
→
0 lim h
→
0
( h
+
2 a
) =
2 x
( ( )
( + x h
) − ( )
(
+
+ x
) h x h
)
−
−
( ) = x
3
; f '
( ) = lim h
→
0
( x h
)
3 − x
3 h
( )
=
=
= lim h
→
0 lim h
→
0 lim h
→
0 c
− c
(
=
)
0 h h
+ x
− x
( h
+ h x h
)
2 −
=
1 x
2
= h lim
→
0 h
2
= h lim
→
0 x
3 +
3 x h
+
3 xh
2 h x
3 h lim 3 h
→
0 x
2 +
3 xh h
2 =
3 x
2
+
2 xh
+ x
2 − x
2 h
=
= lim h
→
0
3 x h
+
3 xh
2 + h
3 h
=
10.2.1
Power Rule
(
Binomial Thm (found on ref page 1 of the book): x
+ y
) n = x n + nx n
−
1 y
+
(
2
−
1
) n
−
2 2 x y
( −
1
) (
1 2 3
 k
Def(The Power Rule):
( ) = x n
, f '
( ) = lim h
→
0
( x
+ h
) n − h x n
=
=
= lim h
→
0 x n + nx n
−
1 h lim h
→
0


 nx n
−
1
+
+
2
( −
1
)
( −
1
)
2 x n
−
2 h x n
−
2 h
2
( −
1
) ( n k 1
)
 k h
( −
1
) ( n k 1
)
 k x n k h k x n k h k
−
1
1
) n k k x y nxh n
−
1 nxh n
−
2
+ nxy n
−
1 h n
−
1



=
+ y n
+ h n − x n nx n
−
1
=
Ex 5.
Ex 7.
d dx x
1234567 =
1234567 x
1234566
Ex 6.
( )
=
=
1 x n
( x n
−
1
) (
; f '
( ) n
−
1
)
=
… = lim h
→
0 n ···
( n
− +
1
) ( x
) (
1

1
+
−
1 h x h x

= h lim
→
0
) = n !
( n
− k
1 h

 x
( x
(
+ x
) h
) 

)
!
( x
) (
= h lim
→
0
) … = n !
0!
x
0
( x
−
+
1
)
= − x
1
2
from the d 1 other side one can do dx x
Ex 8.
Ex 9.
d dx d 1 dx x x
=
( )
'
= lim h
→
0
= −
1
2 x
=
1
+
−
1 x h x h
= lim h
→
0
' 1 x
= −
1
2 x
−
1/ 2
= −
1
2 x
= −
2
1 x
2
1 1
1 x h
− h
(
+ 1 x x
)
2
= h lim
→
0
( + x
) ( + ) h
(
1 x h
+ 1 x
)
= lim h
→
0 h
− h
( )
(
1 x h
+ 1 x
)
= −
1
2 x
3/ 2
Def : For any real number
α
, we define d dx x
α = α x
α −
1

Course: Accelerated Engineering Calculus I Instructor: Michael Medvinsky
10.3
Derivative of
( ) and
( ) ( )
Consider f(x) is differentiable, then for any constant c, the function g(x)=cf(x) have the following derivative:
( ) = h lim
→
0
( + h h
) ( )
= h lim
→
0
( + h h
) ( )
= c h lim
→
0
( + h h
) ( )
= cf '
( )
For any two differentiable functions f(x) and g(x) the following is true:
= d dx
{ h lim
→
0
( + h h
) −
}
= lim h
→
0
( )
±
( h lim
→
0
( +
( h
) (
+ h h
−
+ h h
) )
=
− d dx
( ) ( )
± d dx
=
10.4
Derivative of Exponential function a x (3.1)
Consider
( ) = a x , then f '
( ) = lim h
→
0 a h
− a x
= h lim
→
0 a x
( a h −
1
) h
= a x lim h
→
0 a h h
−
1
= a x ⋅ f which means a x is differentiable if and only if it differentiable at x=0. We can’t prove the existence of this limit by the tools we have learned by now. The curious can use the table method we learned at the beginning and find out that for several values of a it is certainly exists. The other one can be convinced by observing the graph of the function, there is no corners, kinks, jumps, or sharp change, i.e. the tangent line isn’t a vertical line.
The more convenient base a will obviously be the one that satisfy f
( ) =
1 . We already have learned such number; this is e , which we now can redefine as a number that satisfy h lim
→
0 e h h
−
1
=
1 . We actually defined e very similar before, since we required the slope of the tangent line at x=0 to be equal 1, but this exactly what we ask here.
We just showed that d dx e x = e x .
10.5
The Product Rule (3.2)
Unlike the Sum and Difference Rule
'
= ( ⋅ )
'
≠ ⋅ = ⋅ (but we know that d dx
{ }
≠ f '
( ) ( )
, for example
'
=
2 x ). Let us develop the right formula.

Course: Accelerated Engineering Calculus I Instructor: Michael Medvinsky
Let’s f(x) and g(x) be differentiable. Define
= lim h
→
0
⋅ = h h
∆ h f
= ( + h
) − ( )
( ) ( )
+ f '
( ) ( ) and solve for to get d dx
=
= h lim
→
0 h lim
→
0
( + h
) = ( ) + ∆ f
= h lim
→
0
(from now we’ll omit the h in
(
+
) (
+ h
)
−
( ) ( ) h
=
(
+ ∆ f
) ( ( ) g
)
+ ∆ h
( ) + ∆ ( ) h
( ) ( )
= f g
( ) ( )
= h lim
→
0
∆ h ).
∆ ( ) + ∆ ( ) h h lim
→
0 f x g '
∆ g h
( ) ( )
+
+ f ' h lim
→
0
∆ f h
( ) ( )
+ +
∆ f
∆ g h lim lim lim
→
0 h h
→
0 h h
→
0
= h
( + )
=
Ex 1.
Ex 2.
Ex 3.
Ex 4.
( )
'
= ( ⋅ )
'
≠ ⋅ + ⋅ =
2 x d dx
( )
= x 3 d dx e x + e x d dx x 3 = x e x + ⋅
3 x 2 d
2 dx 2
( e x 3
)
= d dx
( x +
3 x
)
= d dx
( x
)
+ d dx
( e x ⋅
3 x 2
)
= x +
3 x + d d
+
3 x
2 e x + e x
3 x
2 = x +
3 x +
3 x e x + e x ⋅
6 x
= x e x +
6 x +
6 xe x dx dx
(
5
+ x
3
) (
2
+ x
)

′ ( x
3
′ ) (
2 x
) ( x
3
) (
2
+ x
) ′ =
3 x
2
(
2 x
) (
5 x
3
)
=
4 x
3 +
6 x
2 +
5 d
 


1 dx
 x
+ x

1 x
− x



=
'
= −
1
+
 x
2
1

1 x
− x
+
 
1 x
+ x

− x
1
2
1

Ex 5.
Ex 6.
Ex 7.
=
(
4
−

1 x
−
( ) ) x
 2 x
−

1 x
+ x
 2
''
=
 4
3 + 4 x f x f
= −
2 x
1
2
+ x x
2
=
− x
3
2
−
′ 
′
= 
12
2 +
4
3 x f x f
(
4
( ) )
=
 12
2 +
8
3
′ + 4 x f x f x f ''
′
=
24 xf
2 x
′  
4
3 ′ + 4 x f x f
+ (
12
+
24
)
2 x f
′
'' 
=
12
2 +
8
3 ′ + 4 x f x f x f
( )
3
''
+ 4 x f x f
''
10.6
The Quotient Rule (3.2)
Let’s f(x) and g(x) be differentiable.
= d
( )
( )
1
( ) h lim
→
0
= h lim
→
0
( f x h
( +
)
)
−
( )
( ) h
= h lim
→
0
( ) ( ) ( ) (
( ) ( )
+ ) h
=
1
( )

 h lim
→
0
( ) ( ) + ∆ ( ) )
−
( ( ) ( ) + ∆ ( )
( + h
)
)
=
(
+ ∆
1
( ) lim h
→
0
) ( ) − ( ) ( )
( + h
)
∆ ( )
(
− ∆
+ h
)
( )
=
+ ∆ g
) 
 =

Course: Accelerated Engineering Calculus I Instructor: Michael Medvinsky
=
=
( )
( )
∆ f lim lim h
→
0 h h
→
0
( ) g x f ' x
1
( ) ( ) ( )
−
(
1
+ h
)
− f x
( )
( )
∆ g lim lim h
→
0 h h
→
0
1 g x g x
=
(
1 g x
+ h
)
=
( ) ( ) g 2
−
( )
( ) ( )
Ex 1.
Ex 2.
Ex 3.
( d 1 dx x
= d 3 x
3 dx x
2 −
3
− x x
2
2
0 1 1 x
2
= − x
1
2 x
2
−
3
+
4 x
=
( x
2 −
3
)(
3 x
3
)(
9 x
2 −
4 x
(
+ x
4
2
)
−
−
3
)
2 x 3 x
2
(
3 −
−
2 x
2 x
2
2
+
+
4
4 x
(
) x x 2
) (
=
−
3
3 x x
2
4
)
2
−
−
) (
31 x
2
−
+ x
3 −
2 x
2
1 2 x
−
12
( x 2 3
)
2
+
4 x
)
=
= d dx

 x x
5
3
+
+ x
3
2
5 x
7 +
2 x
4


+
=
15
(
( x x x
4
3
5 +
+
+
6
3 x x
)
2
2
′ ) (
−
( x
3
3 x
7
(
3 x
+
3
) (
3
+ x
4
3
)
) x
5
2
=
+ x
2
)( x
3 +
3
′ )
2 x
7 +
14 x
4 +
6 x
6 +
6 x
3 +
9
= x
(
5 x
4 +
2 x
)(
( x
3 x
3
3 x
5 + x
2
3 x
+
) (
3
)
2
)
2
=
10.7
Derivatives of Trigonometric functions (3.3) d
Before we start with sin dx x , let remember that lim h
→
0 sin h h
=
1 and do lim h
→
0 cos h
−
1 cos h
+
1 h cos h
+
1
= lim h
→
0 h
( cos
2 h
−
1 cos h
+
1
)
= h lim
→
0 h
(
− sin
2 h cos h
+
1
)
= − lim sin lim h
→
0 h
→
0 sin h h h lim
→
0
(
1 cos h
+
1
)
=
0
( ) 1
=
0
Now we ready:
=
= d dx sin x
= h lim
→
0 sin
( x
+ h h
) − sin lim h
→
0 sin cos h h
− sin x sin x
⋅ + x
⋅ = cos
+ x lim h
→
0 x
= cos sin h h lim h
→
0 sin cos
= x h h
→
+
0 h cos x h h
−
1 h
+
− sin x
= x h
→
0 sin h h
=
Similarly

Course: Accelerated Engineering Calculus I Instructor: Michael Medvinsky d dx cos x
= lim h
→
0
= x h
→
0 cos
( x cos h
−
1
− h
+ h h
) − cos x x h
→
0 h
= sin h h lim
→
0 cos cos h
− x h
− cos h x
= cos x
⋅ − x
⋅ = − sin x
=
Finally, more interesting: d dx tan x
= d sin x dx cos x
= cos x d dx sin x
− sin cos
2 x d x dx cos x
= cos
2 x
+ sin
2 x cos
2 x
2 tan x
=
1 cos
2 x
= sec
2 x
10.8
The Chain Rule (3.4) d dx
= lim h
→
0
( )
(
=
( ( + h
) )
−
( ( ) ) h lim
→
0
+ ∆ h g
)
−
( h
( ) )
∆ h g
∆ h g lim h
→
0 h
=
= lim h
→
0
(
+ ∆ f '
( ( ) ) g '
( ) h h g
)
−
( ( ) )
⋅
∆ h g
∆ h g
=
Thus we got: d dx
 outer function
( inner
( ) function
)
= f '
( derivative of
( ) outer function evaluated at inner
) g inner
( ) function x derivative of function
10.8.1
Combination of Power Rule with the chain rule d
{ n
}
= ( ) n
−
1
'
( ) dx
Quotient Rule using Product and Chain Rules: d
( )
( )
= d dx g x dx
( ) ( ) −
1 = ( ) −
1 d dx
( ) + ( ) d dx
( ) −
1 = f '
( ) ( ) g
2
− f x g
( )
( ) ( )
Derivative of exponential function a x
Ex 1.
d dx a x = d dx e ln a x = d dx e x ln a = e x ln a d dx
{ x ln a
} = e ln a x ln d x a x a ln a dx
= d dx




 1
+ s s 2
 12
=
12
°
1
+ s s 2

′
1
+ s s
2
 11
1
+ −
2 s
2
1
+ s
2
=
12
1
1
− s
2 s
=
(
12 x
11
)
°
1
+ s s
2
 11 s
1
+ s 2
 
*
1
+ s s 2

′
=

Course: Accelerated Engineering Calculus I Instructor: Michael Medvinsky
Ex 2.
d x
3 −
2 dx x
+
1
= d dx


° x x
3 −
2
+
1


=
2
1 x
3 x
−
2
+
1
·
3 x
2
( x 1
) ( x
3 −
2
)
( x
+
1
)
2
=
=
2 x x
3
+
−
1 2
2
· x
(
3 + x
3
+ x
1
)
2
2
+
2
=
2 x
3 x
3
−
+
2
(
3 x x
2
+
+
1
)
2
3/ 2
Ex 3.
=
=
2 d dx cos
ν tan
ν +
1
1 cos
2
ν sin
ν + cos
ν
=
· d dx cos
ν sin
ν
+ cos
ν
1
= d dx
− ν ν ( sin
ν + cos
ν
( cos
2
ν sin
ν + cos
ν
) − sin
ν + cos
ν
= cos
2
ν ( cos
ν − sin
ν
)
2 sin
ν + cos
ν
2 cos
ν
·
−
(
2
ν cos
ν + cos
2
ν sin
ν
(
) sin
ν + cos
ν
+
)
2 cos
3
ν
)
− cos
2
ν sin
ν
=
=
( ν ) cos
2
ν −
(
2 sin
2
ν
ν +
− cos
3c
ν ν
ν )
3/ 2
10.8.2
Tangent lines to Parametric Curves
Recall: parametric curves are given by a pair
( ) =
( ( ) ( ) )
.
Suppose f(t), g(t) are differentiable. Would we able to write the curve in a form y=F(x) we could simply find the slope of the tangent line by using m
= dy
= dx
F '
( )
. Let us assume for a moment that such F exists and use the parameters. We get
( ) y
( ) =
( ( ) )
If we derive the outer part of this equation (
( ) =
( ( ) )
) to get
'
( ) =
F '
( ( ) ) f '
( ) we actually get dy
= dt
'
( ) =
F '
( ( ) ) f '
( ) =
F '
( ) dx
= dy dx dt dx dt
, dx assuming that 0 dt
≠ ,we arrive at dy dx dy
= dt = dy dt dx dt dx
. dt

Course: Accelerated Engineering Calculus I Instructor: Michael Medvinsky
Def : A line normal to a curve at x=a is a line perpendicular line to the tangent line to the curve at x=a. For example, if y=mx+b is tangent to a curve, then y=-x/m+c is it’s normal.
Ex 1.
Find a tangent and normal line to t
)
at t
=
π
4
: m
= dy
= dx
−
3sin t
2 cos t x
= π
/ 4
=
3 1 / 2
= −
3
2
and we arrive at tangent line of the form y
= −
3
+
2 x b , in order to determine b we need to make this line to go through the point
( ) =

π
2 sin , 3cos
4 t
π
4
=
 

2 3
,
2 2


, i.e.
3
2
3 2
2 2 b b
3
2
+
3 2
2
=
3
+
3
=
6
2 2 2 2
Similarly one find a normal line by using the form y
=
2
3
+
3
2
2 2
3 2 c c
3
2
−
2 2
3 2
=
9 4
3 2
=
5
3 2 and substituting