Differentiation for the Beginners
Rules of Differentiation
1. Constant Rule
2. Power Rule
3. Constant Multiple Rule
4. Sum and Difference Rule
5. Product Rule
6. Quotient Rule
7. Chain Rule
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Let’s take
a simple example:
Function = (2x
+ 5)
(2x + 5) is
a function of X, and we can write it as,
f(x) =
(2x + 5)
(Hint: (3t+5)
is a function of t, and we can write it as, f(t)=(3t+5) )
The symbol
for differentiation of X’s function is:
d/dx [f(x)]
(Hint:
The symbol for differentiation of t’s function, we can write it as, d/dt
[f(t)] )
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We can write
the differentiated function as:
f ‘ (X)
It means:
The
function of X is equal to [any function] ( f(X) = [any function] )and
after differentiating it, we can write it as f ‘ (X):
f ‘ (X) = [differentiated any
function]
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Rule
Number 01: THE CONSTANT RULE
d/dx [C] = 0
(C is a Constant)
Ex:
1) f (x) = 5 (5 is a constant)
f ’ (x) = 0
2)
f (x) = 255 (255
is a constant)
f ’ (x) = 0
3)
f (x) = -5 (-5 is
a constant)
f ’ (x) = 0
Rule
Number 02: THE POWER RULE
d/dx [Xn] = n X n-1
(n is any real number)
Ex:
f (X) = X3
f ’ (X) = 3X2
Let’s take the above
example:
Here, as per the
rule: n is equal to 3 and (n-1) is equal to 2.
Xn = X3
nX(n-1) = 3X(3-1)
The answer is 3X(3-1) is equal to 3X2
Ex:
1. f (X) = X4
f ’ (X) = 4.X4-1
= 4X3
2. f (X) = X0
f ’ (X) = 0.X0-1
= 0
3. f (X) = X = X1
f ’ (X) = 1.X1-1
= 1.X0
= 1 ;
(X0 = 1)
Rule
Number 03: THE CONSTANT MULTIPLE RULE
d/dx [C.f(X)] = C.f ’ (X)
(C is a Constant)
Ex:
f (X) = 9 X2
f ‘ (X) = 9.2.X(2-1)
= 18X
Let’s take this example:
[C.f(X)] = 9X2
In here,
C is equal to 9;
f(X) is equal to X2 ;
f ‘ (X) is equal to 2.X(2-1) ;
Then:
d/dx C [f (X)] = C.[ f ’ (X)]
= 9.[ f
’ (X)]
= 9 [2.X(2-1)]
= 9.2. X(1)
= 18X
Rule
Number 04: THE SUM and DIFFERENCE RULE
SUM RULE:
d/dx [ f(X) + g(X) ] = d/dx f(X) + d/dx
g(X)
[ g(X) is also an another function as f(X) ]
f (X) = 6X2 + 5X2
f ‘ (X) = d/dx (6X2)
+ d/dx (5X2)
= 6.2.X(2-1)
+ 5.2.X(2-1)
= 12X + 10X
= 22X
f (X) = 3X2 + 6X1
f ‘ (X) = d/dx (3X2) +
d/dx (6X1)
= 3.2.X(2-1)
+ 6.1.X(1-1)
= 6X + 6X0 ; (X0 = 1)
= 6X + 6.1
= 6X + 6
DIFFERENCE RULE:
d/dx [ f(X) - g(X) ] = d/dx f(X) - d/dx
g(X)
[ g(X) is also an another function as f(X) ]
Ex:
f (X) = 5X2 – 3X
f ‘ (X) = d/dx (5X2)
- d/dx (3X1)
= 5.2.X(2-1) – 3.1.X(1-1)
= 10X1
- 3X0 ; (X0 = 1)
=
10X – 3
f (X) = 2X – 5X3
f ‘ (X) = d/dx (2X1)
- d/dx (5X3)
= 2.1.X(1-1) – 5.3.X(3-1)
= 2X0
- 15X2 ; (X0 = 1)
=
2 – 15X2
Rule
Number 05: THE PRODUCT RULE
d/dx [f
(X) . g (X)] = [ f (X) . g’ (X) ] + [ g (X)
. f ‘ (X) ]
f (X) and g (X) are functions of X.
When the
first function f(X), is multiplied by the second function g(X), the answer
for its differentiation is:
(First
function multiplied by differentiation of second function + Second function multiplied by
differentiation of the first function)
Ex:
f (X) . g (X) = 4X2 .
(5X-2)
d/dx [4X2
. (5X-2)] = [4X2. d/dx (5X1 – 2)] + [(5X1
– 2) . d/dx (4X2)]
f ‘ [4X2 . (5X-2)] = 4X2 . [5.1.X(1-1) - 0]
+ (5X1 – 2) . [4.2.X(2-1)]
= 4X2.[5-0]
+ (5X1 – 2).[8X]
= 20X2
+ 40X2 – 16X
= 60X2
– 16X
f (X) =
9X3 . (2X4 -1)
f ‘ (X) = 9X3 . d/dx [(2X4
-1)] + (2X4 -1). d/dx [9X3]
= 9X3 . [2.4.X(4-1)
- 0] + (2X4 -1) . [9.3.X(3-1)]
= 9X3 (8X3) + (2X4 -1).(27.X2)
= 72X6 + 54X6
– 27X2
= 126X6 – 27X2
Rule
Number 06: THE QUOTIENT RULE
d/dx [ f
(X) / g (X) ] = [ g (X) . f ‘
(X) - f (X) . g ‘ (X) ] / g (X)2
g (X) ≠ 0
f (X) and g (X) are functions of X.
When a
function, f (X), is divided by another function, g (X), the answer for
its differentiation is:
(Denominator
function is multiplied by differentiation of Numerator function) MINUS (Numerator
function is multiplied by differentiation of Denominator function) and the whole answer is DIVIDED by the square value of the denominator function.
Ex:
f (X) = 3X / (X - 1)
f ‘ (X) = [(X - 1). d/dx (3X) ] – [(3X). d/dx (X-1) ] / (X-1)2
= (X - 1).(3.1.X(1-1)) -
3X.(1.X(1-1) - 0) / (X-1)2
= [ (X - 1).3 - 3X.1
] / (X-1)2
= [ (X - 1).3 - 3X ]
/ (X-1)2
Ex:
f (X) =
5X2 / (X2 - 3)
f ‘ (X) = [ (X2 - 3) . d/dx (5X2) - 5X2 . d/dx (X2 - 3) ] / (X2
- 3)2
= [ (X2 - 3). 5.2.X(2-1) - 5X2 . 2.X(2-1) - 0 ] /
(X2 - 3)2
= [ (X2 - 3).10X - 5X2
.2X ] / (X2 - 3)2
= [ 10X3 – 30X – 10X3
] /
(X2 - 3)2
= [ -30X ] / (X2 - 3)2
Rule
Number 07: THE CHAIN RULE
d/dx (y) = d/du (y) . d/dx
(u)
(If y =
f (u) is a differentiable function of u, and u = g(x) is a differentiable
function of x, then y = f (g (x)) is a differentiable function of x)
Ex:
f (X) = (5X -3)3
f ‘ (X) = [3. (5X -3)(3-1)]. d/dx (5X1-3)
= [3.
(5X -3)2 ].
[5.1.X(1-1) - 0]
= [3. (5X -3)2
] . [5]
= 15. (5X -3)2
Ex:
f (X) = (X2 -1)4
f ‘ (X) = [4. (X2 - 1)(4-1)]. d/dx (X2 -
1)
= [4.
(X2 - 1)3 ]. [2.X(2-1) - 0]
= [4. (X2
- 1)3 ] .
[2X]
=
8X. (X2 - 1)3
Simple Examples
for the Beginners:
1. y = ln |(1 + X)|
d/dx y = d/dx . ln |(1 + X)|
=1/(1 + X)
2 Rule no 07: Chain Rule Example
X = t2 + 2
Y = t3
– 1
Find d/dx
Y;
X = t2
+ 2
Here, X
is a function of t. Then;
d/dt X = d/dt
(t2 + 2)
= 2t < 1st answer
Y = t3
- 1
Here, Y
is a function of t. Then;
d/dt Y =
d/dt (t3 - 1)
= 3t2
< 2nd answer
According
to the Chain Rule;
d/dx Y = d/dt Y
. d/dx t
But from
the first answer;
d/dt X is
equal to 2t
Then
inverse it;
d/dx t is
equal to 1/2t
According
to the Chain Rule;
d/dx Y = d/dt Y .
d/dx t
= 3t2
. 1/2t
= (3/2) t
3. Rule no 01, Rule no 02 : Constant Rule and Power Rule
g(X) = 5 . √X + 4
g ‘ (X) = d/dx (5 . √X + 4)
= d/dx (5 . X(1/2)
+ 4)
= 5 . [ (1/2)X(1/2 - 1) ]+ 0
= 5/2 . X(-1/2)
4. Y = 4X(-2) + 2X2
d/dx Y = d/dx 4X(-2) + d/dx (2X2)
= 4.(-2).X(-2-1) +
2.2.X(2-1)
= -8.X(-3) + 4X
5. Rule no 05 : The Product Rule
f (X) = (X3 – 3X) . (2X4
+ 5X)
d/dx (X) =
(X3 – 3X) d/dx (2X4 + 5X1) + (2X4
+ 5X) d/dx (X3 – 3X1)
= (X3 – 3X) . [ 2.4.X(4-1)
+ 5.1.X(1-1) ] + (2X4 + 5X). [ 3X(3-1)
– 3.1.X(1-1) ]
= (X3 – 3X) . [8X3 + 5] + (2X4 + 5X).[3X2
- 3]
6 Rule no 06 : The Quotient Rule
f (X) = (X-3)/(X+4)
d/dx f (X) =
[(X+4).d/dx (X1-3) - (X-3).d/dx (X1+4) ] /
(X+4)2
f ‘ (X) = [(X+4). (1.X(1-1) – 0) -
(X-3). (1.X(1-1) – 0) ] / (X+4)2
= [(X+4).1 -
(X-3).1 ] / (X+4)2
=
[(X+4) -
(X-3)] / (X+4)2
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