Compute the Derivative by Definition: The four step procedure
Given a function f (x), the definition of the derivative of f (x), is
f 0 (x) = lim
h→0
f (x + h) − f (x)
f (a + h) − f (a)
, and at x = a, f 0 (a) = lim
,
h→0
h
h
provided these limits exist. Given a function f (x), we can follow the following steps to find f 0 (x).
Step 1 Evaluate f (x + h) and f (x).
Step 2 Compute f (x+h)−f (x). Combine like terms. If h is a common factor of the terms, factor the expression
by removing the common factor h.
f (x + h) − f (x)
Step 3 Simply
. As h → 0 in the last step, we must cancel the zero factor h in the denominator
h
in Step 3.
f (x + h) − f (x)
Step 4 Compute lim
by letting h → 0 in the simplified expression.
h→0
h
Example 1 Let f (x) = ax2 + bx + c. Compute f 0 (x) by the definition (that is, use the four step process).
Solution: Step 1, write
f (x + h) = a(x + h)2 + b(x + h) + c = a(x2 + 2xh + h2 ) + bx + bh + c = ax2 + 2axh + ah2 + bx + bh + c.
Step 2: Use algebra to single out the factor h.
f (x + h) − f (x) = (ax2 + 2axh + ah2 + bx + bh + c) − (ax2 + bx + c) = 2axh + ah2 + bh = h(2ax + ah + b).
Step 3: Cancel the zero factor h is the most important thing in this step.
f (x + h) − f (x)
h(2ax + ah + b)
=
= 2ax + ah + b.
h
h
Step 4: Let h → 0 in the resulted expression in Step 3.
f 0 (x) = lim
h→0
Example 2 Let f (x) =
f (x + h) − f (x)
= lim 2ax + ah + b = 2ax + 0 + b = 2a + b.
h→0
h
1
. Compute f 0 (x) by the definition (that is, use the four step process).
x+1
Solution: Step 1, write
f (x + h) =
1
1
=
.
(x + h) + 1
x+h+1
Step 2: Use algebra to single out the factor h.
f (x + h) − f (x) =
1
1
(x + 1) − (x + h + 1)
h
−
=
=
.
x+h+1 x+1
(x + h + 1)(x + 1)
(x + h + 1)(x + 1)
Step 3: Cancel the zero factor h is the most important in this step.
f (x + h) − f (x)
1
1
1
1
−h
−1
=
−
=
=
.
h
h x+h+1 x+1
h (x + h + 1)(x + 1)
(x + h + 1)(x + 1)
Step 4: Let h → 0 in the resulted expression in Step 3.
f 0 (x) = lim
h→0
f (x + h) − f (x)
−1
−1
−1
= lim
=
=
.
h→0 (x + h + 1)(x + 1)
h
(x + 0 + 1)(x + 1)
(x + 1)2
1
Example 3 Let f (x) =
√
2x + 5. Compute f 0 (x) by the definition (that is, use the four step process).
Solution: Step 1, write
f (x + h) =
p
√
2(x + h) + 5 = 2x + 2h + 5.
Step 2: Use algebra to single out the factor h. Here we need the identity (A + B)(A − B) = A2 − B 2 to get rid
of the square root so that h can be factored out.
f (x + h) − f (x) =
√
2x + 2h + 5 −
√
(2x + 2h + 5) − (2x + 5)
2h
√
√
.
=√
2x + 5 = √
2x + 2h + 5 + 2x + 5
2x + 2h + 5 + 2x + 5
Step 3: Cancel the zero factor h is the most important thing in this step.
f (x + h) − f (x)
1
2h
2
√
√
√
=√
.
=
h
h
2x + 2h + 5 + 2x + 5
2x + 2h + 5 + 2x + 5
Step 4: Let h → 0 in the resulted expression in Step 3.
f 0 (x) = lim
h→0
2
2
f (x + h) − f (x)
1
√
√
= lim √
=√
=√
.
h→0
h
2x + 0 + 5 + 2x + 5
2x + 5
2x + 2h + 5 + 2x + 5
Find Derivatives by Using Differentiation Rules
Some Differentiation Rules and Higher Order of Derivatives:
d
(1) Derivative of a constant: Let C be a constant, then
C = 0.
dx
(2) Power Rule: For a real number n,
dxn
= nxn−1 .
dx
(3) Linear Property: For a constant c and functions f (x) and g(x),
[f (x) ± g(x)]0 = f 0 (x) ± g 0 (x) and [cf (x)]0 = cf 0 (x).
(4) Higher Order of Derivative: f 00 (x) =
d 0
dx f (x),
f 000 (x) =
d 00
dx f (x),
f (4) (x) =
d 000
dx f (x),
f (5) (x) =
d (4)
(x), · · ·.
dx f
√
3
− 1 + x + 5x4 .
x
Solution: Write every term in the power form so that the power rule can be easily applied:
Example 1 Find the derivative of f (x) =
f (x) =
√
1
3
− 1 + x + 5x4 = 3x−1 − 1 + x 2 + 5x4 .
x
1 1
−1 1 1
Apply the power rules to get f 0 (x) = 3 · (−1)x−1−1 − 0 + x 2 −1 + 5 · 4x4−1 = 2 + x− 2 + 20x3 .
2
x
2
Example 2 Given f (x) =
x2 − x + 1
√
, find f 000 (x).
x
Solution: Write every term in the power form so that the power rule can be easily applied:
f (x) =
−1
3
1
x2 − x + 1
x2
x
1
√
= √ − √ + √ = x2 − x2 + x 2 .
x
x
x
x
Apply the rules to get
f 0 (x) =
3 1
1 1
1 −3
3 −1
1 3
3 −5
3 −3
3 51
15 −7
x 2 − x− 2 − x 2 , f 00 (x) = x 2 + x− 2 + x 2 , f 000 (x) = − x 2 − x− 2 − x 2 .
2
2
2
4
4
4
8
8
8
2