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 the derivative of
f (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.
h(2ax + ah + b)
f (x + h) − f (x)
=
= 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.
1
2h
2
f (x + h) − f (x)
√
√
√
=√
.
=
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 (x + h) − f (x)
2
2
1
√
√
= lim √
=√
=√
.
h→0
h→0
h
2x + 0 + 5 + 2x + 5
2x + 5
2x + 2h + 5 + 2x + 5
f 0 (x) = lim
2
Find an equation of the tangent line (using the 4-step procedure to find slopes)
Given a curve y = f (x) and a point (x0 , y0 ) on it, an equation of the line tangent to the curve y = f (x)
at the point (x0 , y0 ) is
y − y0 = f 0 (x0 )(x − x0 ),
provided the f 0 (x0 ) exists. (Therefore, f 0 (x0 ) is the slope of the tangent line at (x0 , y0 )).
Facts:
(1) If x = a is in the domain of f (x) and f 0 (a) exists, then the line tangent to the curve y = f (x) at (a, f (a))
has equation: y − f (a) = f 0 (a)(x − a).
(2) If y = f (x) does not have a tangent line when x = a, then f 0 (a) does not exist.
(3) If f (x) is differentiable on an interval I, then f (x) is continuous l on
Find derivatives by using differentiation rules
Differentiation Rules:
(1) Derivative of a constant: Let C be a constant, then
(2) Power Rule: For a real number n,
d
C = 0.
dx
dxn
= nxn−1 .
dx
(3) Linear Property: For constant a and b and functions f (x) and g(x),
[af (x) + bg(x)]0 = af 0 (x) + bg 0 (x) and [af (x) − bg(x)]0 = af 0 (x) − bg 0 (x).
(4) Product Rule: [f (x)g(x)]0 = f 0 (x)g(x) + f (x)g 0 (x).
(5) Quotient Rule:
f (x)
f 0 (x)g(x) − f (x)g 0 (x)
=
.
g(x)
[g(x)]2
(6) Generalized Power Rule: for a real number n and a differentiable function f (x),
d
[f (x)]n = n[f (x)]n−1 f 0 (x).
dx
Example 1 Apply differentiation rules to find the derivative of f (x) = (2x2 − 1)(x3 + 2).
Solution: The function f (x) is a product, and each factor is a polynomial. So we first apply Product Rule, and
then apply the linear property and the power rule to get:
f 0 (x)
=
(2(2)x2−1 − 0)(x3 + 2) + (2x2 − 1)(3x3−1 + 0) = 4x(x3 + 2) + 3x2 (2x2 − 1)
=
4x4 + 8x + 6x4 − 3x2 = 10x4 − 3x2 + 8x.
2x2 − 1
.
x3 + 2
Solution 1: The function f (x) is a quotient. So we first apply Quotient Rule, and then apply the linear property
and the generalized power rule to get:
Example 2 Apply differentiation rules to find the derivative of f (x) =
f 0 (x)
=
(2(2)x2−1 − 0)(x3 + 2) − (2x2 − 1)(3x3−1 + 0)
(x3 + 2)2
4x + 8x + 6x − 3x2
x(10x3 − 3x + 8)
=
.
(x3 + 2)2
(x3 + 2)2
4
=
4
3
=
4x(x3 + 2) + 3x2 (2x2 − 1)
(x3 + 2)2
Solution 2: View the function f (x) as a product by using negative exponents.
f (x) = (2x2 − 1)(x3 + 2)−1 .
Then apply Product Rule, and then the linear property and the power rule to get the answer (the answer is
intentionally not simplified to make it easier for a reader to see the computation process).
f 0 (x)
=
(2(2)x2−1 − 0)(x3 + 2)−1 + (2x2 − 1)(−1)(x3 + 2)−1−1 (3x3−1 + 0)
=
4x(x3 + 2)−1 − 3x2 (2x2 − 1)(x3 + 2)−2 .
2x3 − 3x2 + 4x − 5
.
x2
Solution: The function f (x) is a quotient. But we are not in a hurry to apply Quotient Rule, as we observed
that the denominator is just a power of x. Thus we first simplify the fraction.
Example 3 Apply differentiation rules to find the derivative of f (x) =
f (x) =
2x3 − 3x2 + 4x − 5
2x3
3x2
4x
5
=
−
+ 2 − 2 = 2x − 3 + 4x−1 − 5x−2 .
2
2
2
x
x
x
x
x
Then apply the linear property and the power rule.
f 0 (x) = [2x − 3 + 4x−1 − 5x−2 ]0 = 2 − 0 + 4(−1)x−1−1 − 5(−2)x−2−1 = 2 −
Example 4 Write an equation of the line tangent to the curve f (x) =
1
2
− 2
x x
10
4
+ 3.
x2
x
−1
at the point (2, 4/3).
Solution: The equation of this line has the form
y−
4
= f 0 (2)(x − 2).
3
To find the slope f 0 (2), we first compute the derivative f 0 (x). To do that it may be better to simplify the fraction
first
−1 −1 −1
2
1
2x
1
2x − 1
x2
f (x) =
− 2
=
−
=
=
.
2
2
2
x x
x
x
x
2x − 1
Then, apply Quotient Rule
f 0 (x) =
2x(2x − 1) − 2x2
4x2 − 2x − 2x2
2x2 − 2x
=
=
.
(2x − 1)2
(2x − 1)2
(2x − 1)2
Hence f 0 (2) = 49 , and so the equation is
y−
4
4
= (x − 2).
3
9
4