Week 6 – Derivative
The slope of the curve y = f (x) at the point x0 is:
m = lim
h→0
f (x0 + h) − f (x0 )
h
if the limit exists. The tangent line through the point
(x0 , f (x0 )) has this slope.
Derivative: The derivative of the function f (x) is
the function f 0 (x) defined by
f 0 (x) = lim
h→0
f (x + h) − f (x)
h
We can think of the derivative as the rate of change
of a function f or the slope of the curve of y = f (x).
We will use
y0,
f 0 (x),
dy
,
dx
d
f (x)
dx
to denote derivatives and
dy 0
f (a),
dx x=a
to denote their values at a certain point. Note that
derivative is a function, its value at a point is a
number.
Example: Instantaneous velocity is defined as
v = lim
∆t→0
x(t + ∆t) − x(t)
∆t
In mechanics, velocity is the derivative of distance
and acceleration is the derivative of velocity.
Exercise 6-1: Find the derivative of f (x) = x2 using the definition.
Solution: f 0 (x) = lim
h→0
(x + h)2 − x2
h
= lim
x2 + 2xh + h2 − x2
h
= lim
2xh + h2
h
h→0
h→0
= lim 2x + h
h→0
= 2x
Exercise 6-2: Find the derivative of f (x) =
using the definition.
√
√
x+h− x
0
Solution: f (x) = lim
h→0
h
√
x
x+h−x
√
√ h→0 h
x+h+ x
1
= lim √
√
h→0
x+h+ x
= lim
=
1
√
2 x
1
Exercise 6-3: Find the derivative of f (x) = using
x
the definition.
Differentiation Formulas: Using the definition of
derivative, we obtain:
• Derivative of a constant is zero, i.e.
dc
=0
dx
• Derivative of f (x) = x is
f 0 (x) = 1
• Derivative of f (x) = x2 is
f 0 (x) = 2x
• Derivative of f (x) = xn (where n is a positive
integer) is:
f 0 (x) = nxn−1
• Derivative of f (x) =
√
x is:
1
f 0 (x) = √
2 x
• If f is a function and c is a constant, then
(cf )0 = cf 0
• If f and g are functions, then
(f + g)0 = f 0 + g 0
(Derivative is a linear operator)
Exercise 6-4: Evaluate the derivatives of the following functions:
a) f (x) = 7x3 + 3x2 − 8
√
1
b) f (x) = 2 x − x4
4
√
3
7x − 18x + x
c) f (x) =
x
Exercise 6-5: Find the equation of the line that is
tangent to y = x8 + x4 + x2 at the point (1, 3)
Exercise 6-6: Find a horizontal tangent to
y = x3 + x2 − x + 1
Exercise 6-7: Find the equation of the tangent line
to the graph at the given point. Then, sketch the
function and the tangent line on the same coordinate
system:
1
at (1, 1)
a) f (x) =
x
b) f (x) =
√
x at (9, 3)
c) f (x) = x3
at (0, 0)
d) f (x) = x2
at (1, 1)
e) f (x) = x3
at (1, 1)
Differentiability and Continuity: If f is differentiable at a, then it is continuous at a, but if it is
continuous at a, it is not necessarily differentiable.
If the graph of a function has no unique tangent at
a point, then the function has no derivative at that
point. For example, f (x) = |x| is not differentiable
at x = 0. (Its derivative does not exist)
Higher Order Derivatives: We can find the derivative of the derivative of a function. It is called second
derivative and denoted by:
y 00 ,
f 00 (x),
d2 y
dx2
For third derivative, we use f 000 but for fourth and
higher derivatives, we use the notation f (4) (x)
Exercise 6-8: Let f (x) = 7x3 − 18x. Find f 0 , f 00
and f 000 .
Vertical Tangents: The curve y = f (x) has a
vertical tangent at (a, f (a)) if f is continuous at a
and if |f 0 (x)| → ∞ as x → a.
Exercise 6-9: Find the equation of the tangent line
√
to f (x) = x at (0, 0) and sketch the function and
the tangent line on the same coordinate system.
Product Rule: If f and g are differentiable at x,
then f g is differentiable at x and
d
df
dg
(f g) =
g+f
dx
dx
dx
or more briefly:
(f g)0 = f 0 g + f g 0
Reciprocal Rule: If f is differentiable at x and if
f (x) 6= 0 then:
0
−f 0
1
= 2
f
f
Quotient Rule: If f and g are differentiable at x,
f
and g(x) 6= 0 then is differentiable at x and:
g
0
f
f 0g − g0f
=
g
g2
Exercise 6-10: Evaluate the derivatives of the following functions:
x2 − 2
x2 + 2
x4 − 7x2 + 5
b) f (x) =
x
1
c) f (x) =
1 − x−3
√
x+ x
d) f (x) = 2 √
x − x
a) f (x) =
Derivatives of Exponential and Trigonometric
Functions:
We state without proof (at this stage) that:
dex
= ex
dx
Actually, this is what make the number e so special.
It is the only nonzero function whose derivative is
itself.
The derivative of sin x is:
d sin x
sin(x + h) − sin x
= lim
h→0
dx
h
sin x cos h + cos x sin h − sin x
h→0
h
cos h − 1
sin h
+ cos x lim
= sin x lim
h→0
h→0
h
h
2
1 − 2 sin (h/2) − 1
= sin x lim
+ cos x · 1
h→0
h
= cos x
= lim
We can show that
d cos x
= − sin x
dx
similarly.
Exercise 6-11: Prove that
d tan x
= sec2 x
dx
d cot x
= − csc2 x
dx
d sec x
= sec x tan x
dx
d csc x
= − csc x cot x
dx
Exercise 6-12: Evaluate the derivatives of the following functions:
x − cos x
a) y =
x + cos x
b) y = xn ex
c) y = sin 2θ
d) s = θ2 sec θ
e) p =
t2 sin t
t − tan t
Exercise 6-13: Find the following limits:
tan t − 1
a) lim
t→π/4 t − π/4
b) lim
t→π/3
cos t − 1/2
t − π/3
Chain Rule: If f and g are differentiable then
f (g(x)) is also differentiable and
[f (g(x))]0 = f 0 (g(x)) · g 0 (x)
or more briefly
dy
dy du
=
dx
du dx
Power Rule for Rational Powers: If r is a rational
number and f (x) = xr then f 0 = rxr−1
p
Let r = . Then y q = xp .
q
Derivative of both sides gives
qy q−1 y 0 = pxp−1
pxp−1
0
y =
qx
p
(q−1)
q
=
p pq −1
x
q
Exercise 6-14: Find y 0
a) y = (x2 − 2)2
b) y = [x + (x3 + 5)2 ]4
2
1
c) y = t −
t
p
√
d) y = 1 + 1 + x
2
e) y = te−3t
f) y = 3t
g) y = 53t
h) y = es
2 (1−s)3
i) y = cos2 θ
j) y = 2 cos3 (5θ)
k) y = sin sin(πθ)
Review Exercises
Exercise 6-15: Find the derivatives of the following
functions:
√
t
√
a) s =
3−2 t
b) y = x5 cos2 (3x4 )
√
c) r = tan 3θ − 2
d) r =
e) p =
cos θ
1 − θ sin θ
(q −
q2
2
1
− cos2 q)3
f) y = x−5/3 cos(3x − 1) −
√ x−√2
2e
Exercise 6-16: Evaluate the derivatives of the following functions:
√
a) f = sin2 x
x
b) f =
cos 4x
c) f = cos(sin x)
p √
d) f = cos x
1
e) f = x2 cos
x
1
f) f =
2
sin x + cos2 x
— End of WEEK —
Author: Dr. Emre Sermutlu
Last Update: November 9, 2016