Derivatives /
Sections 2.7, 2.8, 3.1-3.6, 3.9
Cagri Haciyusufoglu
October 18, 2025
Derivative of a function
Definition: We define the derivative of a function f (x) at x as follows provided
the limit exists:
f (x + h) − f (x)
f ′ (x) = lim
h→0
h
Example: Find f ′ (x) if f (x) = x2 .
Solution: f (x) = x2 is differentiable for all x:
f (x + h) − f (x)
(x + h)2 − x2
= lim
h→0
h→0
h
h
2xh + h2
= lim (2x + h) = 2x
= lim
h→0
h→0
h
f ′ (x) = lim
Alternative Notations for Derivatives: Let y = f (x).
f ′ (x) = y ′ =
For instance,
df
dy
d
=
=
f (x)
dx
dx
dx
d
x2 = 2x.
dx
1
d √
1
( x) = √
dx
2 x
d
dx
1
1
=− 2
x
x
√
Example: Find f ′ (x) if f (x) = x.
√
Solution: f (x) = x is differentiable for all x > 0:
√
√
f (x + h) − f (x)
x+h− x
= lim
f ′ (x) = lim
h→0
h→0
h
h
√
√ √
√ x+h− x x+h+ x
√
= lim
√
h→0
h
x+h+ x
= lim
h→0 h(
h
1
1
√
√ = lim √
√ = √
h→0
2 x
x + h + x)
x+h+ x
Example: Find f ′ (x) if f (x) = 1/x.
Solution: f (x) = 1/x is differentiable for all x ̸= 0:
1
− x1
f (x + h) − f (x)
= lim x+h
h→0
h→0
h
h
−h
1
1
= lim
= lim −
=− 2
h→0 hx(x + h)
h→0
x(x + h)
x
f ′ (x) = lim
2
Interpretations of Derivative: Tangent Lines
To interpret derivatives an equivalent definition is helpful:
f ′ (x) = lim
f (x + h) − f (x)
h
z =x+h
f ′ (x) = lim
f (z) − f (x)
z−x
z → x ⇐⇒ h → 0
h→0
z→x
Definition: The tangent line to the curve y = f (x) at the point P (a, f (a)) is
the line through P with slope m = f ′ (a) provided that f is differentiable at a.
m = f ′ (a) = lim
x→a
f (x) − f (a)
x−a
(a)
Here, f (x)−f
is the slope of the
x−a
secant line through the points
P (a, f (a)) and Q(x, f (x)).
An equation of the tangent line to the curve y = f (x) at P (a, f (a)) is
y − f (a) = f ′ (a)(x − a)
3
Infinite Derivatives: Vertical Tangent Lines
√
Example: Show that f (x) = 3 x is not differentiable at 0.
Solution: The limit defining the derivative does not exist:
f ′ (0) = lim
x→0
f (x) − f (0)
x−0
√
3
= lim
x→0
x
1
=∞
= lim √
x→0 3 x2
x
Vertical lines in the xy-plane are said to have infinite slope. Therefore, the
√
tangent line to the curve y = 3 x at (0, 0) is the vertical line through (0, 0),
which is the y-axis.
Example: Find an equation of the tangent line to the curve xy = 1 at the
point (1/2, 2).
d
Solution: Remember that we showed dx
(1/x) = −1/x2 . Writing y = 1/x, the
slope of the tangent line is −4. Therefore, an equation of the tangent line is
y − 2 = −4(x − 1/2)
=⇒
y + 4x = 4
4
Interpretations of Derivative: Instantaneous rate of change
• The difference quotient below is called the average rate of change of f
between x and x + h:
f (x + h) − f (x)
h
• The derivative f ′ (x) is called the instantaneous rate of change of f at x,
if the limit exists.
f (x + h) − f (x)
f ′ (x) = lim
h→0
h
• For instance, if s(t) is the position function of a particle moving along a
straight line, then s′ (a) is called the instantaneous velocity; the limit of
average velocities on the time interval between a and a + h as h → 0.
• If C(x) is the cost of producing x units, then C ′ (x) is called marginal
cost. For instance, C ′ (1000) = 9 means that after 1000 units have been
produced, the cost of production is increasing 9 times as fast as the
additional production.
• If b(t) is the number of bacteria after t hours in a laboratory experiment,
then b′ (10) = −1000 means that after 10 hours, the number of bacteria
is decreasing 1000 times as fast as the additional time spent.
5
Differentiability implies continuity but not vice versa
Theorem: If f (x) is differentiable at a, then f (x) is continuous at a.
Proof: Assume that f (x) is differentiable at x = a. We need to show that
lim f (x) = f (a)
x→a
⇐⇒
lim [f (x) − f (a)] = 0
x→a
f (x) − f (a)
We write f (x) − f (a) =
(x − a) and apply the product law:
x−a
f (x) − f (a)
lim [f (x) − f (a)] = lim
lim (x − a) = f ′ (a) × 0 = 0
x→a
x→a
x→a
x−a
Remark: The opposite statement is not true. For instance, let f (x) = |x|.
|x| is continuous at zero but |x| is not differentiable at zero.
lim
x→0
|x|
f (x) − f (0)
= lim
x→0 x
x−0

|x|
x

 lim
= lim
= 1
x
x→0+ x
x→0+

 lim |x| = lim − x = −1
x
x→0− x
x→0+
6
The Sum/Difference Rule for derivatives
Theorem: If f (x) and g(x) are differentiable at x then, the sum f + g is also
differentiable at x and (f + g)′ = f ′ + g ′ .
Proof: Let h = f + g.
h(z) − h(x)
h (x) = lim
= lim
z→x
z→x
z−x
′
= lim
z→x
f (z) − f (x)
g(z) − g(x)
+
z−x
z−x
f (z) − f (x)
g(z) − g(x)
+ lim
= f ′ (x) + g ′ (x)
z→x
z−x
z−x
The proof of the difference rule is similar: (f − g)′ = f ′ − g ′ .
7
The Power Rule
Theorem:
d
(xa ) = axa−1
dx
Proof: The formula is true for all a ∈ R. For now, we will prove it only for
positive integers. Therefore, assume that a ∈ N:
d a
z a − xa
x = lim
z→x z − a
dx
(z − x)(z a−1 + xz a−2 + x2 z a−3 + · · · + xa−2 z + xa−1 )
z→x
z−x
= lim
= lim (z a−1 + xz a−2 + x2 z a−3 + · · · + xa−2 z + xa−1 ) = axa−1
z→x |
{z
}
a many terms
Example:
d
d 3
d −3
x3 + x−3 =
(x ) +
(x ) = 3x2 − 3x−4
dx
dx
dx
8
The Product Rule
Theorem: If f (x) and g(x) are differentiable at x then, the product f g is also
differentiable at x and (f g)′ = f g ′ + gf ′ .
Proof: Let h(x) = f (x)g(x).
h(z) − h(x)
f (z)g(z) − f (x)g(x)
d
h(x) = lim
= lim
z→x
z→x
dx
z−x
z−x
= lim
z→x
f (z)g(z)−f (z)g(x) + f (z)g(x) − f (x)g(x)
z−x
g(z) − g(x)
f (z) − f (x)
f (z)
+ g(x)
z→x
z−x
z−x
= lim
= lim f (z) lim
z→x
z→x
f (z) − f (x)
g(z) − g(x)
+ g(x) lim
z→x
z−x
z−x
= f (x)g ′ (x) + g(x)f ′ (x)
Remark: Since f (x) is differentiable at x, f (x) is also continuous at x:
lim f (z) = f (x)
z→x
9
The Constant Multiple Rule / Derivatives of Polynomials
Theorem: If f (x) is differentiable at x then the constant multiples cf are also
differentiable at x and (cf )′ = cf ′ for c ∈ R.
d
Proof: This follows from the product rule and that dx
(c) = 0:
df
d
df
d
(cf ) = c
+ f (c) = c
dx
dx
dx
dx
We can now differentiate any polynomial using the sum rule, the product rule,
and the constant multiple rule:
Example: Find f ′ (x) if f (x) = 5x7 − 4x6 + 10x3 − 2x + 23.
d
f ′ (x) =
5x7 − 4x6 + 10x3 − 2x + 23
dx
=5
d 7
d
d
d
d
(x ) − 4 (x6 ) + 10 (x3 ) − 2 (x) +
(23)
dx
dx
dx
dx
dx
= 35x6 − 24x5 + 30x2 − 2
10
The Quotient Rule for derivatives
Theorem: If f (x) and g(x) are differentiable at x then, the quotient f /g is
also differentiable at x and (f /g)′ = (gf ′ − f g ′ )/g 2 .
f (z)/g(z) − f (x)/g(x)
g(x)f (z) − g(z)f (x)
d f (x)
= lim
= lim
z→x
z→x
dx g(x)
z−x
g(z)g(x)(z − x)
= lim
z→x
g(x)f (z)−g(x)f (x) + g(x)f (x) − g(z)f (x)
g(z)g(x)(z − x)
= lim
z→x
= lim
f (x) g(x) − g(z)
1 f (z) − f (x)
+
g(z)
z−x
g(z)g(x)
z−x
1
lim
z→x g(z) z→x
=
f (z) − f (x)
f (x)
g(x) − g(z)
+ lim
lim
z→x g(z)g(x) z→x
z−x
z−x
f (x) ′
g(x)f ′ (x)−f (x)g ′ (x)
1 ′
f (x) − 2
g (x) =
g(x)
g (x)
g 2 (x)
Remark: Since g(x) is differentiable at x, g(x) is also continuous at x:
lim f (z) = f (x)
z→x
11
Derivatives of Rational Functions
Now that we can differentiate polynomials, we can also differentiate rational
functions using the quotient rule.
Example: Find y ′ if y =
t3 + 3t
.
t2 − 4t + 3
Solution: We have y ′ =
gf ′ − f g ′
where f = t3 + 3t and g = t2 − 4t + 3.
g2
g
f′
f
g′
(t2 − 4t + 3)(3t2 + 3) − (t3 + 3t)(2t − 4)
dy
=
dt
(t2 − 4t + 3)2
g2
=
t4 − 8t3 + 17t2 + 9
(t2 − 4t + 3)2
12
The Chain Rule: Derivative of a composition is the product of
derivatives
Theorem: If g is differentiable at x and f is differentiable at g(x), then the
composite function F = f ◦ g defined is differentiable at x:
F ′ (x) = g ′ (x)f ′ (g(x))
Proof: What is the gap in this proof?
F ′ (x) = lim
z→x
f (g(z)) − f (g(x))
z−x
= lim
z→x
f (g(z)) − f (g(x)) g(z) − g(x)
g(z) − g(x)
z−x
= f ′ (g(x))g ′ (x)
13
The Power Rule combined with the Chain Rule
d
(f (x))n = n(f (x))n−1 f ′ (x)
dx
Example:
Example:
7
d
4t3 − 2t2 + 5 = 7(4t3 − 2t2 + 5)6 (12t2 − 4t)
dt
f
f′
1
d p3
1
t + 4t − 3 = (t3 + 4t − 3)− 2 (3t2 + 4)
dx
2
f
f′
3t2 + 4
= √
2 t3 + 4t − 3
Example:
d
1
= −2(t3 + 4t − 3)−3 (3t2 + 4)
dx (t3 + 4t − 3)2
f
f′
=−
6t2 + 8
(t3 + 4t − 3)3
14
Some useful exponential and logarithmic identities
• a = bc ⇐⇒ c = logb a
•
ax and loga x are inverse functions.
c
ab = abc
• ab ac = ab+c
• loga (bc ) = c loga b
(loga bc ) ̸= c loga b
• loga (bc) = loga b + loga c
• loga (b/c) = loga b − loga c
• ln x = loge x
15
d x
e = ex
dx
Derivative of the natural exponential function:
eh − 1
= 1.
h→0
h
Definition: The number e is defined as the number such that lim
f (x + h) − f (x)
d x
(e ) = lim
h→0
dx
h
ex+h − ex
eh − 1
= ex lim
= ex
h→0
h→0
h
h
= lim
Examples: By the Chain Rule. we have
d f (x)
e
= f ′ (x)ef (x) .
dx
•
d 2x
d
e = e2x (2x) = 2e2x
dx
dx
•
e
d xe f (x)=xe xe d e
e ======= e
x = exe−1 ex
dx
dx
•
x
d ex f (x)=ex ex d x
e ======= e
e = ex ee
dx
dx
•
2
d x2 +2x
e
= (2x + 2)ex
dx
16
d x
a = ax ln a
dx
Derivatives of Exponential Functions:
x
Since x = eln x we have ax = eln a = ex ln a . Differentiating by the Chain Rule,
d x ln a
d
d x
a =
e
= ex ln a (x ln a) = ax ln a
dx
dx
dx
Example: By the Chain Rule,
d f (x)
a
= f ′ (x)af (x) ln a
dx
d exe f (x)=exe exe d xe
3 ======= 3
e ln 3
a=3
dx
dx
f (x)=xe
xe
====== 3e ex
a=e
xe
e
d e
x ln 3
dx
e
= 3e ex xe−1 e ln 3
17
Derivative of the natural logarithm:
y = ln x =⇒ x = ey =⇒
=⇒
d
1
ln |x| =
dx
x
d
d y ⋆
dy
x=
e =⇒ 1 = ey
dx
dx
dx
dy
1
1
d
1
= y =
=⇒
ln x = ,
dx
e
x
dx
x
x>0
We applied the Chain Rule in ⋆ because y depends on x.
d
1 d
1
ln(−x) = −
(−x) = we have
dx
x dx
x
(
ln x,
if x > 0
d
1
ln |x| =
ln |x| =
=⇒
dx
x
ln(−x), if x < 0
Remark: Since
Example: By the Chain Rule, we have
ln |f (x)| =
f ′ (x)
.
f (x)
d
2x − 2
ln |x2 − 2x − 3| = 2
dx
x − 2x − 3
18
Proof of the Power Rule / Logarithmic Differentiation
Previously, we proved the power rule for natural number powers a ∈ N. Now,
we will prove the formula for any real number a ∈ R:
y = xa =⇒ |y| = |x|a =⇒ ln |y| = a ln |x|
=⇒
y′
a
ay
dy
=
=⇒ y ′ =
=⇒
= axa−1
y
x
x
dx
Example: Find y ′ if y = xx .
d x
x ̸= xxx−1 = x.
dx
Instead, we use Logarithmic Differentiation by first taking the logarithm:
Solution: The Power Rule cannot be applied directly:
y = xx =⇒ ln y = x ln x =⇒
d
d
ln y =
(x ln x)
dx
dx
=⇒
y′
= 1 + ln x =⇒ y ′ = y(1 + ln x)
y
=⇒
d x
x = xx (1 + ln x)
dx
19
When there are too many products/quotients/powers...
Example: Find y ′ if y =
√
x3/4 x2 + 1
.
(3x + 2)5
Solution: First take the logarithm to soften it up because the logarithm
• transforms products into sums,
• transforms quotients into differences, and
• transforms powers into products.
This makes differentiation easier; it is called Logarithmic Differentiation.
y=
√
x3/4 x2 + 1
(3x + 2)5
=⇒
d/dx
=⇒
ln y =
3
1
ln x + ln(x2 + 1) − 5 ln(3x + 2)
4
2
y′
3
x
15
=
+ 2
−
y
4x
x +1
3x + 2
√
x
15
x3/4 x2 + 1 3
y =
+ 2
−
(3x + 2)5
4x
x +1
3x + 2
′
20
Derivatives of logarithmic functions:
y = loga x =⇒ x = ay
=⇒ 1 = ay
=⇒
=⇒
d
1
loga x =
dx
x ln a
d
d y
x=
a
dx
dx
dy
1
1
dy
ln a =⇒
= y
=
dx
dx
a ln a
x ln a
d
1
loga x =
dx
x ln a
Example: By the Chain Rule,
f ′ (x)
d
[loga f (x)] =
.
dx
f (x) ln a
•
f (x)=ln(x2 +1) d ln(x2 + 1)
d
log5 (ln(x2 + 1)) =========== dx 2
a=5
dx
ln(x + 1) ln 5
f (x)=x2 +1
2x + 1
========= 2
a=e
(x + 1) ln(x2 + 1) ln 5
21
Derivative of sine and cosine:
d
d
sin x = cos x
cos x = − sin x
dx
dx
sin(x + h) − sin x
d
sin x cos h + sin h cos x − sin x
(sin x) = lim
= lim
h→0
h→0
dx
h
h
cos h − 1
sin h
= sin x lim
+ cos x lim
= cos x
h→0
h→0
h
h
cos h − 1
cos h − 1 cos h + 1
sin2 h
= lim
= lim −
h→0
h→0
h→0
h
h
cos h + 1
h(cosh +1)
sin h sin h
= lim −
=1×0=0
h→0
h cos h + 1
lim
22
Derivatives of Trigonometric Functions
d
(cos x) = − sin x is an exercise.
dx
The rest of the formulas below can be obtained easily.
The proof of the fact
•
d
(sin x) = cos x
dx
•
d
(cos x) = − sin x
dx
•
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
Example:
d sin x
cos2 x + sin2 x
1
d
tan x =
=
=
= sec2 x
dx
dx cos x
cos2 x
cos2 x
Example:
d
d 1
sin x
sec x =
=
= sec x tan x
dx
dx cos x
cos2 x
f ′ (x)
d 1
=− 2
dx f (x)
f (x)
23
Examples
•
f (x)=x2
d
(sin x2 ) ======= 2x cos x2
dx
d
sin(f (x)) = f ′ (x) cos f (x)
dx
•
d
(sin2 x) = 2 sin x cos x
dx
d 2
f (x) = 2f (x)f ′ (x)
dx
•
d
(tan5 x) = 5 tan4 x sec2 x
dx
d 5
f (x) = 5f 4 (x)f ′ (x)
dx
•
f (x)=tan(3x)
d
(sec(tan(3x))) ========== 3 sec2 (3x) sec(tan(3x)) tan(tan(3x))
| {z } |
{z
}
dx
f′
d
sec x = sec x tan x
dx
=⇒
sec(f (x)) tan(f (x))
d
sec(f (x)) = f ′ (x) sec(f (x)) tan(f (x))
dx
24
More Examples
•
d
dx
sin x − x ln x cos x
sin2 x
(f /g)′ = (gf ′ − f g ′ )/g 2
•
d
sin x
(sin x ln x) =
+ cos x ln x
dx
x
(f g)′ = f g ′ + gf ′
•
d sin x
(e
) = cos xesin x
dx
d f (x)
e
= f ′ (x)ef (x)
dx
•
d
sec2 x
ln | tan x| =
dx
tan x
f ′ (x)
d
ln |f (x)| =
dx
f (x)
ln x
sin x
=
25
Derivative of a composition is the product of derivatives
d
d
(h ◦ g ◦ f )(x) =
h(g(f (x))) = f ′ (x) g ′ (f (x)) h′ (g(f (x))
dx
dx
f (x) = 4x
x
d 34x 4x
34
x
2
•
=4
ln 4} 3
| {z
| {zln 3} |2 {zln 2}
dx
′
′
′
f (x)
g (f (x)) h (g(f (x)))
g(x) = 3x
h(x) = 2x
•
d
1
ln(ln(ln x)) =
dx
x ln x ln(ln x)
•
d √
1
√
ln x =
dx
2x ln x
•
d
(x sin x tan x) = sin x tan x + x cos x tan x + x sin x sec2 x
dx
f (x) = g(x) = h(x) = ln x
f ′ (x)
d p
f (x) = p
dx
2 f (x)
(f gh)′ = f ′ gh + f g ′ h + f gh′
26
Derivatives of Inverse Trigonometric functions
Principle: Derivative of f −1 can be obtained from the derivative of f .
• Let y = arctan x. We know
x = tan y
d/dx
=⇒
1 = y ′ sec2 y =⇒ y ′ =
=⇒
d
1
(arctan x) =
dx
1 + x2
• Let y = arcsin x. We know
x = sin y
d
tan x = sec2 x:
dx
1
1 + tan2 y
d
sin x = cos x:
dx
=⇒
1
1 = y ′ cos y =⇒ y ′ = p
1 − sin2 y
=⇒
d
1
(arcsin x) = √
dx
1 − x2
d/dx
27
Derivatives of Inverse Trigonometric Functions
One can obtain the following derivatives similarly:
•
d
1
(arcsin x) = √
dx
1 − x2
•
d
1
(arccos x) = − √
dx
1 − x2
•
d
1
(arctan x) =
dx
1 + x2
•
d
1
(arccot x) = −
dx
1 + x2
•
d
1
(arcsec x) = √
dx
x x2 − 1
•
d
1
(arccsc x) = − √
dx
x x2 − 1
28
Examples
f ′ (x)
d
arctan f (x) =
dx
1 + f 2 (x)
•
f (x)=x2
d
2x
(arctan x2 ) =======
dx
1 + x4
•
f (x)=ln x
d
1
(arcsin(ln x)) ======== p
dx
x 1 − (ln x)2
f ′ (x)
d
arcsin f (x) = p
dx
1 − f 2 (x)
•
f (x)=arctan x
d
1
ln | arctan x| ==========
dx
(1 + x2 ) arctan x
f ′ (x)
d
ln |f (x)| =
dx
f (x)
29
Implicit Differentiation
Example: Find an equation of the tangent line to the curve x3 + y 3 = 6xy at
the point (3, 3).
Solution: Note that the derivative dy/dx gives the slope of the tangent line at
x = 3. However, it is not possible to solve for y explicitly. Therefore, we
consider y = f (x) implicitly. Then, by the Chain Rule, we have
d
f (y) = y ′ f ′ (y)
dx
We differentiate the curve with respect to x:
x3 + y 3 = 6xy
dy/dx
=⇒
3x2 + 3y 2 y ′ = 6y + 6xy ′
x=y=3
9 + 9y ′ = 6 + 6y ′ =⇒
=⇒
dy
= −1
dx x=3
Therefore, an equation for the tangent line is
y − 3 = (−1)(x − 3) =⇒ x + y = 6
30
Implicit Differentiation
Example: Find y ′ if sin(x + y) = y 2 cos x.
Solution: Again, it is not possible to solve y in terms of x explicitly. Therefore,
we assume implicitly y = f (x) where f (x) is not given explicitly.
sin(x + y) = y 2 cos x
d/dx
=⇒
(1 + y ′ ) cos(x + y) = 2yy ′ cos x − y 2 sin x
=⇒
y′ = −
cos(x + y) + y 2 sin x
cos(x + y) − 2y cos x
31
Higher Order Derivatives
Notations: Let y = f (x).
Use the prime notation only for derivatives of order less than four.
• Second derivative: f ′′ =
d2 f
d2 y
=
= y ′′
2
dx
dx2
• Fourth derivative: f (4) =
d4 y
d4 f
=
= y (4)
4
dx
dx4
• n-th derivative: f (n) =
dn f
dn y
=
= y (n)
n
dx
dxn
f n (x) = (f (x))n is the n-th power where as f (n) (x) =
dn f
is the n-the derivative.
dxn
For instance, don’t write f ′′′′ but f (4) .
32
Higher Order Derivatives
Example: Show that sin x is a solution of the equation f (4) (x) = f (x) and
find the 27-th derivative of sin x.
Solution: Let f (x) = sin x. Then f (4) (x) = f (x). Therefore,
f (n+4) (x) = f (n) (x)
df
(sin x) =
dx
cos x
d2 f
(sin x) = − sin x
dx2
d3 f
(sin x) = − cos x
dx3
d4 f
(sin x) =
dx4
=⇒
d27 f
d3 f
=
= − cos x
dx27
dx3
sin x
Example: Find the n-th derivative of f (x) = xex .
f ′ (x) = ex + xex = (x + 1)ex =⇒ f ′′ (x) = ex + (x + 1)ex = (x + 2)ex
···
f ′′′ (x) = ex + (x + 2)ex = (x + 3)ex =⇒ f (n) (x) = (x + n)ex
33
Related Rates Problems
Example: Air is being pumped into a spherical balloon so that its volume
increases at a rate of 100 cm3 /s. How fast is the radius of the balloon
increasing when the diameter is 50 cm?
Solution: Let V (t) be the volume and r(t) be the radius of the balloon. Let t0
be the time at which r(t0 ) = 25cm.
V (t) =
4 3
πr (t)
3
=⇒
d/dt
V ′ (t) = 4πr2 (t)r′ (t)
t=t0
100 = 4π(25)2 r′ (t0 ) =⇒ r′ (t0 ) =
relation
=⇒
1
cm/s
25π
34
Related Rates Problems
Example: A ladder 10 ft long rests against a vertical wall. If the bottom of the
ladder slides away from the wall at a rate of 1 ft/s, how fast is the top of the
ladder sliding down the wall when the bottom of the ladder is 6 ft from the
wall?
Solution: Let x(t) be the distance from the bottom of the ladder to the
vertical wall and y(t) be the distance from the top of the ladder to the ground.
Let t0 be the time at which x(t0 ) = 6 ft.
x2 (t) + y 2 (t) = 100
=⇒
d/dt
x(t)x′ (t) + y(t)y ′ (t) = 0
t=t0
=⇒
6 × 1 + 8 × y ′ (t0 ) = 0
=⇒
y ′ (t0 ) = −
relation
3
ft/s
4
35
Related Rates Problems
Example: A water tank has the shape of an inverted circular cone with a base
radius 2 m and height 4 m. If water is being pumped into the tank at a rate of
2 m3 /min, find the rate at which the water level is rising when the water is 3 m
deep.
Solution: Let V (t) be the volume of the water in the tank and h(t) be the
water level and r(t) be the base radius of the part filled with water. From the
h(t)
similarity of triangles we obtain r(t) =
. Let t0 be the time at which
2
h(t0 ) = 3 m.
V (t) =
π 2
π 3
r (t)h(t) =
h (t)
3
12
d/dt
=⇒
V (t) =
relation
t=t0
π 2
h (t)h′ (t)
4
9π ′
h (t0 )
4
=⇒
2=
=⇒
h′ (t0 ) =
8
m/s
9π
36
Related Rates Problems
Example: At noon, ship A is 100 km west of ship B. Ship A is sailing south at
35 km/h and ship B is sailing north at 25 km/h. How fast is the distance
between the ships changing at 4 : 00 pm?
Solution: Let L be the horizontal west-east line. Let a(t) be the vertical
distance from ship A to L and b(t) be the vertical distance from ship B to L.
Note that the horizontal distance between ship A and ship B is equal to 100
km for all times. Let t0 = 4 : 00 pm. Let d(t) be the distance between the
ships. We have a′ (t) = 35, a(t0 ) = 140, b′ (t) = 25, b(t0 ) = 100, d(t0 ) = 260.
From the right triangle, we obtain the relation:
1002 + (a(t) + b(t))2 = d2 (t)
=⇒
d/dt
(a(t) + b(t))(a′ (t) + b′ (t)) = d(t)d′ (t)
t=t0
=⇒
(140 + 100) × (35 + 25) = 260d′ (t0 )
=⇒
d′ (t0 ) =
relation
720
km/h
13
37
Related Rates Problems
Example: A man walks along a straight path at a speed of 4 ft/s. A
searchlight is located on the ground 20 ft from the path and is kept focused on
the man. At what rate is the searchlight rotating when the man is 15 ft from
the point on the path closest to the searchlight?
Solution: Let L be the vertical from the searchlight to the path. Let x(t) be
the distance from the man to the point on the path closest to the searchlight.
Let θ(t) be the angle of rotation of the searchlight as it follows the man. Let t0
be the time at which x(t0 ) = 15 ft. Below x′ (t) = 4 ft/s and sec θ(t0 ) = 5/4.
tan θ(t) =
x(t)
20
=⇒
d/dt
θ′ (t) sec2 θ(t) =
t=t0
=⇒
1
25 ′
θ (t0 ) =
16
5
=⇒
θ′ (t0 ) =
x′ (t)
20
16
rad/s
125
38