7.1 Inverse Functions and their Derivatives
2 groups of functions:
1. Algebraic- (ex. polynomial) functions obtained by addition, subtraction,
multiplication, division, or by taking powers and roots.
2. Transcendental – not algebraic (ex. trigonometric, exponential, logarithmic)
Inverse of f = a function that undoes or inverts
** a function is a rule that assigns a value from its range to each element in its domain
 some functions assign the same range value to more than 1 element in the
domain
Ex. f(x) = x² assigns the same value 1 to -1 and 1
 other functions assume each value in their range no more than once …
One to One function = a function f(x) is one to one on a domain, D, if
f(x1) ≠ f(x2) whenever x1 ≠ x2 in D
*functions take on any one value in their range exactly once
Horizontal Line Test for 1 to 1 Functions: a function y=f(x) is 1:1 if and only if its
graph intersects each horizontal line at most once
Inverse Functions - suppose that f is 1:1 functions on its domain D, with range R. Then
The inverse f-1 is defined by: f-1(a) = b if f(b) = a
 The domain of f-1 = range of f(x)
 The range of f-1 = domain of f(x)
If you compose a function with its inverse (f-1◦f)(x) = x
 only a 1:1 function can have an inverse
Functions with Inverses:
1. function that is increasing on an interval is 1:1 (makes f(x2)>f(x1), when x2>x1)
2. Decreasing functions
3. Functions that have positive derivatives at all x are increasing  so have inverses
4. Functions with negative derivatives at all x are decreasing  so have inverses
5. Functions that are neither decreasing nor increasing may still be 1:1 and have an
inverse (ex. sec-1x)
Process of passing from f to f-1 can be summarized as a 2- step process:
#1. Solve the equation y = f(x) for x  this gives a formula x=f-1(y) whose x is
expressed as a function of y.
#2. Interchange x and y, obtaining a formula y=f-1(x) where f-1 is expressed in the
conventional format with x as the independent variable and y is the dependent variable
Derivatives of Inverses of Differentiable Functions
*Derivatives of f(x) and f-1(x) will be reciprocals of each other*
(when graphed – slopes are reciprocals of each other at corresponding points)
Theorem 1: The Derivative Rule for Inverses
*If f has an interval I as domain and f-1(x) exists and is never 0 on I, then f-1 is
differentiable at every point in its domain. The value of (f-1)′ at a point, b, in the domain
of f-1 is the reciprocal of the value of f′ at the point a = f-1(b)
(f-1)′(b) = 1__
f′(f-1(b))
or df-1|
dx |x=b
1
df |
dx|x=f-1(b)
**What says:
--when y=f(x) is differentiable at x=a, and we change x by a small amount dx, the
corresponding change in y is approximately dy=f′(a) dx
--means that y changes @ f′(a) times as fast as x when x=a and that x changes @ 1/f′(a)
times as fast as y when y=b.
--It is reasonable that the derivative of f-1 at b= the reciprocal of the derivative of f at a.
**The derivative of f evaluated at a point and the derivative of the inverse evaluated at
the point at which f was originally was evaluated are derivatives:
Let f(x) = x³ -2. Find the value of df-1/dx at x = 6 = f(2)
Df/dx = 3x²|x=2 = 12
Now: df-1/dx = 1
Df/dx|x=2
=1
12
Ex’s..
7.2 Natural Logarithms
Exponential Form: f(x) = ax – easy to define when x is an integer/rational #
--when irrational the meaning is not as clear
Now we will study the Natural Log function defined as an integral through the
Fundamental Theorem of Calculus.
The Natural Logarithm Function As an Integral:
x
ln x = ∫ 1/t dt x>0
1
a) if x>1 then ln x is the area under the curve y = 1/t from t=1 to t=x
b) for 0<x<1, ln x gives the negative of the area under the curve from x to 1
c) for x≤0 function is not defined
1
Zero Width Interval Rule
ln 1 = ∫ 1/t dt = 0
1
The Number e = that # in the domain of the natural logarithm satisfying ln e = 1
**Geometrically – corresponds to: the point on the x-axis for which the area under the
graph of y=1/t and above the interval [1,e] is the exact area of the unit square
Derivative of y = ln x
for every positive value of x, we have:
d ln x = 1
dx
x
** derivative is always +  so the natural logarithm is an increasing function
Therefore, it is 1:1 and invertible
** If u is a differentiable function of x whose values are +, so that ln u is defined  then
apply the chain rule:
d ln u = 1 du
if u>0
dx
u dx
Ex. d ln 2x = 1 (2) = 1
dx
2x
x

y=ln 2x has the same derivative as y = lnx  this is true of y =
ln ax for any positive #a  because constant ends up
cancelling out.
Ex. d ln 4x = 1 (4) = 1
dx
4x
x
Properties of Logs – for any #’s a>0 and x>0, the natural log satisfies the following:
1. Product Rule:
ln ax = lna + lnx
2. Quotient Rule:
ln a/x = lna – lnx
3. Reciprocal Rule: ln 1/x = -lnx
4. Power Rule :
lnxr = rlnx
Ex. ln 10 = ln(5∙2) = ln 5 + ln2
Ex. ln2 – ln3 =
Ex. ln1/5 =
Applying the properties
Ex. ln 6 + ln cos x =
Ex. ln (5x+2) =
(3x-1)
Ex. ln √x+2
5
The graph and range of ln x
a) the 1st derivative of ln x = 1/x is positive for x>0 and is increasing
b) the 2nd derivative of ln x = -1/x² so graph is concave down
lim ln x = ∞
x∞
and
lim ln x = -∞
x0+
Domain: set of all positive real #’s
Range = entire real line giving the graph of lnx
The Integral ∫1/u du
** If u is a differentiable function that is never 0  ∫1/u du = ln|u| + C
Ex.:
** use above formula only if matches that form. – if not do normal integration
Integrals of tan x and cot x
∫tan x dx = ∫sinx dx = ∫-du
cosx
u
=-∫du = -ln|u| + c
u
= -ln|cos x| + C  ln
1 +C
|cos x|
= ln |sec x | + C
Or:
∫tan u du = -ln|cos u| + c = ln|sec u| + c
∫cot u du = ln|sin u| + c = -ln|csc u| + c
Ex.
Logarithmic Differentiation = Using the law of log to simplify the formulas before
differentiating: (think back to implicit differentiation)
Ex. find dy/dx of
y = (x²+1)(x+3)1/2
x-1
x>1
steps:
HI!!!
7.3 Exponential function
x=ex (the inverse of ln x)
**Remember ln x  is an increasing function of x with domain (0,∞) with range (-∞,∞)
 it has an inverse (ln-1) x with domain (-∞,∞) with range (0,∞)
domain: lim ln-1x = ∞
range: lim ln-1x = 0
x∞
ln e = 1
e² = e∙e
y = ex
x-∞
e = 2.718281828459045…
so e = ln-1(1)
e-2 = 1/e²
e1/2=√e
**er has a logarithm  ln er = rlne = r∙1 = r
The natural Exponential Function = for every real # x, ex = ln-1x = exp x
Inverse Equations for ex and ln x
a) eln x = x
(all x>0)
b) ln(ex) = x (all x)
Domain of ex = (-∞, ∞)
Range of ex = (0, ∞)
Ex.
General Exponential Functions – for any #’s a>0 and x, the exponential function with
base a is ax = exlna
ex. 2√3
2π
Laws of Exponents for ex
1.
2.
3.
4.
ex1∙ex2 = ex1+x2
e-x = 1/ex
ex1/ex2 = ex1-x2
(ex1)x2 = ex1x2 = (ex2)x1
Derivative and Integral of ex = exponential function is differentiable b/c it is the inverse
of differentiable function whose derivative is never = 0
*The natural exponential function ex is its own derivative
d ex = ex
dx
Chain Rule of ex = d eu = eu du
dx
dx
Ex…
The number e expressed as a limit – can be calculated as the limit e = lim (1+x)1/x
x0
Power Rule (general form) – if u is a positive differentiable function of x and n is any
real #, then un is a differentiable function of x and
d un = nun-1 du
dx
dx
Ex.
Integral of ex = ∫eu du = eu + c
Ex.
Initial Value Problems:
Ex. ey dy/dx = 2x
Steps:
x>√3
y(2)=0
7.4 ax and logax
The derivative of au
** If a>0 and u is a differentiable function of x, then au is a differentiable function of x
and:
d au = auln a du
dx
dx
ex. d/dx 3x
= 3xln 3
ex. d/dx 3sinx
The derivative of ax is positive if a>1, and negative if 0<a<1
**increasing if >1 and decreasing if 0<a<1
** ax is 1 to 1
** 2nd derivative d²/dx² (ax) = d/dx (axln a) = (ln a)²ax
 is positive of all x  so the graph of ax is concave up on each interval
The Integral of au
∫au du = au + c
ln a
-- if a ≠ 1 so that ln a ≠ 0 then
ex.
Logarithms with Base a
**If a is any positive # other than 1, the function ax is one to one and has a nonzero
derivative at every point  therefore, has a differentiable inverse
Inverse is  logarithm of x with base a  denoted by logax
logax is the inverse function of ax (graph is the reflection of ax across line y=x)
Inverse Equations for ax and logax
1. alogax = x for x >0
ex. 2log23 = 3
x
2. loga(a ) = x
ex. log2(25) = 5
Change of Base Formula:
*logax = 1 ∙ ln x = ln x
ln a
ln a
ex. log102 = ln2
1n10
** All properties also apply to logarithms  product/power/quotient
Derivative of logax -- must convert to natural logs to find
d/dx (logau) = d ln u = 1 ∙ 1 du
dx ln a
ln a
u dx
Ex. d/dx log10(3x+1)
Ex. Log25ex – log5√x
Integral of logax - convert to change of base: ln x then 1 ∫ u du
Ln a
ln a
ex. ∫log2x dx
x
7.5 Exponential Growth and Decay
**Increases or decreases in a function with changes in the independent variable
**In real world situations a quantity y, increases/decreases at a rate proportional to size
at a given time t.
Law of Exponential Change: y = y0ekt
y0 = initial condition k>0 growth
k<0 decay k=rate constant of equation
**says that the rate of change is proportional to what has already been accumulated
**if we derived the above  it shows that the only factors that are their own
derivatives are constant multiples of the exponential function
Compounding Interest Continuously = A(t) = A0ert A0 = initial amount
t=years r = continuous interest rate
Half-Life = the time required for half-life of the radioactive nuclei present in a sample to
Decay.
** a constant – that doesn’t depend on the # of radioactive nuclei initially present
in the sample, but only on the radioactive substance
Half-life = ln2
k
7.7 Inverse Trig. Functions
Inverse Trig Functions:
1. are not 1:1  their values repeat periodically so therefore do not have an inverse
2. However, can restrict their domain to intervals on which they are 1:1
Original Function
Sin x
Domain
[-π/2,π/2]
Range
[-1,1]
Cos x
[0,π]
[-1,1]
Tan x
(-π/2,π/2)
(-∞,∞) has asymptotes
Cot x
(0,π)
(-∞,∞)
Sec x
[0,π/2)U(π/2,π]
Csc x
[-π/2,0)U(0,π/2] (-∞,-1]U[1,∞)
(-∞,-1]U[1,∞)
Inverses are denoted by:
Y = sin-1x
or
y= arcsin x
Y= cos-1x
or
y= arccos x
Y = tan-1x
or
y = arctan x
Y = cot-1x
or
y = arccot x
Y= sec-1x
or
y = arcsec x
Y = csc-1x
or
y = arccsc x
** -1 in notation means inverse  NOT RECIPROCAL
#1. Arcsin x= the angle in the restricted domain: [-π/2,π/2] whose sine is x
a) Sin-1x domain: [-1,1] and range is [-π/2,π/2]
b) arcsin is symmetric about the origin  so odd function
c) values come from Q1 and Q4 b/c of specified domain (Q4 is the – radians of
Q1)
ex. sin-1½ (saying: “what radian value in my interval gives me a sine of ½)
ex. sin -1(-√2/2)
ex. sin-1(-1)
ex. sin-1(sin π/6)
#2 Arccos x = the angle in the restricted domain [0.π] whose cosine is x
a) cos-1x domain: [-1,1] and range is [0,π]
b) there is no symmetry
c) Values come from Q1 and Q2 b/c of specified domain
Ex. cos-1½
Ex. cos-10
Ex. cos-1(-√2/2)
Ex. cos-1(cos 5π/4)
#3. Arctan x = the angle in the restricted domain (-π/2, π/2) whose tan is x
a) we use open intervals ( ) b/c cotangent and tangent have asymptotes
b) tan-1x domain: (-∞,∞) and range is: (-π/2,π/2)
c) graph of tan is symmetric about origin and odd
d) Values come from Q1 and Q4 – Q4 values are the – radians in Q1
Ex. tan-11
Ex. tan-1 (-√3)
#4 Arccot x = the angle in the restricted domain (0,π) whose cot is x
a) no symmetry in this graph
b) cot-1 x domain: (-∞,∞) and range: (0,π)
c) values are in Q1 and Q2
ex. cot-1(-1)
ex. cot-1(√3)
ex. cot-1(-1/√3)
#5. Arcsec x = the angle in the restricted domain [0,π/2)U(π/2,π] whose sec is x
a) there is no general agreement about how to define sec-1x for negative values
of x, so we agree to use Q2 because it makes cos-11/x = sec-1x. It also makes
sec-1x an increasing function
b) sec-1x domain: |x| ≥ 1 and range: [0,π/2)U(π/2,π]
c) Values come from Q1 and Q2
ex. sec-1√2
ex. sec-1(-2/√3)
#6 Arccsc x = the angle in the restricted domain[-π/2,0)U(0,π/2]
a) 0 is an asymptote
b) Csc-1x domain: |x|≥ 1 and range: [-π/2,0)U(0,π/2]
c) Values come from Q1 and Q4 – Q4 are the – radians of Q1
Ex. csc-1(-√2)
Ex. csc-1(2/√3)
In situations when the inverses do not match radian values on the unit circle, we rely on
the Pythagorean Theorem to solve:
Ex. find the cos x, tanx, sec x, csc x, cot x, if x = sin-12/3
** remember sin = o/h so draw a right triangle and go from there
Derivatives of Inverse Trigonometric Functions:
#1. Derivative of y = sin-1x

Differentiable on interval [-π/2,π/2]

Its derivative –cosine is positive on this interval

So it applies to theorem 1 in sect. 7.1 will have an inverse so is differentiable on --1<x<1

Not differentiable at -1 and 1 b/c tangents to the graph are vertical at these points
d/dx (sin-1u) =
1
du/dx
|u| < 1
∙ 2x =
2x
√ 1- u²
Ex. d/dx (sin-1x²) =
1
√1- (x²)²
√1-x4
Derivative of y = tan-1u
d/dx (tan-1u) = 1 du/dx
1 + u²
Ex. tan-1√t
=
1
∙ ½t-1/2
1 + (√t)²
Derivative of sec-1u – derivative is positive for 0<x<π/2 and π/2<x<π so is differentiable
by theorem 7.1 (having an inverse of a function)
d/dx (sec-1u) =
∙ du
dx
1
|u|√u²-1
ex. d/dx sec-1(5x4) =
1
∙
4
4
|5x |√(5x )² - 1
|u|>1
(20x³)
= 4
x√25x8 -1
Other derivatives: use Inverse Functions  Inverse Cofunction Identities:
Cos-1x = π/2 – sin-1x
Cot-1x = π/2 – tan-1x
Csc-1x = π/2 – sec-1x
** the derivatives of the inverse cofunctions are the negatives of the derivatives of the
corresponding inverse functions **
d/dx cos-1u =
- 1 ∙ du
√1 - u² dx
d/dx cot-1u =
- 1 ∙ du
1 + u² dx
d/dx csc-1u =
-1
∙ du
|u|√u²-1
dx
Integration Formulas: - 3 useful formulas – hold for any constant a ≠0
#1.
#2.
#3.
∫ du
√a²-u²
u
= sin a + C
-1
(valid for u²<a²)
∫ du
a²+ u² =
1
u
a tan-1 a + C
(valid for all u)
∫ du
u√u²-a² =
1
|u |
a sec-1| a | + C
(valid for |u| > a > 0)