Derivatives of Exponential and
Inverse Trig Functions
Objective: To derive and use
formulas for exponential and Inverse
Trig Functions
Differentiability
• Geometrically, a function is differentiable at
those points where its graph has a nonvertical
1
y

f
( x)
tangent line. Since the graph of
is the reflection of the graph of y  f (x) about
1
f
the line y = x, it follows that the points where
is not differentiable are reflections of the points
where the graph of f has a horizontal tangent
1
f
line. Algebraically,
will fail to be
differentiable at a point (b, a) if f / (a)  0.
Differentiability
• We know that the equation of the tangent line to
the graph of f at the point (a, b) is
y  b  f (a)( x  a)
/
Differentiability
• We know that the equation of the tangent line to
the graph of f at the point (a, b) is
y  b  f (a)( x  a)
/
• To find the reflection on the line y = x we switch
the x and the y, so it follows that the tangent line
1
to the graph of f at the point (b, a) is
x  b  f / (a)( y  a)
Differentiability
• We will rewrite this equation to make it
x  b  f (a)( y  a)
/
1
( y  a)  /
( x  b)
f (a)
Differentiability
• We will rewrite this equation to make it
1
( y  a)  /
( x  b)
f (a)
x  b  f (a)( y  a)
/
1 /
(
f
) (b)
• This equation tells us that the slope
of the tangent line to the graph of f 1 at (b, a) is
1
( f ) (b)  /
f (a)
1 /
If
f (a )  b, then
or
1 /
( f ) (b) 
f 1 (b)  a
1
f / ( f 1 (b))
Theorem 3.3.1
• The derivative of the inverse function of f is
defined as:
d
1
1
[ f ( x)]  / 1
dx
f ( f ( x))
Example
• Find the derivative of the inverse for the
following function.
2
f ( x) 
x3
Example
• Find the derivative of the inverse for the
following function.
2
y
2
f ( x) 
x3
x3
2
x
y3
2
y3
x
2
1
f ( x)   3
x
Example
• Find the derivative of the inverse for the
following function.
2
y
2
f ( x) 
x3
x3
2
x
y3
2
y3
x
d
2
2
1
1
[ f ( x)]  2
f ( x)   3
dx
x
x
Example
• Find the derivative of the inverse for the
following function.
2
f ( x) 
x3
d
1
[ f 1 ( x)]  / 1
dx
f ( f ( x))
Example
• Find the derivative of the inverse for the
following function.
2
f ( x) 
x3
d
1
[ f 1 ( x)]  / 1
dx
f ( f ( x))
( x  3)(0)  2(1)
2
f ( x) 

2
( x  3)
( x  3) 2
/
Example
• Find the derivative of the inverse for the
following function.
2
f ( x) 
x3
d
1
[ f 1 ( x)]  / 1
dx
f ( f ( x))
( x  3)(0)  2(1)
2
f ( x) 

2
( x  3)
( x  3) 2
/
1
 ( x  3) 2

/
f ( x)
2
Example
• Find the derivative of the inverse for the
following function.
2
f ( x) 
x3
d
1
[ f 1 ( x)]  / 1
dx
f ( f ( x))
( x  3)(0)  2(1)
2
f ( x) 

2
( x  3)
( x  3) 2
/
1
 ( x  3) 2

/
f ( x)
2
2
f 1 ( x )   3
x
 ( 2x  3  3) 2  2

 2
/
1
f ( f ( x))
2
x
1
Theorem 3.3.1
• The derivative of the inverse function of f is
defined as:
d
1
1
[ f ( x)]  / 1
dx
f ( f ( x))
1
• We can state this another way if we let y  f ( x)
then
d
1
1
1
[ f ( x)]  /

dx
f ( y ) dx / dy
Theorem 3.3.1
• We can state this another way if we let y  f 1 ( x)
then
d
1
1
1
[ f ( x)]  /

dx
f ( y ) dx / dy
y  x3  2 x  4
x  y  2y  4
3
dx
dy
 3y2  2
1
1
 2
dx / dy 3 y  2
Theorem 3.3.1
• We can state this another way if we let y  f 1 ( x)
then
d
1
1
1
[ f ( x)]  /

dx
f ( y ) dx / dy
y  x3  2 x  4
x  y3  2 y  4
1  3y
2 dy
dx
1

2
3y  2
2
dy
dx
dy
dx
1
1
 2
dx / dy 3 y  2
One-to-One
• We know that a graph is the graph of a function
if it passes the vertical line test.
• We also know that the inverse is a function if the
original function passes the horizontal line test.
One-to-One
• We know that a graph is the graph of a function
if it passes the vertical line test.
• We also know that the inverse is a function if the
original function passes the horizontal line test.
• If both of these conditions are satisfied, we say
that the function is one-to-one. In other words,
for every y there is only one x and for every x
there is only one y.
Increasing/Decreasing
• If a function is always increasing or always
decreasing, it will be a one-to-one function.
• Theorem 3.3.2 Suppose that the domain of a
function f is an open interval I on which f / ( x)  0
/
1
f
(
x
)

0
f
or on which
. Then f is one-to-one, ( x)
is differentiable at all values of x in the range of f
and the derivative is given by
d
1
1
[ f ( x)]  / 1
dx
f ( f ( x))
Example 2
5
f
(
x
)

x
 x 1 .
• Consider the function
a) Show that f is one-to-one.
b) Show that f 1 is differentiable on the interval
(,)
c) Find a formula for the derivative of f 1 .
d) Compute ( f 1 ) / (1) .
Example 2
5
f
(
x
)

x
 x 1 .
• Consider the function
a) Show that f is one-to-one.
f / ( x)  5 x 4  1 which is always positive, so the
function is one-to-one.
Example 2
5
f
(
x
)

x
 x 1 .
• Consider the function
a) Show that f is one-to-one.
b) Show that f 1 is differentiable on the interval
(,)
Since the range of f is (,) , this is the
domain of f 1 and from Theorem 3.3.2 is
differentiable at all values of the range of f .
Example 2
5
f
(
x
)

x
 x 1 .
• Consider the function
a) Show that f is one-to-one.
b) Show that f 1 is differentiable on the interval
(,)
c) Find a formula for the derivative of f 1 .
x  f ( y)  y 5  y  1
dx
 5y4 1
dy
Example 2
5
f
(
x
)

x
 x 1 .
• Consider the function
a) Show that f is one-to-one.
b) Show that f 1 is differentiable on the interval
(,)
c) Find a formula for the derivative of f 1 .
x  f ( y)  y 5  y  1
dx
 5y4 1
dy
d
1
1
1
[ f ( x)] 
 4
dx
dx / dy 5 y  1
Implicit
• Using implicit differentiation, we get
x  y5  y 1
dy
1  (5 y  1)
dx
4
1
dy

4
5 y 1 dx
Example 2
5
f
(
x
)

x
 x 1 .
• Consider the function
a) Show that f is one-to-one.
b) Show that f 1 is differentiable on the interval
(,)
dy
c) Find a formula for f 1 .
1 /
( f ) (1) 
1 /
dx
d) Compute ( f ) (1)

x 1
1
5 y 4  1 x 1
Example 2
5
f
(
x
)

x
 x 1 .
• Consider the function
a) Show that f is one-to-one.
b) Show that f 1 is differentiable on the interval
(,)
dy
1
c) Find a formula for f 1 .
1 /
( f ) (1) 
 4
1 /
dx x 1 5 y  1 x 1
d) Compute ( f ) (1)
• Since (0, 1) is a point on the function, the point
(1, 0) is on the inverse function.
Example 2
5
f
(
x
)

x
 x 1 .
• Consider the function
a) Show that f is one-to-one.
b) Show that f 1 is differentiable on the interval
(,)
dy
1
c) Find a formula for f 1 .
1 /
( f ) (1) 
 4
1 /
dx x 1 5 y  1 x 1
d) Compute ( f ) (1)
• Since (0, 1) is a point on the function, the point
(1, 0) is on the inverse function.
1 /
( f ) (1) 
1
5 y  1 y 0
4
1
Example 2A
• This is how we will use the other equation.
• If f -1 is the inverse of f, write an equation of the
tangent line to the graph of y = f -1(x) at x = 6.
Example 2A
• This is how we will use the other equation.
• If f -1 is the inverse of f, write an equation of the
tangent line to the graph of y = f -1(x) at x = 6.
• If f(1) = 6, f -1(6) = 1.
y  1  m( x  6)
Example 2A
• This is how we will use the other equation.
• If f -1 is the inverse of f, write an equation of the
tangent line to the graph of y = f -1(x) at x = 6.
• If f(1) = 6, f -1(6) = 1.
y  1  14 ( x  6)
1
1
( f )  / 1
 / 1
 / 
f ( f ( x)) f ( f (6)) f (1) 4
1 /
1
1
You Try
• This is how we will use the other equation.
• If f -1 is the inverse of f, write an equation of the
tangent line to the graph of y = f -1(x) at x = 9.
Example 2A
• This is how we will use the other equation.
• If f -1 is the inverse of f, write an equation of the
tangent line to the graph of y = f -1(x) at x = 9.
• If f(2) = 9, f -1(9) = 2.
y  2  m( x  9)
Example 2A
• This is how we will use the other equation.
• If f -1 is the inverse of f, write an equation of the
tangent line to the graph of y = f -1(x) at x = 9.
• If f(2) = 9, f -1(9) = 2.
y  2  12 ( x  9)
1
1
( f )  / 1
 / 1
 /

f ( f ( x)) f ( f (9)) f (2) 2
1 /
1
1
Derivatives of Exponential
Functions
• We will use our knowledge of logs to find the
derivative of y  b x. We are looking for dy/dx.
Derivatives of Exponential
Functions
• We will use our knowledge of logs to find the
derivative of y  b x. We are looking for dy/dx.
x
• We know that y  b
is the same as x  log b y.
Derivatives of Exponential
Functions
• We will use our knowledge of logs to find the
derivative of y  b x. We are looking for dy/dx.
• We know that y  b is the same as x  log b y.
• We will take the derivative with respect to x and
simplify.
x
x  log b y.
1 dy
1

y ln b dx
dy
y ln b 
dx
Derivatives of Exponential
Functions
• We will use our knowledge of logs to find the
derivative of y  b x. We are looking for dy/dx.
• We know that y  b is the same as x  log b y.
• We will take the derivative with respect to x and
simplify.
x
x  log b y.
1 dy
1

y ln b dx
• Remember that y  b
x
so
dy
y ln b 
dx
dy
 b x ln b
dx
Derivatives of Exponential
Functions
• This formula, d [b x ]  b x ln b works with any
dx
base, so if the base is e, it becomes
d x
[e ]  e x ln e
dx
but remember
ln e  1 , so
d x
[b ]  b x ln b
dx
d x
[e ]  e x
dx
Derivatives of Exponential
Functions
• With the chain rule these formulas become:
d u
du
u
[b ]  b ln b 
dx
dx
d u
u du
[e ]  e 
dx
dx
Example 3
• Find the following derivatives:
d x
[2 ]
dx
Example 3
• Find the following derivatives:
d x
[2 ]  2 x ln 2
dx
Example 3
• Find the following derivatives:
d x
[2 ]  2 x ln 2
dx
d 2 x
[e ]
dx
Example 3
• Find the following derivatives:
d x
[2 ]  2 x ln 2
dx
d 2 x
[e ]  2e  2 x
dx
Example 3
• Find the following derivatives:
d x
[2 ]  2 x ln 2
dx
d 2 x
[e ]  2e  2 x
dx
d x3
[e ]
dx
Example 3
• Find the following derivatives:
d x
[2 ]  2 x ln 2
dx
d 2 x
[e ]  2e  2 x
dx
d x3
2 x3
[e ]  3 x e
dx
Example 3
• Find the following derivatives:
d x
[2 ]  2 x ln 2
dx
d x3
2 x3
[e ]  3 x e
dx
d 2 x
[e ]  2e  2 x
dx
d cos x
[e ]
dx
Example 3
• Find the following derivatives:
d x
[2 ]  2 x ln 2
dx
d x3
2 x3
[e ]  3 x e
dx
d 2 x
[e ]  2e  2 x
dx
d cos x
[e ]   sin xecos x
dx
Example 4
• Use logarithmic differentiation to find
d
[( x 2  1) sin x ]
dx
Example 4
• Use logarithmic differentiation to find
• Let
y  [( x 2  1)sin x ]
d
[( x 2  1) sin x ]
dx
Example 4
• Use logarithmic differentiation to find
• Let
y  [( x 2  1)sin x ]
ln y  ln[( x 2  1)sin x ]
d
[( x 2  1) sin x ]
dx
Example 4
• Use logarithmic differentiation to find
• Let
y  [( x 2  1)sin x ]
ln y  ln[( x 2  1)sin x ]
ln y  sin x ln[( x 2  1)]
d
[( x 2  1) sin x ]
dx
Example 4
• Use logarithmic differentiation to find
ln y  sin x ln[( x 2  1)]
1 dy
2x
  sin x  2
 ln[( x 2  1)] cos x
y dx
x 1
d
[( x 2  1) sin x ]
dx
Example 4
• Use logarithmic differentiation to find
d
[( x 2  1) sin x ]
dx
ln y  sin x ln[( x 2  1)]
1 dy
2x
  sin x  2
 ln[( x 2  1)] cos x
y dx
x 1
dy
2x
 [sin x  2
 ln[( x 2  1)] cos x][( x 2  1)sin x ]
dx
x 1
Derivatives of Inverse Trig
Functions
• We want to find the derivative of
y  sin 1 x.
Derivatives of Inverse Trig
Functions
• We want to find the derivative of
y  sin 1 x.
• We will rewrite this as sin y  x and take the
derivative.
Derivatives of Inverse Trig
Functions
• We want to find the derivative of
y  sin 1 x.
• We will rewrite this as sin y  x and take the
derivative.
dy
cos y 
1
dx
Derivatives of Inverse Trig
Functions
• We want to find the derivative of
y  sin 1 x.
• We will rewrite this as sin y  x and take the
derivative.
dy
cos y 
1
dx
dy
1

dx cos y
Derivatives of Inverse Trig
Functions
• We want to find the derivative of
y  sin 1 x.
• We will rewrite this as sin y  x and take the
derivative.
dy
cos y 
1
dx
dy
1

dx cos(sin 1 x)
dy
1

dx cos y
Derivatives of Inverse Trig
Functions
• We need to simplify
dy
1

dx cos(sin 1 x)
• We will construct a triangle to help us do that.
1
Remember that sin x represents an angle q
where sin q = x.
Derivatives of Inverse Trig
Functions
• We need to simplify
dy
1

dx cos(sin 1 x)
• The cosine is the adjacent over the hypotenuse.
dy
1

dx
1 x2
Special Triangle
• Find
sin(cos 1 x)
Special Triangle
• Find
sin(cos 1 x)
• Again, we will construct a triangle where the
cos q = x to help solve this problem.
sin(cos 1 x)  sin q  1  x 2
Derivatives of Inverse Trig
Functions
d
1
du
1
[sin u ] 

2
dx
1  u dx
d
1
du
1
[cos u ] 

dx
1  u 2 dx
d
1
du
1
[tan u ] 

2
dx
1  u dx
d
 1 du
1
[cot u ] 

2
dx
1  u dx
d
1
du
1
[sec u ] 

2
dx
| u | u  1 dx
d
1
du
[csc 1 u ] 

dx
| u | u 2  1 dx
Example 5
• Find dy/dx if:
y  sin 1 ( x 3 )
y  sec 1 (e x )
Example 5
• Find dy/dx if:
y  sin 1 ( x 3 )
dy
1

 3x 2
dx
1  ( x3 )2
dy
3x 2

dx
1  x6
y  sec 1 (e x )
Example 5
• Find dy/dx if:
y  sin 1 ( x 3 )
y  sec 1 (e x )
dy
1

 3x 2
dx
1  ( x3 )2
dy
1

 ex
dx | e x | (e x ) 2  1
dy
3x 2

dx
1  x6
dy
1

dx
e2 x 1
Homework
•
•
•
•
•
Section 3.3
Pages 201-202
1, 5, 7, 9
15-27 odd
37-51 odd