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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 tangent line. Since the graph of
is the reflection of the graph of about
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 line. Algebraically, will fail to be differentiable at a point (b, a) if
)(1 xfy
)(xfy 1f
1f
.0)(/ af
Differentiability
• We know that the equation of the tangent line to the graph of f at the point (a, b) is
))((/ axafby
Differentiability
• We know that the equation of the tangent line to the graph of f at the point (a, b) is
• To find the reflection on the line y = x we switch the x and the y, so it follows that the tangent line to the graph of at the point (b, a) is
))((/ axafby
))((/ ayafbx
1f
Differentiability
• We will rewrite this equation to make it
))((/ ayafbx )()(
1)(
/bx
afay
Differentiability
• We will rewrite this equation to make it
• This equation tells us that the slope
of the tangent line to the graph of at (b, a) is
or
If then
))((/ ayafbx )()(
1)(
/bx
afay
)()( /1 bf
1f
)(
1)()(
//1
afbf
))((
1)()(
1//1
bffbf
,)( baf abf )(1
Theorem 3.3.1
• The derivative of the inverse function of f is defined as:
))((
1)]([
1/1
xffxf
dx
d
Example
• Find the derivative of the inverse for the following function.
3
2)(
x
xf
Example
• Find the derivative of the inverse for the following function.
3
2)(
x
xf
32
)(
23
3
23
2
1
xxf
xy
yx
xy
Example
• Find the derivative of the inverse for the following function.
3
2)(
x
xf
32
)(
23
3
23
2
1
xxf
xy
yx
xy
21 2)]([
xxf
dx
d
Example
• Find the derivative of the inverse for the following function.
))((
1)]([
1/1
xffxf
dx
d
3
2)(
x
xf
Example
• Find the derivative of the inverse for the following function.
))((
1)]([
1/1
xffxf
dx
d
3
2)(
x
xf
22/
)3(
2
)3(
)1(2)0)(3()(
xx
xxf
Example
• Find the derivative of the inverse for the following function.
))((
1)]([
1/1
xffxf
dx
d
3
2)(
x
xf
22/
)3(
2
)3(
)1(2)0)(3()(
xx
xxf
2
)3(
)(
1 2
/
x
xf
Example
• Find the derivative of the inverse for the following function.
))((
1)]([
1/1
xffxf
dx
d
3
2)(
x
xf
22/
)3(
2
)3(
)1(2)0)(3()(
xx
xxf
2
)3(
)(
1 2
/
x
xf 2
22
1/
2
2
)33(
))((
1
xxffx
32
)(1
xxf
Theorem 3.3.1
• The derivative of the inverse function of f is defined as:
• We can state this another way if we let
then
))((
1)]([
1/1
xffxf
dx
d
)(1 xfy
dydxyfxf
dx
d
/
1
)(
1)]([
/1
Theorem 3.3.1
• We can state this another way if we let
then )(1 xfy
dydxyfxf
dx
d
/
1
)(
1)]([
/1
23
42
42
2
3
3
y
yyx
xxy
dydx
23
1
/
12
ydydx
Theorem 3.3.1
• We can state this another way if we let
then )(1 xfy
dydxyfxf
dx
d
/
1
)(
1)]([
/1
dxdy
dxdy
dxdy
y
y
yyx
xxy
23
1
231
42
42
2
2
3
3
23
1
/
12
ydydx
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
or on which . Then f is one-to-one,
is differentiable at all values of x in the range of f and the derivative is given by
0)(/ xf0)(/ xf )(1 xf
))((
1)]([
1/1
xffxf
dx
d
Example 2
• Consider the function .
a)Show that f is one-to-one.
b)Show that is differentiable on the interval
c)Find a formula for the derivative of .
d)Compute .
1)( 5 xxxf
1f
),( 1f
)1()( /1f
Example 2
• Consider the function .
a)Show that f is one-to-one.
which is always positive, so the function is one-to-one.
1)( 5 xxxf
15)( 4/ xxf
Example 2
• Consider the function .
a)Show that f is one-to-one.
b)Show that is differentiable on the interval
Since the range of f is , this is the domain of and from Theorem 3.3.2 is differentiable at all values of the range of f .
1)( 5 xxxf
1f
),( ),(
1f
Example 2
• Consider the function .
a)Show that f is one-to-one.
b)Show that is differentiable on the interval
c)Find a formula for the derivative of .
1)( 5 xxxf
1f
),( 1f
1)( 5 yyyfx
15 4 ydy
dx
Example 2
• Consider the function .
a)Show that f is one-to-one.
b)Show that is differentiable on the interval
c)Find a formula for the derivative of .
1)( 5 xxxf
1f
),( 1f
1)( 5 yyyfx
15 4 ydy
dx15
1
/
1)]([
41
ydydxxf
dx
d
Implicit
• Using implicit differentiation, we get
15 yyx
dx
dyy )15(1 4
dx
dy
y
15
14
Example 2
• Consider the function .
a)Show that f is one-to-one.
b)Show that is differentiable on the interval
c)Find a formula for .
d)Compute
1)( 5 xxxf
1f
),( 1f
14
1
/1
15
1)1()(
xx ydx
dyf
)1()( /1f
Example 2
• Consider the function .
a)Show that f is one-to-one.
b)Show that is differentiable on the interval
c)Find a formula for .
d)Compute• Since (0, 1) is a point on the function, the point
(1, 0) is on the inverse function.
1)( 5 xxxf
1f
),( 1f
14
1
/1
15
1)1()(
xx ydx
dyf
)1()( /1f
Example 2
• Consider the function .
a)Show that f is one-to-one.
b)Show that is differentiable on the interval
c)Find a formula for .
d)Compute• Since (0, 1) is a point on the function, the point
(1, 0) is on the inverse function.
1)( 5 xxxf
1f
),( 1f
115
1)1()(
04
/1
yy
f
)1()( /1f 14
1
/1
15
1)1()(
xx ydx
dyf
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.
)6(1 xmy
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.
)6(1 41 xy
4
1
)1(
1
))6((
1
))((
1)(
/1/1//1
fffxfff
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.
)9(2 xmy
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.
)9(2 21 xy
2
1
)2(
1
))9((
1
))((
1)(
/1/1//1
fffxfff
Derivatives of Exponential Functions
• We will use our knowledge of logs to find the derivative of . We are looking for dy/dx. xby
Derivatives of Exponential Functions
• We will use our knowledge of logs to find the derivative of . We are looking for dy/dx.
• We know that is the same as xby
xby
.log yx b
Derivatives of Exponential Functions
• We will use our knowledge of logs to find the derivative of . We are looking for dy/dx.
• We know that is the same as• We will take the derivative with respect to x and
simplify.
xby
xby
.log yx b
.log yx bdx
dy
by
ln
11
dx
dyby ln
Derivatives of Exponential Functions
• We will use our knowledge of logs to find the derivative of . We are looking for dy/dx.
• We know that is the same as• We will take the derivative with respect to x and
simplify.
• Remember that so
xby
xby
.log yx b
.log yx bdx
dy
by
ln
11
dx
dyby ln
xby bbdx
dy x ln
Derivatives of Exponential Functions
• This formula, works with any
base, so if the base is e, it becomes
but remember , so
bbbdx
d xx ln][
eeedx
d xx ln][
bbbdx
d xx ln][ xx ee
dx
d][
1ln e
Derivatives of Exponential Functions
• With the chain rule these formulas become:
dx
dubbb
dx
d uu ln][dx
duee
dx
d uu ][
Example 3
• Find the following derivatives:
]2[ x
dx
d
Example 3
• Find the following derivatives:
2ln2]2[ xx
dx
d
Example 3
• Find the following derivatives:
2ln2]2[ xx
dx
d
][ 2xedx
d
Example 3
• Find the following derivatives:
2ln2]2[ xx
dx
d
xx eedx
d 22 2][
Example 3
• Find the following derivatives:
2ln2]2[ xx
dx
d
xx eedx
d 22 2][
][3xe
dx
d
Example 3
• Find the following derivatives:
2ln2]2[ xx
dx
d
xx eedx
d 22 2][
33 23][ xx exedx
d
Example 3
• Find the following derivatives:
2ln2]2[ xx
dx
d
xx eedx
d 22 2][
33 23][ xx exedx
d
][ cos xedx
d
Example 3
• Find the following derivatives:
2ln2]2[ xx
dx
d
xx eedx
d 22 2][
33 23][ xx exedx
d
xx xeedx
d coscos sin][
Example 4
• Use logarithmic differentiation to find ])1[( sin2 xxdx
d
Example 4
• Use logarithmic differentiation to find
• Let ])1[( sin2 xxy
])1[( sin2 xxdx
d
Example 4
• Use logarithmic differentiation to find
• Let ])1[( sin2 xxy
])1[( sin2 xxdx
d
])1ln[(ln sin2 xxy
Example 4
• Use logarithmic differentiation to find
• Let ])1[( sin2 xxy
])1[( sin2 xxdx
d
])1ln[(ln sin2 xxy
)]1ln[(sinln 2 xxy
Example 4
• Use logarithmic differentiation to find
])1[( sin2 xxdx
d
)]1ln[(sinln 2 xxy
xxx
xx
dx
dy
ycos)]1ln[(
1
2sin
1 22
Example 4
• Use logarithmic differentiation to find
])1[( sin2 xxdx
d
)]1ln[(sinln 2 xxy
xxx
xx
dx
dy
ycos)]1ln[(
1
2sin
1 22
])1][(cos)]1ln[(1
2[sin sin22
2xxxx
x
xx
dx
dy
Derivatives of Inverse Trig Functions
• We want to find the derivative of .sin 1 xy
Derivatives of Inverse Trig Functions
• We want to find the derivative of
• We will rewrite this as and take the derivative.
.sin 1 xy
xy sin
Derivatives of Inverse Trig Functions
• We want to find the derivative of
• We will rewrite this as and take the derivative.
.sin 1 xy
xy sin
1cos dx
dyy
Derivatives of Inverse Trig Functions
• We want to find the derivative of
• We will rewrite this as and take the derivative.
.sin 1 xy
xy sin
1cos dx
dyy
ydx
dy
cos
1
Derivatives of Inverse Trig Functions
• We want to find the derivative of
• We will rewrite this as and take the derivative.
.sin 1 xy
xy sin
1cos dx
dyy
ydx
dy
cos
1
)cos(sin
11 xdx
dy
Derivatives of Inverse Trig Functions
• We need to simplify
• We will construct a triangle to help us do that. Remember that represents an angle where sin = x.
x1sin
)cos(sin
11 xdx
dy
Derivatives of Inverse Trig Functions
• We need to simplify
• The cosine is the adjacent over the hypotenuse.
)cos(sin
11 xdx
dy
21
1
xdx
dy
Special Triangle
• Find )sin(cos 1 x
Special Triangle
• Find
• Again, we will construct a triangle where the cos = x to help solve this problem.
)sin(cos 1 x
21 1sin)sin(cos xx
Derivatives of Inverse Trig Functions
dx
du
uu
dx
d
21
1
1][tan
dx
du
uu
dx
d
2
1
1
1][sin
dx
du
uu
dx
d
2
1
1
1][cos
dx
du
uuu
dx
d
1||
1][sec
2
1
dx
du
uu
dx
d
2
1
1
1][cot
dx
du
uuu
dx
d
1||
1][csc
2
1
Example 5
• Find dy/dx if:
)(sin 31 xy )(sec 1 xey
Example 5
• Find dy/dx if:
)(sin 31 xy )(sec 1 xey
2
233
)(1
1x
xdx
dy
6
2
1
3
x
x
dx
dy
Example 5
• Find dy/dx if:
)(sin 31 xy )(sec 1 xey
2
233
)(1
1x
xdx
dy
6
2
1
3
x
x
dx
dy
x
xxe
eedx
dy
1)(||
12
1
12
xedx
dy
Homework
• Section 3.3
• Pages 201-202
• 1, 5, 7, 9
• 15-27 odd
• 37-51 odd