antiderivative of lnx

Using implicit di↵erentiation for good: Inverse functions.
Warmup: Calculate
1. e y = xy
dy
dx
if
2. cos(y ) = x + y
Every time:
d
(1) Take dx
of both sides.
(2) Add and subtract to get the
(3) Factor out
dy
dx
dy
dx
terms on one side and everything else on the other.
and divide both sides by its coefficient.
The Derivative of y = ln x
Remember:
(1) y = e x has a slope through the point (0,1) of 1.
(2) The natural log is the inverse to e x , so
y = ln x
=)
ey = x
y=ex
y=ln(x)
m=1
m=1
The Derivative of y = ln x
To find the derivative of ln(x), use implicit di↵erentiation!
Rewrite
y = ln x as e y = x
Take a derivative of both sides of e y = x to get
dy y
e =1
dx
dy
1
= y
dx
e
so
Problem: We asked “what is the derivative of ln(x)?” and got
back and answer with y in it!
Solution: Substitute back!
dy
1
1
1
= y = ln(x) =
dx
e
x
e
d
dx
ln(x) =
1
x
d
dx
ln(x) =
1
x
Does it make sense?
1
f (x) = ln(x)
1
2
3
4
1
2
3
4
-1
3
2
f (x) =
1
x
1
Examples
Calculate
1.
d
dx
ln x 2
2.
d
dx
ln(sin(x 2 ))
3.
d
dx
log3 (x)
[hint: loga x =
ln x
ln a ]
Quick tip: Logarithmic di↵erentiation
sin(x)
Example: Calculate dy
dx if y = x
Problem: Both the base and the exponent have the variable in
them! So we can’t use
d a
x = ax a
dx
1
or
d x
a = ln(a)ax .
dx
Fix: Take the log of both sides and use implicit di↵erentiation:
ln(y ) = ln(x sin(x) ) = sin(x) ⇤ ln(x)
b
(using ln(a ) = b ln(a))
Taking the derivative of both sides gives
1 dy
1
= cos(x) ln(x) + sin(x)
y dx
x
Then solving for
dy
=y
dx
✓
dy
dx ,
1
cos(x) ln(x) + sin(x)
x
◆
= x
sin(x)
✓
1
cos(x) ln(x) + sin(x)
x
◆
.
Back to inverses
In the case where y = ln(x), we used the fact that ln(x) = f
where f (x) = e x , and got
d
1
ln(x) = ln(x) .
dx
e
In general, calculating
(1) Rewrite y = f
d
dx f
1 (x):
1 (x)
as f (y ) = x.
(2) Use implicit di↵erentiation:
f 0 (y ) ⇤
dy
=1
dx
so
dy
1
1
= 0
= 0
.
dx
f (y )
f (f 1 (x))
Examples
Just to check, use the rule
d
1
f 1 (x) = 0
dx
f (f 1 (x))
to calculate
d
dx
ln(x) (the inverse of e x )
In the notation above, f 1 (x) = ln(x) and f (x) = e x .
We’ll also need f 0 (x) = e x . So
d
1
ln(x) = ln(x) ,
dx
e
p
d
2. dx
x (the inverse of x 2 )
p
In the notation above, f 1 (x) = x and f (x) = x 2 .
We’ll also need f 0 (x) = 2x. So
d p
1
p
x=
,
dx
2 ⇤ ( x)
1.
1 (x),
Inverse trig functions
Two notations:
f (x)
sin(x)
cos(x)
tan(x)
sec(x)
csc(x)
cot(x)
f 1 (x)
1
(x) = arcsin(x)
1 (x) = arccos(x)
1 (x) = arctan(x)
1 (x) = arcsec(x)
1 (x) = arccsc(x)
1 (x) = arccot(x)
sin
cos
tan
sec
csc
cot
There are lots of points we know on these functions...
Examples:
1. Since sin(⇡/2) = 1, we have arcsin(1) = ⇡/2
2. Since cos(⇡/2) = 0, we have arccos(0) = ⇡/2
Etc...
In general:
arc
( - ) takes in a ratio and spits out an angle:
!
θ
#
"
cos(✓) = a/c
so
arccos(a/c) = ✓
sin(✓) = b/c
so
arcsin(b/c) = ✓
tan(✓) = b/a
so
arctan(b/a) = ✓
Domain problems:
sin(0) = 0,
sin(⇡) = 0,
sin(2⇡) = 0,
So which is the right answer to arcsin(0), really?
sin(3⇡) = 0, . . .
Domain:
Domain/range
Domain/range
1x 1
Domain:
1x 1
2
Range:
- "!/
- "!/
!/2
y = arcsin(x)
1x 1
-
2
Domain/range
Domain:
y = sin(x)
y = arcsin(x)
-
!/2
y = arcsin(x)
y = sin(x)
Domain/range
⇡/2  y  ⇡/2
Domain:
Domain/range
Domain/range
1x 1
!
0
-
y = arccos(x)
y = cos(x)
Domain:
Domain/range
Domain:
Domain/range
1x 1
!
0
-
Range: 0  y  ⇡
y = arccos(x)
1x 1
y = cos(x)
y = arccos(x)
Domain:
Domain/range
Domain/range
1x 1
Domain:
1x 1
2
Range:
- "!/
- "!/
!/2
y = arctan(x)
1x 1
-
2
Domain/range
Domain:
y = tan(x)
y = arctan(x)
-
!/2
y = arctan(x)
y = tan(x)
Domain/range
⇡/2 < y < ⇡/2
Domain: x 
Domain/range
Domain/range
0
-
1 and 1  x
!
y = arcsec(x)
y = sec(x)
Domain: x 
Domain/range
Domain: x 
Domain/range
0
-
1 and 1  x
!
y = arcsec(x)
1 and 1  x
y = sec(x)
y = arcsec(x)
Range: 0  y  ⇡
Domain: x 
Domain/range
Domain/range
1 and 1  x
Domain: x 
1 and 1  x
- "!/
- "!/
!/2
Range:
2
y = arccsc(x)
1 and 1  x
-
2
Domain/range
Domain: x 
y = csc(x)
y = arccsc(x)
-
!/2
y = arccsc(x)
y = csc(x)
Domain/range
⇡/2  y  ⇡/2
Domain:
Domain/range
Domain/range
0
-
1x 1
!
y = arccot(x)
y = cot(x)
Domain:
Domain/range
Domain:
Domain/range
0
-
1x 1
!
Range: 0 < y < ⇡
y = arccot(x)
1x 1
y = cot(x)
y = arccot(x)
Back to Derivatives
Use implicit di↵erentiation to calculate the derivatives of
1. arcsin(x)
2. arctan(x)
Use the rule
d
1
f 1 (x) = 0
dx
f (f 1 (x))
to check your answers, and then to calculate the derivatives of the
other inverse trig functions:
1.
2.
3.
4.
d
dx arccos(x)
d
dx arcsec(x)
d
dx arccsc(x)
d
dx arccot(x)
Recall:
f (x)
f 0 (x)
sin(x)
cos(x)
cos(x)
sin(x)
tan(x)
sec2 (x)
sec(x)
sec(x) tan(x)
csc(x)
csc(x) cot(x)
cot(x)
csc2 (x)
Using implicit di↵erentiation to calculate
d
dx
arcsin(x)
If y = arcsin(x) then x = sin(y ).
Take
d
dx
of both sides of x = sin(y ):
d
x =1
dx
dy
dy
d
sin(y ) = cos(y )⇤
= cos(arcsin(x))⇤
dx
dx
dx
Left hand side:
Right hand side:
So
dy
1
=
.
dx
cos(arcsin(x))
Simplifying cos(arcsin(x))
Call arcsin(x) = ✓.
sin(✓) = x
1
θ
!
!
Key: This is a simple triangle to write down
whose angle ✓ has sin(✓) = x
Simplifying cos(arcsin(x))
Call arcsin(x) = ✓.
sin(✓) = x
1
!
θ
√1"#"$²
p
cos( arcsin(x)) = 1
So
x2
d
1
1
arcsin(x) =
=p
.
dx
cos(arcsin(x))
1 x2
So
Calculating
!
d
dx
arctan(x).
We found that
d
1
arctan(x) =
=
2
dx
sec (arctan(x))
✓
Simplify this expression using
!
arctan(x)
!
1
1
sec(arctan(x))
◆2
To simplify the rest, use the triangles
1
!
arccos(x)
arcsec(x)
!
!
!
!
1
1
arccsc(x)
!
1
arccot(x)
!
!