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Inverse Functions
Review from Last Time: The Derivative of y = ln x
Last time we saw that
THEOREM 22. 0. 1. The natural log function is differentiable and
d dx (ln x) =
x
More generally, the chain rule version is
d dx (ln u) =
u
du dx
EXAMPLE 22. 0. 2. Determine Dx
[
(ln( 2 x + 1 ))^5
]
SOLUTION. Use the chain rule.
Dx [(ln( 2 x + 1 ))^5 ] = Dx (u^5 ) = 5 u^4 du dx = 5 (ln( 2 x + 1 ))^4 ·
2 x + 1
10 [ln( 2 x + 1 )]^4 2 x + 1
An Important Special Case. Since we can only take logs of positive numbers, often
times we use the log of an absolute value, e.g., ln |x|. We can find the derivatives of
such expressions as follows.
Dx [ln |x|] =
Dx [ln x] if x > 0,
Dx [ln(−x)] if x < 0
1
x if^ x^ >^ 0,
1
−x (−^1 ) =^
1
x if^ x^ <^0
x
if x 6 = 0
In other words, we get the ‘same rule’ as without the absolute value:
THEOREM 22. 0. 3. For x 6 = 0, Dx (ln |x|) =
x The chain rule version when u is a function of x is
d dx (ln |u|) =
u
du dx
EXAMPLE 22. 0. 4. Here’s one that involves a number of log properties:
Dt
[
ln
∣∣^ e
t (^) cos t √ t^2 + 1
]
= Dt
[
ln et^ + ln | cos t| − ln
t^2 + 1
= Dt
[
t + ln | cos t| − 12 ln |t^2 + 1 |
]
cos t · (− sin t) − 2 t 2 (t^2 + 1 )
= 1 − tan t − t t^2 + 1
YOU TRY IT 22. 1. Try finding these derivatives. Use log rules to simplify the functions before taking the derivative.
(a) Dx
[
ln | 6 x^3 sin x|
]
(b) Dx
[
6 x^3 ln | sin x|
]
(different)
(c) Dx
[
ln
x^4 − 1 x^2 + 1
]
(d) Dt
[
ln(t(e t (^) ) )
]
(e) Dx
[
ln 3
3 x^3 + x + 1
]
(f ) Ds
[
ln ( 5 ln^ s)
]
and log properties, then simplify. 3. 0. 22. Use Theorem 1. answer to you try it 22
xD )a( [
|x sin 3 x 6 | ln ]
xD = [
|x sin | ln + | 3 x 6 | ln ]
= 3 x ) 18b( x cot +^ x x cot 3 x 6 + |x sin | ln 3
xD )c ( [
ln ∣∣ x∣∣ 1 − 4
1 + 2 x ∣∣
∣∣ ]
xD = [
ln ∣∣
1 − 4 x∣ ∣∣
ln − ∣ ∣∣
1 + 2 x|∣ ∣∣
∣ ]
= 3 x^4
−^1 −^4 x x 2
tD )d(^1 +^2 x [ e(t(^ ln ) t ) ]
tD = [
t ln te ]
1 t
xD )e ( [ 3 ln √
1 + x + 3 x 3 ]
xD = [ 1
3 3 ln^ · √
1 + x + 3 x 3 ] 1 =
3 1 + 2 x 9
1 + x + 3 x 3 sD ) f( [
) s ln 5 ( ln ] ln 5^ ] =ln 5^ s^ ln[^ sD^ =
s
In part (f), remember that ln 5 is just a constant.
EXAMPLE 22. 0. 6. Find the derivatives of the following functions.
(a) y = 2 x^ (b) z = 15 ( 3 t/10) (c) y = 5 t (^3) sin t (d) y = 4 x^ tan( 4 x)
Solution. Using Theorem 22. 0. 5 ,
(a)
d
dx
( 2 x^ ) = 2 x^ ln 2.
(b) This time we use the chain rule:
d
dt
15 ( 3 t/10)
= 15 ( 3 t/10) ln 3 ·
d
dt (^
t
1
10 =^
3
2 (^3
t/10) ln 3.
(c) Use the chain rule in combination with Theorem 22. 0. 5 ,
d
dt
u
t^3 sin t))^ = 5 t^3 sin t ln 5 ·
d
dt (t
3 sin t)
( 3 t^2 sin t − t^3 cos t).
(d) Use the product rule:
d
dx
( 4 x^ tan( 4 x)) = 4 x^ ln 4 · tan( 4 x) + 4 x^ sec^2 ( 4 x) · 4 = 4 x^ (ln 4 tan( 4 x) + 4 sec^2 ( 4 x)).
Logarithmic Differentiation
There are still types of functions that we have not tried to differentiate yet. Some-
times we can make use of our existing techniques and clever algebra to find
derivatives of very complicated functions. Logarithmic differentiation refers to
the process of first taking the natural log of a function y = f (x), then solving for
the derivative
dy
dx
. On the surface of it, it would seem that logs would only make a
complicated function more complicated. But remember that logs turn powers into
products and products into sums. That’s the key.
Here’s a neat problem to illustrate the idea.
EXAMPLE 22. 0. 7. Use the chain rule and implicit differentiation along with logs to find the derivative of y = f (x) = xx.
Solution. We begin by taking the natural log of both sides and simplifying using
log properties.
ln y = ln xx^ Powers = x ln x.
Remember we want to find
dy
dx
, so take the derivative of both sides (implicitly on
the left).
d
dx
(ln y) =
d
dx
(x ln x) ⇒
y
dy
dx
= 1 · ln x + x ·
x
= ln(x) + 1
dy
dx
Solve
= y[ln(x) + 1 ]
dy
dx
Substitute
= xx^ [ln(x) + 1 ]
In other words, we have shown that
d
dx
(xx^ ) = xx^ [ln(x) + 1 ]. Neat! Easy!
Here are a couple more.
EXAMPLE 22. 0. 8. Find the derivative of y = ( 1 + x^2 )tan^ x.
Solution. Take the natural log of both sides and simplify using log properties.
ln y = ln( 1 + x^2 )tan^ x^
Powers
= tan x ln( 1 + x^2 ).
Take the derivative of both sides (implicitly on the left) and solve for
dy
dx
d
dx
(ln y) =
d
dx
tan x ln( 1 + x^2 )
y
dy
dx
= sec^2 x ln( 1 + x^2 ) + tan x ·
2 x
1 + x^2
dy
dx
Solve
= y
[
sec^2 x ln( 1 + x^2 ) +
2 x tan x
1 + x^2
]
dy
dx
Substitute
= ln( 1 + x^2 )tan^ x
[
sec^2 x ln( 1 + x^2 ) +
2 x tan x
1 + x^2
]
So
d
dx
ln( 1 + x^2 )tan^ x^
= ln( 1 + x^2 )tan^ x
[
sec^2 x ln( 1 + x^2 ) +
2 x tan x
1 + x^2
]
. Not bad!
EXAMPLE 22. 0. 9. Find the derivative of y = (ln x)x 3 .
Solution. Be careful. This function is NOT the same as ln(xx
3
) which would
equal x^3 ln x. Instead, take the natural log of both sides and simplify using log
properties.
ln y = ln(ln x)x
(^3) Powers
= x^3 ln(ln x).
Take the derivative of both sides (implicitly on the left) and solve for
dy
dx
Do you see the difference when com- pared to ln(xx^3 )
y
dy
dx
= 3 x^2 ln(ln x) + x^3 ·
ln x
x
y
dy
dx
Solve = y
[
3 x^2 ln(ln x) +
x^3
x ln x
]
dy
dx
Substitute
= (ln x)x
3
[
3 x^2 ln(ln x) +
x^3
x ln x
]
Logs can also be used to simplify products and quotients.
EXAMPLE 22. 0. 10. Find the derivative of y = (x^2 − 1 )^5
1 + x^2 x^4 + 4
Solution. Use logarithmic differentiation to avoid a complicated quotient rule
derivative Take the natural log of both sides and then simplify using log proper-
ties.
ln y = ln
(x^2 − 1 )^5
1 + x^2
x^4 + 4
Log Prop
= ln(x^2 − 1 )^5 + ln( 1 + x^2 )1/2^ − ln(x^4 + 4 )
Log Prop
= 5 ln(x^2 − 1 ) +
ln( 1 + x^2 ) − ln(x^4 + 4 ).
Take the derivative of both sides and solve for
dy
dx
