• We cannot apply the exponential rule, because the base is not a constant. So it’s not xx (ln(x)).
We have y = xx so ln(y) = x ln(x). We can differentiate both sides with respect to x, remembering to use the
chain rule when differentiating ln(y):
1
y′
= ln(x) + x
y
x
or
y ′ = y(ln(x) + 1) = xx (ln(x) + 1)
which is somehow a funny compromise between the two rules we couldn’t use!
ALWAYS apply the logarithm to differentiate a function of the form f (x)g(x) .
The power rule and the exponential rule only apply when either the power
or the base is a constant.
Remark 3.47. We can now prove the power rule for any n: set y = xn . Then ln(y) = n ln(x) so
1
y
y′
= n ⇒ y ′ = n = nxn /x = nxn−1 .
y
x
x
3.5
Lecture 10: More on logarithms, as well as inverse trigonometric functions
Example 3.48. Find the derivative of y = (x2 + 1)sin(x) .
Solution: apply ln to both sides and differentiate using the chain rule. See Section 3.7 in Stewart for similar
examples.
Now let’s look at an example where the logarithm is useful. Our motivation is:
Logarithms are friendly functions. Remember the rules of logarithms:
ln(ab ) = b ln(a)
and
ln(ab) = ln(a) + ln(b)
3/4 √
1+x
. Find h′ (x).
Example 3.49. Let h(x) = ln x x2 +1
Solution: Well, we could just go ahead an use
′
h (x) = 1
√
x3/4 1+x
x2 +1
·
!′
√
x3/4 1 + x
x2 + 1
but let’s use the rules of logarithms to simplify, first. Namely, we have
√
h(x) = ln(x3/4 ) + ln( 1 + x) − ln(x2 + 1)
1
3
= ln x + ln(1 + x) − ln(x2 + 1)
4
2
whence
h′ (x) =
3
1
2x
+
−
4x 2(1 + x) x2 + 1
which is much less painful.
49
2/3
2
3
7
+1)(x +1)
√
Example 3.50. Suppose y = x (x
. Find y ′ .
x3 +x+2
Well, we could use the quotient rule and product rule; that’s certainly an option. But remember that logarithms
simplify expressions. So here’s an approach:
We have
!
x2/3 (x2 + 1)(x3 + 1)7
√
ln(y) = ln
x3 + x + 2
2
1
= ln(x) + ln(x2 + 1) + 7 ln(x3 + 1) − ln(x3 + x + 2)
3
2
So we can differentiate both sides with respect to x (remembering that y is a function of x so we need to use the
chain rule!):
2
2x
21x2
3x2 + 1
1 ′
y =
+ 2
+ 3
−
y
3x x + 1 x + 1 2(x3 + x + 2)
whence
y′ =
2x
21x2
3x2 + 1
2
+ 2
+ 3
−
3x x + 1 x + 1 2(x3 + x + 2)
x2/3 (x2 + 1)(x3 + 1)7
√
x3 + x + 2
!
which is still ugly but at least we got to an answer mostly painlessly.
Example 3.51. Set f (x) = ln |x|. Find f ′ (x).
Solution: This function is a piecewise-defined function, given by
(
ln(x)
if x > 0;
f (x) = ln |x| =
ln(−x) if x < 0,
and undefined at x = 0. Its graph is like ln(x) but including the reflection across the y-axis; it has a vertical
asymptote at x = 0.
Its derivative is
(
(ln(x))′ = x1
if x > 0;
′
f (x) =
1
1
′
(ln(−x)) = −x · (−1) = x if x < 0.
Wow! The nice-looking function y = 1/x is the derivative of ln |x| everywhere!
1
d
ln |x| =
dx
x
d
ln |x| 6=
Note that dx
is always positive.
Example 3.52.
1
|x|
— sketch the graphs and see that ln |x| should have negative slope for x < 0 and
Z
1
dx = ln |x| + c.
x
1
|x|
That is, the most
general anti-derivative of 1/x is ln |x| and we prefer it because it has a bigger domain so is more useful.
3.5.1
Inverse trigonometric functions
See [S, Ch 3.6].
We are interested in the inverse trigonometric functions for two reasons:
1. We use them whenever we want to solve an equation like cos(x) = 21 .
2. Their derivatives, as we’ll see, are rather incredibly useful functions; so inverse trig functions occur very
frequently in integration problems!
50
3.5.2
The inverse sine function: arcsin(x) or sin−1 (x)
Both notations are acceptable but you must recall that sin−1 (x) means the inverse function of sine, NOT csc(x),
DESPITE the suggestive −1. The “−1” is intended to evoke “inverse function” NOT reciprocal.
We’re write y = arcsin(x) when sin(y) = x.
So we sketch the graph of y = sin(x); this is not one-to-one; therefore, like we did for y = x2 , we have to agree
on a portion of the domain of y = sin(x) to which we can restrict the function. We have universally agreed on
[−π/2, π/2]. So
sine is one-to-one on domain [−π/2, π/2] with image [−1, 1]; so
arcsine is defined on domain [−1, 1] with image [−π/2, π/2].
They are related by:
y = sin(x) ⇔ x = arcsin(y) for all x ∈ [−π/2, π/2]
Example 3.53. Find all x such that sin(x) = 0.
Solution: exactly one solution is given by x = arcsin(0) = 0. To find all the others, we look at the graph of
y = sin(x) and see that kπ, for any integer k, is also a solution.
Example 3.54. Find all x such that sin(x) = 21 .
Solution: exactly one solution is given by x = arcsin( 21 ) = π/6. To find all others, we look at the graph of
y = sin(x), or we use the identities:
sin(x + 2πk) = sin(x) for any integer k
and
sin(x) = sin(π − x).
We see that these account for all the possible solutions, so our final answer is:
x = π/6 + 2πk,
or x = (π − π/6) + 2πk = 5π/6 + 2πk
for any integer k.
3.5.3
The derivative of arcsin(x)
We consider the identity
sin(arcsin(x)) = x
for all x ∈ [−1, 1]. Differentiating both sides, we get
cos(arcsin(x))
or
d
(arcsin(x)) = 1
dx
1
d
(arcsin(x)) =
.
dx
cos(arcsin(x))
This looks somewhat hideous: but let’s simplify it.
So arcsin(x) is the angle theta, with −π/2 ≤ θ ≤ π/2, such that sin(θ) = x. Now we know
sin2 (θ) + cos2 (θ) = 1
and moreover, on −π/2 ≤ θ ≤ π/2, cos(θ) ≥ 0, so we can conclude
q
p
cos(θ) = 1 − sin2 (θ) = 1 − x2 .
51
Therefore:
1
d
(arcsin(x)) = √
dx
1 − x2
A quick reality check: indeed, this function is defined only for −1 < x < 1, which is what you’d expect for the
derivative of arcsin(x).
Example 3.55. Let y = arcsin(ex + x2 ). Then
3.5.4
1
· (ex + 2x)
y′ = p
1 − (ex + x2 )2
The inverse tangent function: arctan(x) or tan−1 (x)
Sketching the graph of y = tan(x), we see that our natural choice for restricting the domain is similar to that for
sin(x); except we must exclude endpoints because of the vertical asymptotes.
tan is one-to-one on domain (−π/2, π/2) with image R; so
arctan is defined on domain R with image (−π/2, π/2).
They are related by:
y = tan(x) ⇔ x = arctan(y) for all x ∈ (−π/2, π/2)
Example 3.56. Find all solutions to tan(x) = 1.
One solution is arctan(1) = π/4. The graph of y = tan(x) is periodic with period π, and so in fact, we simply
have that all solutions are
x = π/4 + πk
for some integer k.
3.5.5
The derivative of arctan(x)
Suppose y = arctan(x). Then tan(y) = x, so by differentiating both sides with respect to x, we have
sec2 (y)y ′ = 1
1
which gives y ′ = cos2 (y) = sec2 (arctan(x))
.
Now if arctan(x) = θ, then we deduce (by a trig identity):
sec2 (θ) = 1 + tan2 (θ) = 1 + tan(arctan(x))2 = 1 + x2
so
d
1
arctan(x) =
dx
1 + x2
Again, this is reasonable; this function is defined on all of R and goes to 0 as x goes to ±∞, as you’d expect
from the graph of arctan(x).
Example 3.57. Let y =
1
arctan(x) .
Then y = (arctan(x))−1 so
y ′ = − arctan(x)
52
1
.
1 + x2
3.5.6
The inverse cosine function: arccos(x) or cos−1 (x)
We can do the same with arccos(x).
cosine is one-to-one on domain [0, π] with image [−1, 1]; so
arccosine is defined on domain [−1, 1] with image [0, π].
They are related by:
y = cos(x) ⇔ x = arccos(y) for all x ∈ [0, π]
√
Example 3.58. Find all solutions
to
cos(x)
=
3/2.
√
One solution is arccos( 3/2) = π/6; this is the only solution in the interval [0, π] since cos(x) is one-to-one
there. To find all other solutions, we use the identities:
cos(x + 2πk) = cos(x) for all integers k
and
cos(−x) = cos(x).
We see from the graph that these give us all other solutions. Therefore the answer is
x = π/6 + 2πk,
or x = −π/6 + 2πk
for any integer k.
3.5.7
The derivative of arccos(x)
We could repeat the argument used above, but instead we might stare at the graph of arccos(x) and realize that
it has a very similar shape to that of arcsin(x), because they are each portions of a sinusoidal curve in y. How can
we use this?
We can relate the portions of the sine and cosine graphs on which we took the inverse functions. We know
that for 0 ≤ x ≤ π,
cos(x) = sin(π/2 − x)
with −π/2 ≤ π/2 − x ≤ π/2. So
y = arccos(x) ⇒ x = cos(y) = sin(π/2 − y) ⇒ π/2 − y = arcsin(x)
so that
arccos(x) = π/2 − arcsin(x).
It follows that
−1
d
(arccos(x)) = √
dx
1 − x2
Interesting, but this did not add to our store of interesting functions whose anti-derivatives we know; we
typically use arcsin(x) if we have a choice.
Example 3.59.
Z
√
1
dx = arcsin(x) + c
1 − x2
53