Today we will discuss material that is not going to be covered on the math
110 exam, but that is important to those who will go on to take integral calculus.
The first topic we will discuss today are the inverse trigonometric functions
arcsin x, arccos x, and arctan x. Occasionally, people instead use the notation
sin−1 x, cos−1 x, and tan−1 x.
The most important property that these functions have is their range: Traditionally, the range of the arcsin function is [−π/2, π/2]; the range of the arccos
function is [0, π], and the range of the arctan function is (−π/2, π/2). This
means that expressions such as arcsin sin x may not behave the way you might
expect. You might think that arcsin sin x should be equal to x for every x: however, this is impossible because the range of the arcsin
is (−π/2, π/2).
√ function
2π
3
So, for example, arcsin sin 3 is equal to arcsin 2 , which is π3 . It’s very
important when dealing with inverse trigonometric functions to be careful about
the range.
Note that we have that arcsin x and arccos x each have domain [−1, 1]: these
are the only values that sin x and cos x can take. arctan x has domain (−∞, ∞),
is increasing, and has two horizontal asymptotes:
π
lim arctan x = −
x→−∞
2
π
lim arctan x = .
x→∞
2
The other topic that you will be expected to know for integral calculus that
has not been covered in math 110 is so called “logarithmic differentiation”.
This is a trick for quickly differentiating a messy product, quotient, or power of
functions.
′
(x)
. So if log f (x) is
The trick works like this: The derivative of log f (x) is ff (x)
significantly easier to differentiate than f (x), we can simply differentiate log f (x)
and multiply the result by f (x).
Here’s an example of this technique in action: suppose f (x) is given by
f (x) =
x sin x
.
(x + 1)ex
You can differentiate f (x) using the quotient rule, but it might be easier to use
logarithmic differentiation:
log f (x) = log x + log sin x − log(x + 1) − log(ex )
let’s do a little cleaning up: log ex is equal to x:
log f (x) = log x + log sin x − log(x + 1) − x
so we get
1
1
1
f ′ (x)
= +
· cos x −
−1
f (x)
x sin x
x+1
therefore:
f ′ (x) =
x sin x
(x + 1)ex
1 cos x
1
+
−
−1 .
x
sin x
x+1
1