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2.6. DERIVATIVES OF INVERSE FUNCTIONS
−
π
2
π
2
−
π
2
π
2
f
f −1
(
π
2 , 1)
(1,
π
2 )
Figure 2.10: A graph of f (x) = sin(x) (in blue), restricted to the domain [−
π
2 ,
π
2 ], along
with its inverse, f −1 (x) = arcsin(x) (in magenta).
y.” For example, we say that
π
6 is the angle whose sine is
1
2 , which can be written more
concisely as arcsin(
1
2 ) =
π
6 , which is equivalent to writing sin(
π
6 ) =
1
2 .
Next, we determine the derivative of the arcsine function. Letting h(x) = arcsin(x),
our goal is to find h ′ (x). Since h(x) is the angle whose sine is x, it is equivalent to write
sin(h(x)) = x.
Differentiating both sides of the previous equation, we have
d
dx
[sin(h(x))] =
d
dx
[x],
and by the fact that the righthand side is simply 1 and by the chain rule applied to the left
side,
cos(h(x))h
′ (x) = 1.
Solving for h ′ (x), it follows that
h
′ (x) =
1
cos(h(x))
.
Finally, we recall that h(x) = arcsin(x), so the denominator of h ′ (x) is the function
cos(arcsin(x)), or in other words, “the cosine of the angle whose sine is x.” A bit of right
triangle trigonometry allows us to simplify this expression considerably.
Let’s say that θ = arcsin(x), so that θ is the angle whose sine is x. From this, it follows
that we can picture θ as an angle in a right triangle with hypotenuse 1 and a vertical
leg of length x, as shown in Figure 2.11. The horizontal leg must be
√
1 − x 2 , by the
Pythagorean Theorem. Now, note particularly that θ = arcsin(x) since sin(θ) = x, and