Derivatives of inverse functions
1. Prove that
d
dx
arcsin(x) =
√ 1
.
1−x2
Proof: Set y = arcsin(x) so that we want to show
the definition of arcsine we know
x = sin(y)
(!)
and
− π2 ≤ y ≤
π
2
dy
dx
=
√ 1
.
1−x2
By
(!!)
Differentiating both sides of (!) with respect to x gives
d
d
(x) =
(sin(y))
dx
dx
=⇒
1=
dy
1
=
.
dx
cos(y)
dy
=⇒
dx
Now let’s use the pythagorean identity:
=⇒
2
1 = cos(y)
2
sin (y)+cos (y) = 1
2
!
cos(y) = ± 1 − sin2 (y)
2
cos (y) = 1−sin (y) =⇒
"
By (!!) we know cos(y) > 0 so that cos(y) = 1 − sin2 (y). So we
have
1
dy
1
1
(")
= √
.
=
="
2
dx
cos(y)
1 − x2
1 − sin (y)
2. Prove that
d
dx
=⇒
d
dy
(sin(y))
dy
dx
arccos(x) =
√ −1 .
1−x2
Proof: Set y = arccos(x) so that we want to show
the definition of arccosine we know
x = cos(y)
(!)
and
0≤y≤π
dy
dx
=
√ −1 .
1−x2
By
(!!)
Differentiating both sides of (!) with respect to x gives
d
d
(x) =
(cos(y))
dx
dx
=⇒
1=
dy
=⇒
dx
Now let’s use the pythagorean identity:
=⇒
1 = − sin(y)
sin2 (y)+cos2 (y) = 1
d
dy
(cos(y))
dy
dx
dy
−1
=
.
dx
sin(y)
"
sin2 (y) = 1−cos2 (y) =⇒ sin(y) = ± 1 − cos2 (y)
"
By (!!) we know sin(y) > 0 so that sin(y) = 1 − cos2 (y). So we
have
−1
−1
−1
dy
(")
.
= √
=
="
2
dx
sin(y)
1 − x2
1 − cos (y)
=⇒
d
dx
3. Prove that
arctan(x) =
1
.
1+x2
Proof: Set y = arctan(x) so that we want to show
definition of arctangent we know
x = tan(y)
dy
dx
=
1
.
1+x2
By the
(!)
[Note: it is also true that − π2 ≤ y ≤ π2 , but we will not need this fact
in the proof.] Differentiating both sides of (!) with respect to x gives
d
d
(x) =
(tan(y))
dx
dx
=⇒
1 = sec2 (y)
=⇒
dy
dx
1=
=⇒
d
dy
(tan(y))
dy
dx
dy
1
=
.
dx
sec2 (y)
Now let’s use the pythagorean identity:
sin2 (y)+cos2 (y) = 1
=⇒
sin2 (y) cos2 (y)
1
+
=
cos2 (y) cos2 (y)
cos2 (y)
=⇒
tan2 (y)+1 = sec2 (y)
So we have
1
dy
1
1
(")
=
=
=
.
dx
sec2 (y)
1 + x2
tan2 (y) + 1
4. Prove that
d
dx
ln(x) = x1 .
Proof: Set y = ln(x) so that we want to show
definition of the natural logarithm we know
x = ey
dy
dx
=
1
x.
By the
(!)
Differentiating both sides of (!) with respect to x gives
d
d y
(x) =
(e )
dx
dx
=⇒
5. Prove that
d
dx
logb (x) =
=⇒
1=
dy
1
= y
dx
e
d y dy
(e )
dy
dx
(")
=⇒
=⇒
1 = ey
dy
dx
dy
1
= .
dx
x
1
x ln(b) .
Proof: Set y = logb (x) so that we want to show
definition of a logarithm we know
x = by
(!)
dy
dx
=
1
x ln(b) .
By the
Differentiating both sides of (!) with respect to x gives
d
d y
(x) =
(b )
dx
dx
=⇒
=⇒
1=
d y dy
(b )
dy
dx
dy
1
= y
dx
b ln(b)
=⇒
1 = by ln(b)
dy
dx
dy
1
=
.
dx
x ln(b)
(")
=⇒
6. Let f be a differentiable function with inverse f −1 which is also differentiable. Prove that
1
d −1
f (x) = # −1
dx
f (f (x))
as long as the denominator is not zero.
Proof: Set y = f −1 (x) so that we want to show
the definition of inverse functions we know
x = f (y)
dy
dx
=
1
.
f ! (f −1 (x))
By
(!)
Differentiating both sides of (!) with respect to x gives
d
d
(x) =
(f (y))
dx
dx
=⇒
=⇒
1=
1
dy
= #
dx
f (y)
d
dy
(f (y))
dy
dx
=⇒
=⇒
1 = f # (y)
dy
1
= # −1
.
dx
f (f (x))
dy
dx