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22 Inverse Functions Derivatives
In this section we find the derivatives of several inverse functions we have already met, including: ln(x), arcsin(x), arccos(x) and arctan(x). By the end of this section, you should have the following skills:
- An understanding of how to calculate the derivative of an inverse func- tion, including the natural logarithm.
- Find derivatives of inverse trigonometric functions such as inverse sine, cosine and tan.
22.1 Derivative of the inverse function
Let y = f (x) be a function which has inverse function y = g(x). We now show how to obtain the derivative of g(x).
It follows from the definition of an inverse function that
f (g(x)) = x.
Hence on differentiating both sides of this equation using the Chain Rule we obtain
gโฒ(x)f โฒ(g(x)) = 1 โ
gโฒ(x) =
f โฒ(g(x))
We apply this result to find the derivatives of our list of inverse functions:
22.2 Example The Natural Logarithm
The natural logarithm, g(x) = ln(x), is the inverse function of f (x) = ex^ = exp(x). So applying the formula for the derivative of an inverse function we have: d(ln(x)) dx
= gโฒ(x) =
f โฒ(g(x))
exp(ln(x))
x
22.2.1 Inverse sine: g(x) = arcsin(x).
This is the inverse function of f (x) = sin(x). So applying the formula for the derivative of an inverse function we have:
d(arcsin(x)) dx
= gโฒ(x) =
f โฒ(g(x))
cos(arcsin(x))
Now t = arcsin(x) โ sin(t) = x โ cos(t) = ยฑ
1 โ sin^2 (t) = ยฑ
1 โ x^2. Hence we have found that
d(arcsin(x)) dx
cos(arcsin(x))
1 โ x^2
What sign should we choose? The function arcsin(x) is increasing in the range โ 1 โค x โค 1, so by our earlier work on increasing and decreasing functions, the derivative has to be positive. Hence we choose the positive square root i.e.
d(arcsin(x)) dx
1 โ x^2
Graph of arcsin(x), an increasing function.
22.2.3 Inverse tan: g(x) = arctan(x).
This is the inverse function of f (x) = tan(x). So applying the formula for the derivative of an inverse function we have:
d(arctan(x)) dx
= gโฒ(x) =
f โฒ(g(x))
sec^2 (arctan(x))
Now t = arctan(x) โ tan(t) = x โ sec^2 (t) = 1 + tan^2 (t) = 1 + x^2. Hence we have found that
d(arctan(x)) dx
1 + x^2
Note that 1/(1 + x^2 ) > 0 for all x. Hence arctan(x) is an increasing function on its domain as we can see from its graph:
Graph of arctan(x).
Example 1 Find the derivatives of:
(a) ln(1 + x).
(b) ln(x^3 + 3x + 1).
(c)
ln(x + 1) x^2 + 2x + 1
(d) ln
x x + 1
(e) arcsin(2x โ 1).
(f ) arctan(x + 3).
(g) arctan(x^2 โ 1).
Solutions.
(a) 1 /(1 + x).
(b) Let f (x) = ln(x^3 + 3x + 1). Using the Chain rule and letting u = x^3 + 3x + 1 โ f (u) = ln(u) we get
df dx
df du
du dx
3 x^2 + 3 u
3 x^2 + 3 x^3 + 3x + 1
(f )
f (x) = arctan(x + 3) โ
f โฒ(x) =
1 + (x + 3)^2
=
10 + 6x + x^2
(g)
f (x) = arctan(x^2 โ 1) โ
f โฒ(x) =
2 x 1 + (x^2 โ 1)^2
=
2 x x^4 โ 2 x^2 + 2
Exercise 1 Find the derivatives of:
(a) ln(2 โ x).
(b) ln(x^3 + x โ 1).
(c)
ln(x^2 + 1) 2 x + 1
(d) ln
x^2 x^3 + 1
(e) arcsin(3x โ 2).
(f) arctan(
x โ 1).
Solutions to exercise 1
(a) d(ln(2 โ x)) dx
= โ 1 /(2 โ x).
(b) Let f (x) = ln(x^3 + x โ 1). Using the Chain rule and letting u = x^3 + x โ 1 โ f (u) = ln(u) we get
df dx
df du
du dx
3 x^2 + 1 u
3 x^2 + 1 x^3 + x โ 1
(c) Use the Quotient Rule:
Let f (x) =
ln(x^2 + 1) 2 x + 1
f โฒ(x) =
((2x + 1) ร (2x/(x^2 + 1)) โ 2 ร ln(x^2 + 1) (2x + 1)^2
=
2 x(2x + 1) โ 2(x^2 + 1) ln(x^2 + 1) (2x + 1)^2 (x^2 + 1)
(d) Remember that ln(a/b) = ln(a) โ ln(b). So y = ln
x^2 x^3 + 1
= ln(x^2 ) โ ln(x^3 + 1)
and yโฒ^ =
x
3 x^2 x^3 + 1
2 โ x^3 x^3 + 1
(e)
f (x) = arcsin(3x โ 2) โ
f โฒ(x) = 3
1 โ (3x โ 2)^2
=
12 x โ 9 x^2 โ 3
22.3 Videos
Using inverse functions to differentiate log of a quadratic
This video shows how to differentiate y = ln(x^2 โ x + 1). Note that using the inverse function for ln we have ey^ = x^2 โ x + 1 and then the video shows how to implicitly differentiate both sides with respect to x and then obtain dy/dx.
Using inverse functions to differentiate log of a sine
The function y = ln(sin(x + x^2 )) is differentiated using ey^ = sin(x^2 + x). Implicit differentiation and the Chain Rule are then used to finish the exam- ple
Using inverse functions to differentiate arctan of a function
The function to be differentiated is y = arctan(
1 + x) and since arctan is the inverse function of tan this gives tan(y) =
1 + x. The example is then finished by using implicit differentiation and the trigono- metric identity 1 + tan^2 (y) = sec^2 (y).
