Kenneth Kuttler

5.3. DERIVATIVES OF INVERSE FUNCTIONS 125

Proof: This is left to you. Use the chain rule and the product rule.Higher order derivatives are defined in the obvious way. f ′′ ≡ ( f ′)′ etc. Also the Leibniz

notation is defined by dydx = f ′ (x) where y = f (x) and the second derivative is denoted as

d2ydx2 with various other higher order derivatives defined similarly. When people write y(n)

they mean the nth derivative. Similarly f (n) (x) refers to the nth derivative.The chain rule has a particularly attractive form in Leibniz’s notation. Suppose y= g(u)

and u = f (x) . Thus y = g◦ f (x) . Then from the above theorem

(g◦ f )′ (x) = g′ ( f (x)) f ′ (x) = g′ (u) f ′ (x)

or in other words, dydx = dy

dududx . Notice how the du cancels. This particular form is a very

useful crutch and is used extensively in applications.

5.3 Derivatives of Inverse FunctionsIt happens that if f is a differentiable one to one function defined on an interval, [a,b] ,and f ′ (x) exists and is non zero then the inverse function f−1 has a derivative or one sidedderivative at the point f (x) .

Theorem 5.3.1 Let f : [a,b] → R be continuous and one to one. Suppose f ′ (x)exists for some x ∈ [a,b] and f ′ (x) ̸= 0, a one sided derivative at the end points. Then(

f−1)′( f (x)) exists and is given by the formula,

(f−1)′( f (x)) = 1

f ′(x) .

Proof: By Lemma 4.4.3, and Corollary 4.5.1 on Page 105 f is either strictly increasingor strictly decreasing and f−1 is continuous on an interval f ([a,b]). Constrain h to have theappropriate sign if at an endpoint of f ([a,b]) , and letting |h| be sufficiently small otherwise,let x be a point where f ′ (x) ̸= 0 and f (x) = y

h = f(

f−1 (y+h))− f

(f−1 (y)

f ′ (x)(

f−1 (y+h)− f−1 (y))+o(

f−1 (y+h)− f−1 (y))

(∗)

By continuity of f−1,∣∣o( f−1 (y+h)− f−1 (y)

)∣∣ < 12 | f

′ (x)|∣∣ f−1 (y+h)− f−1 (y)

∣∣ if h issmall enough and so, from the triangle inequality in ∗,

|h| ≥ 12

∣∣ f ′ (x)∣∣ ∣∣ f−1 (y+h)− f−1 (y)∣∣ ,∣∣o( f−1 (y+h)− f−1 (y)

)∣∣|h|

≤2∣∣o( f−1 (y+h)− f−1 (y)

)∣∣| f ′ (x)| | f−1 (y+h)− f−1 (y)|

showing that o(

f−1 (y+h)− f−1 (y))= o(h) . From ∗,

1f ′ (x)

h+o(h) = f−1 (y+h)− f−1 (y) = f−1 ( f (x)+h)− f−1 ( f (x))

Which proves the theorem.This is one of those theorems which is very easy to remember if you neglect the

difficult questions and simply focus on formal manipulations. Consider the following.f−1 ( f (x)) = x. Now use the chain rule to write

(f−1)′( f (x)) f ′ (x) = 1, and then divide

both sides by f ′ (x) to obtain(

f−1)′( f (x)) = 1

f ′(x) . Of course this gives the conclusion ofthe above theorem rather effortlessly and it is formal manipulations like this which aid inremembering formulas such as the one given in the theorem.

5.3. DERIVATIVES OF INVERSE FUNCTIONS 125Proof: This is left to you. Use the chain rule and the product rule. JHigher order derivatives are defined in the obvious way. f” = (f")’ etc. Also the Leibniznotation is defined by 7° dy — f!(x) where y = f (x) and the second derivative is denoted asay with various other higher order derivatives defined similarly. When people write y”)they mean the n‘” derivative. Similarly f) (x) refers to the n' derivative.The chain rule has a particularly attractive form in Leibniz’s notation. Suppose y = g (u)and u = f (x). Thus y = go f (x). Then from the above theorem(sof) @=8 FM) FO) =8' (wf)or in other words, 2 = Be du Notice how the du cancels. This particular form is a veryuseful crutch and is used extensively in applications.5.3. Derivatives of Inverse FunctionsIt happens that if f is a differentiable one to one function defined on an interval, [a,b],and f” (x) exists and is non zero then the inverse function f—! has a derivative or one sidedderivative at the point f (x).Theorem 5.3.1 Let f : [a,b] > R be continuous and one to one. Suppose f" (x)exists for some x € [a,b] and f' (x) #0, a one sided derivative at the end points. Then(f-')'(£@) exists and is given by the formula, (f-')' (£0) = FO:Proof: By Lemma 4.4.3, and Corollary 4.5.1 on Page 105 f is either strictly increasingor strictly decreasing and f—! is continuous on an interval f ({a,b]). Constrain h to have theappropriate sign if at an endpoint of f ({a,b]) , and letting |A| be sufficiently small otherwise,let x be a point where f’ (x) 4 0 and f (x) =h=f(f-'+h))-f(F'0)) =f’ (x) fee a nie (*)By continuity of f~!,|o(f~! (y+h) — (v))| <a)0 fo! +h) — 1 (y)| if hissmall enough and so, from the triangle Ae in *,?nl > SIP LO + A) —F"0)lo(f +h -f oN . 2\o(f-' (y+A)—f- 0)|h| If’ (x) F- @+h)—f-1)|showing that o (f~!(y+h) — f7! (y)) =0(h). From «,Fgh toll) =F FMF") =F (FO) +8) =F)Which proves the theorem. jfThis is one of those theorems which is very easy to remember if you neglect thedifficult questions and simply focus on formal manipulations. Consider the following.f7! (f(x) =x. Now use the chain rule to write (f~')' (f(x)) f” (x) = 1, and then divideboth sides by f’ (x) to obtain (f “1! (f (x)) = Foy: Of course this gives the conclusion ofthe above theorem rather effortlessly and it is formal manipulations like this which aid inremembering formulas such as the one given in the theorem.