126 CHAPTER 5. THE DERIVATIVE

Example 5.3.2 Let f (x)= 8+x2+x3+7x. Show that f has an inverse and find(

f−1)′(8) .

I am not able to find a formula for the inverse function. This is typical in useful ap-plications so you need to get used to this idea. The methods of algebra are insufficient tosolve hard problems in analysis. You need something more. The question is to determinewhether f has an inverse. To do this, f ′ (x) = 2x+3x2+7 > 0 for all x. By Corollary 5.11.5on Page 139, this function is strictly increasing on R and so it has an inverse function al-though I have no idea how to find an explicit formula for this inverse function. However, Ican see that f (0) = 8 and so by the formula for the derivative of an inverse function,

(f−1)′ (8) = ( f−1)′ ( f (0)) =

1f ′ (0)

=17.

In practice, we typically don’t bother with the mathematical details. We have f (x) = yand the inverse function is of the form x = f−1 (y) . Thus it involves finding dx

dy (y) ≡(f−1)′(y) . The existence of the derivative of the inverse function exists by the above argu-

ment. Therefore, all that remains is to use the chain rule. Take ddy of both sides of f (x) = y,

f ′ (x) dxdy = 1. Thus

dxdy

(y)≡(

f−1)′ (y) = 1f ′ (x)

=1

f ′ ( f−1 (y))=

1f ′ (x)

which is the same as obtained earlier. You know the inverse has a derivative and so itsuffices to use the chain rule.

5.4 Circular Functions and InversesHere in this section, the derivatives of the circular functions are derived and then the deriva-tives of their inverse functions are considered.

Theorem 5.4.1 sin′ (x) = cos(x) ,cos′ (x) =−sin(x).

Proof: Consider the picture where here h is small

h

(cos(h),sin(h))

(1− cos(h))

From Corollary 2.3.9

(1− cos(h))+ sin(h)≥ h ≥ sin(h)

It follows that

sin(h)1+ cos(h)

=1− cos2 (h)

sin(h)(1+ cos(h))=

1− cos(h)sin(h)

+1 ≥ hsin(h)

≥ 1 (5.5)