Calculus Volume 1

Chapter 3 | Derivatives

301

Solution The function g ( x ) = x 3 is the inverse of the function f ( x ) = x 3 . Since g ′( x ) = 1 f ′ ⎛ ⎝ g ( x )

⎞ ⎠ , begin by finding

f ′( x ). Thus,

2

⎛ ⎝ x 3

⎞ ⎠

=3 x 2/3 .

f ′( x ) =3 x 2 and f ′ ⎛

⎞ ⎠ =3

⎝ g ( x )

Finally,

x −2/3 .

g ′( x ) = 1

= 1 3

3 x 2/3

3.43

Find the derivative of g ( x ) = x 5 by applying the inverse function theorem.

From the previous example, we see that we can use the inverse function theorem to extend the power rule to exponents of the form 1 n , where n is a positive integer. This extension will ultimately allow us to differentiate x q , where q is any rational number.

Theorem 3.12: Extending the Power Rule to Rational Exponents The power rule may be extended to rational exponents. That is, if n is a positive integer, then

d dx (1/ n )−1 . Also, if n is a positive integer and m is an arbitrary integer, then ⎞ ⎠ = 1 n x ⎛ ⎝ x 1/ n

(3.20)

⎛ ⎝ x m / n

⎞ ⎠ = m

d dx

(3.21)

( m / n )−1 .

n x

Proof The function g ( x ) = x 1/ n is the inverse of the function f ( x ) = x n . Since g ′( x ) = 1 f ′ ⎛ ⎝ g ( x )

⎞ ⎠ , begin by finding f ′( x ).

Thus,

n ) n −1 = nx ( n −1)/ n .

⎞ ⎠ = n ( x 1/

f ′( x ) = nx n −1 and f ′ ⎛

⎝ g ( x )

Finally,

(1− n )/ n = 1

(1/ n )−1 .

g ′( x ) = 1

= 1 n x

n x

nx ( n −1)/ n

m and apply the chain rule. Thus,

To differentiate x m / n we must rewrite it as ⎛ ⎝ x 1/ n ⎞ ⎠

⎛ ⎝ ⎛

m ⎞

m −1

⎛ ⎝ x m / n

⎞ ⎠ = d

⎞ ⎠

⎛ ⎝ x 1/ n

⎞ ⎠

d dx

(1/ n )−1 = m

( m / n )−1 .

⎝ x 1/ n

⎠ = m

· 1 n x

n x

dx

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