Calculus Volume 1

300

Chapter 3 | Derivatives

Theorem 3.11: Inverse Function Theorem Let f ( x ) be a function that is both invertible and differentiable. Let y = f −1 ( x ) be the inverse of f ( x ). For all x satisfying f ′ ⎛ ⎝ f −1 ( x ) ⎞ ⎠ ≠0, dy dx = d dx ⎛ ⎝ f −1 ( x ) ⎞ ⎠ = ⎛ ⎝ f −1 ⎞ ⎠ ′( x ) = 1 f ′ ⎛ ⎝ f −1 ( x ) ⎞ ⎠ .

Alternatively, if y = g ( x ) is the inverse of f ( x ), then

g '( x ) = 1 f ′ ⎛

.

⎞ ⎠

⎝ g ( x )

Example 3.60 Applying the Inverse Function Theorem Use the inverse function theorem to find the derivative of g ( x ) = x +2

x . Compare the resulting derivative to that

obtained by differentiating the function directly.

Solution The inverse of g ( x ) = x +2 x

is f ( x ) = 2

Since g ′( x ) = 1 f ′ ⎛

⎞ ⎠ , begin by finding f ′( x ). Thus,

x −1 .

⎝ g ( x )

x 2

and f ′ ⎛

⎞ ⎠ = −2 ⎛

f ′( x ) = −2

⎝ g ( x )

= −2 ⎛ ⎝ x +2

2 = −

2 .

( x −1) 2

⎠ 2

⎝ g ( x )−1 ⎞

⎞ ⎠

x −1

Finally,

g ′( x ) = 1 f ′ ⎛

= − 2 x 2

.

⎞ ⎠

⎝ g ( x )

We can verify that this is the correct derivative by applying the quotient rule to g ( x ) to obtain g ′( x ) = − 2 x 2 .

3.42

Use the inverse function theorem to find the derivative of g ( x ) = 1 x +2 .

Compare the result obtained

by differentiating g ( x ) directly.

Example 3.61 Applying the Inverse Function Theorem

Use the inverse function theorem to find the derivative of g ( x ) = x 3 .

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