Chapter 3 | Derivatives
325
h ( x ) = ln ⎛
⎞ ⎠ is given by
⎝ g ( x )
(3.31)
h ′( x ) = 1
g ′( x ).
g ( x )
Proof If x >0 and y = ln x , then e y = x . Differentiating both sides of this equation results in the equation e y dy dx =1.
dy dx
Solving for
yields
dy dx = 1 e y .
Finally, we substitute x = e y to obtain
dy dx = 1 x . We may also derive this result by applying the inverse function theorem, as follows. Since y = g ( x ) = ln x is the inverse of f ( x ) = e x , by applying the inverse function theorem we have dy dx = 1 f ′ ⎛ ⎝ g ( x ) ⎞ ⎠ = 1 = 1 x .
e ln x
Using this result and applying the chain rule to h ( x ) = ln ⎛ ⎝ g ( x ) ⎞
⎠ yields
h ′( x ) = 1
g ′( x ).
g ( x )
□ The graph of y = ln x and its derivative dy
dx = 1 x
are shown in Figure 3.35 .
Figure 3.35 The function y = ln x is increasing on (0, +∞). Its derivative y ′ = 1 x is greater than zero on (0, +∞).
Example 3.77 Taking a Derivative of a Natural Logarithm
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