07Flogderivative

toc = Derivative of log e (x) =

Recall that log e (x) is only defined for x > 0

math \\ \text{For } \; f:R^+\rightarrow{R},\;f(x)=log_e(x) \qquad .\\. \\ . \qquad f':R^+\rightarrow{R},\;f'(x)=\dfrac{1}{x} math

math \\ \text{For }\;f:R^+\rightarrow{R},\;f(x)=log_e(kx) \qquad. \\ . \\ . \qquad f':R^+\rightarrow{R},\;f'(x)=\dfrac {k}{kx} = \dfrac{1}{x} \qquad. math

Note: This rule only applies to log(base e).

We will not study the derivative of logs with other bases. (But they can be found by using the change of base rule to convert to base e).



The graph show y = ln(x) and its derivative. Notice
 * Gradient (derivative) is always positive
 * Gradient (derivative) starts at positive infinity near x = 0 and then reduces
 * Gradient (derivative) approaches zero as x approaches positive infinity.


 * CAUTION: ** The brackets inside a log are important.
 * log e (x + 3) is NOT the same as log e x + 3
 * ALWAYS use brackets inside a log function to avoid any possible confusion.

** Example 1 **
Find the derivative of each of the following with respect to x: math \\ . \qquad (a) \quad y = \log_e \big( 5x \big) \quad x > 0 \qquad. \\ . \\ . \qquad (b) \quad y = \log_e \big( 4x+3 \big) \quad x > -\frac{3}{4} \qquad. math


 * __Solution:__**

math \\ . \qquad y = \log_e \big( 5x \big) \qquad. \\ .\\ . \qquad \dfrac{dy}{dx}=\dfrac{1}{x} \quad x > 0 \qquad. math
 * (a) **

math . \qquad y = \log_e \big( 4x+3 \big) \qquad. math
 * (b) **

... ... ... Using the chain rule: math \\ . \qquad \text{Let } u = 4x+3 \qquad \text{ then } \qquad y = \log_e \big( u \big) \qquad. \\ . \\ . \qquad \dfrac{du}{dx}=4 \qquad \text{ and } \quad \dfrac{dy}{du}=\dfrac{1}{u} \quad u > 0 \qquad. math

math . \qquad \dfrac{dy}{dx}=\dfrac{dy}{du}.\dfrac{du}{dx} \qquad. math

math \\ . \qquad \dfrac{dy}{dx}=\dfrac{4}{u} \qquad u > 0 \qquad. \\ . \\ . \qquad \dfrac{dy}{dx}=\dfrac{4}{4x+3} \qquad x > -\frac{3}{4} \qquad. math

Chain Rule Version
The above example shows that if: math \\ . \qquad y=log_e \big[ g(x) \big] \quad \text{ then } \qquad. \\ . \\ . \qquad \dfrac{dy}{dx}=\dfrac{g'(x)}{g(x)} \qquad g(x) > 0 \qquad. math

** Example 2 **
math . \qquad \text{Find } \dfrac{dy}{dx} \text{ given that } y=log_e(x^2+2) \qquad. math


 * __Solution:[[image:07Feg2.gif width="292" height="175" align="right"]]__**

math . \qquad y=log_e(x^2+2) \qquad. math

math \\ . \qquad \text{Let } g(x)=x^2+2, \text{ then } y=log_e \big[ g(x) \big] \qquad. \\ . \\ . \qquad \quad \quad g'(x)=2x \qquad. math

math . \qquad \dfrac{dy}{dx}=\dfrac{g'(x)}{g(x)} \quad \text{ so } \quad \dfrac{dy}{dx}=\dfrac{2x}{x^2+2} \quad x \in R \qquad. math

math . \qquad \text{Note: Domain: } \; x^2 + 2 > 0 \; \text{ is true for all } \; x \in R \qquad. math


 * Note: ** The basic rule for the derivative of log e (x) is simply a special case of this Chain Rule version.

math . \qquad y = log_e (x) \\. \\ . \qquad g(x) = x \; \text{ so } \; g'(x) = 1 \qquad. \\ . \\ . \qquad \dfrac{dy}{dx}=\dfrac{g'(x)}{g(x)} \quad \text{ so } \quad \dfrac{dy}{dx}=\dfrac{1}{x} \quad x > 0 \qquad. math


 * Derivative of log e |x| **

(not in course)

If x __<__ 0 then log e (x) is undefined.

But |x| > 0 for all x in R\{0}.

So log e |x| is defined for all x in R\{0}

So we get:

math . \qquad \dfrac{d}{dx} \Big( log_e \; |x| \; \Big) = \dfrac{1}{x} \quad x \in R \backslash \{ 0 \} \qquad. math

OR

math . \qquad \dfrac{d}{dx} \Big( log_e \; \big| g(x) \big| \; \Big) = \dfrac{g'(x)}{g(x)} \quad g(x) \neq 0 \qquad. math



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