09 1derivln

# Derivative of loge(x)

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Recall that $\log_e(x)$ is only defined for $x > 0$

… For $f(x)=log_e(x) \qquad \longrightarrow \qquad f'(x)=\dfrac{1}{x} \qquad x > 0$

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… For $f(x)=log_e(kx) \qquad \longrightarrow \qquad f'(x)=\dfrac {k}{kx} = \dfrac{1}{x} \qquad x > 0$

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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).

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This graph shows $y = ln(x)$ and its derivative $\dfrac{dy}{dx} = \dfrac{1}{x} \text{ for } x > 0$.

Notice

• Gradient (derivative) is always positive
• Gradient (derivative) starts at positive infinity near $x = 0$ and then reduces as x gets larger.
• Gradient (derivative) approaches zero as x approaches positive infinity.

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CAUTION: The brackets inside a log are important.

• $\log_e(x + 3)$ is NOT the same as $\log_ex + 3$
• ALWAYS use brackets inside a log function to avoid any possible confusion.

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### Example 1

Find the derivative of each of the following with respect to x:

a) .. $y = \log_e \big( 5x \big) \quad x > 0$

b) .. $y = \log_e \big( 4x+3 \big) \quad x > -\dfrac{3}{4}$

Solution:

a) .. $y = \log_e \big( 5x \big) \quad x > 0$

… … … $\dfrac{dy}{dx}=\dfrac{1}{x} \quad x > 0$
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b) .. $y = \log_e \big( 4x+3 \big) \quad x > -\dfrac{3}{4}$

… … … $y = \log_e \big( 4x+3 \big)$

… … … Using the chain rule:

… … … Let $u = 4x+3 \qquad \text{ then } \qquad y = \log_e \big( u \big)$

… … … $\dfrac{du}{dx}=4 \qquad \text{ and } \quad \dfrac{dy}{du}=\dfrac{1}{u} \quad u > 0$

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… … … $\dfrac{dy}{dx}=\dfrac{dy}{du}.\dfrac{du}{dx}$

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… … … $\dfrac{dy}{dx}=\dfrac{4}{u} \qquad u > 0$

… … … $\dfrac{dy}{dx}=\dfrac{4}{4x+3} \qquad x > -\dfrac{3}{4}$

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## Chain Rule Version

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The above example shows that if:

… … $y=log_e \big( g(x) \big)$

then

… … $\dfrac{dy}{dx}=\dfrac{g'(x)}{g(x)} \qquad g(x) > 0$

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### Example 2

Find $\dfrac{dy}{dx}$ given that $y=log_e\big(x^2+2\big)$

Solution:

… … $y=log_e\big(x^2+2\big)$

… … Let $g(x)=x^2+2, \text{ then } y=log_e \big( g(x) \big)$

… … $g'(x)=2x$

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… … $\dfrac{dy}{dx}=\dfrac{g'(x)}{g(x)} \quad \text{ so } \quad \dfrac{dy}{dx}=\dfrac{2x}{x^2+2} \quad x \in R$
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… …Note: Domain: $x^2 + 2 > 0$ is true for all $x \in R$

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Note: The basic rule for the derivative of $\log_e(x)$ is simply a special case of this Chain Rule version.

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… … $y = log_e (x)$

… … $g(x) = x \; \text{ so } \; g'(x) = 1$

… … $\dfrac{dy}{dx}=\dfrac{g'(x)}{g(x)} \quad \text{ so } \quad \dfrac{dy}{dx}=\dfrac{1}{x} \quad x > 0$

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## Derivative of loge|x|

(In Specialist Maths Course, not in Maths Methods Course)

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If $x < 0$ then $\log_e(x)$ is undefined.

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

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So $\log_e|x|$ is defined for all $x \in R\backslash\{0\}$

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So we get:

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

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OR

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

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