Derivatives of Logarithmic Functions

Derivatives of logarithmic functions are mainly based on the chain rule. However, we can generalize it for any differentiable function with a logarithmic function. The differentiation of log is only under the base $e,$ but we can differentiate under other bases, too.

$\frac{d}{dx} \ln {x} = \frac{1}{x}$

Now we will prove this from first principles:

From first principles, $\frac{d}{dx} f(x) = \displaystyle \lim_{h \rightarrow 0} {\dfrac{f(x+h)-f(x)}{h}}$.

Now let $f(x) = \ln{x},$ then

$\begin{aligned} \dfrac{\text{d}}{\text{d}x}f(x) & = \lim_{h \rightarrow 0} {\dfrac{\ln(x+h) - \ln{x}}{h}} \\ & = \lim_{h \rightarrow 0} {\dfrac{\frac{x}{h}\ln\left(1 + \frac{h}{x}\right)}{x} } \\ & = \lim_{h \rightarrow 0} {\dfrac{\ln{\left(1 + \frac{h}{x}\right)^{\frac{x}{h}}}}{x}} \\ & = \lim_{h \rightarrow 0} {\dfrac{\ln{e}}{x}} \\ & = \dfrac{1}{x}.\ _\square \end{aligned}$

Find the derivative of $\ln {x}$ at $x = 2$.

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We have

$\dfrac{\text{d}}{\text{d}x} \ln {x} = \dfrac{1}{x}.$

Hence

$\dfrac{\text{d}}{\text{d}x} \ln x \Bigg |_{x=2}= \dfrac{1}{2}.\ _\square$

Differentiate $\ln 5x$

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Solution 1: Use the chain rule.

Let $f(x) = \ln x$ and $g(x) = 5x$. Then we are asked to find $( f \circ g ) '$.

Using chain rule, we know that $( f \circ g ) ' = ( f' \circ g) \times g' .$

Since $f' \circ g = \frac{1}{5x}$ and $g'(x) = 5,$ we have $( f \circ g ) ' = \frac{1}{5x} \times 5 = \frac{1}{x}.\ _\square$

Solution 2: Use properties of logarithms.

We know the property of logarithms $\log_a b + \log_a c = \log_a bc$. Using this property,

$\ln 5x = \ln x + \ln 5.$

If we differentiate both sides, we see that

$\dfrac{\text{d}}{\text{d}x} \ln 5x = \dfrac{\text{d}}{\text{d}x} \ln x$

since differentiation of $\ln 5$ which is a constant is $0.$

We have seen that $\frac{\text{d}}{\text{d}x} \ln x = \frac{1}{x}$, and this is the answer to this question. $_\square$

Generalization: For any positive real number $p$, we can conclude $\frac{\text{d}}{\text{d}x} \ln px = \frac{1}{x}$. Note that the derivative is independent of $p$. This can be proven by writing $p$ instead of $5$ in the above solutions.

If $a$ is a positive real number and $a \neq 1$, then

$\dfrac{\text{d}}{\text{d}x}\log_{a} {x} = \dfrac{1}{x \ln {a}}.$

We will use base-changing formula to change the base of the logarithm to $e:$

$\log_{a}{x} = \dfrac{\ln{x}}{\ln{a}} \\ \dfrac{\text{d}}{\text{d}x}\log_{a}x = \dfrac{\text{d}}{\text{d}x} \dfrac{\ln{x}}{\ln{a}}.$

Since $\frac{1}{\ln{a}}$ is a constant,

$\dfrac{\text{d}}{\text{d}x} \dfrac{\ln x}{\ln a} = \dfrac{1}{\ln a} \dfrac{\text{d}}{\text{d}x} \ln x = \dfrac{1}{x \ln{a}}.\ _\square$

For any other type of log derivative, we use the base-changing formula.

Since this is a composite function, we can differentiate it using chain rule.

$\dfrac{\text{d}}{\text{d}x}\ln\big(f(x)\big) = \dfrac{f'(x)}{f(x)}$

Now we will start with $g(x) = \ln \big(f(x)\big).$ To find its derivative, we will substitute $u = f(x).$ Now the derivative changes to $g(x) = \log{u}.$ So,

$\begin{aligned} g'(x) &= \frac{d}{dx}\log{u} \\ &= \frac{du}{dx} \times \frac{d}{du} \ln{u} \\ &= \dfrac{f'(x)}{f(x)}.\ _\square \end{aligned}$

Find the derivative of $\ln(x^2 + 4)$.

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Using the theorem, the derivative of $\ln\big(f(x)\big)$ is $\frac{f'(x)}{f(x)}$. In this problem, $f(x) = x^2 +4,$ so $f'(x) = 2x$.

Hence $\frac{d}{dx}\log\big(x^2 + 4\big) = \frac{2x}{x^2 +4}.\ _\square$