Derivative of the Logarithm
The derivative of the logarithm with base a>0 at a point x>0 is given by $$ D[ \log_a x ] = \frac{1}{x} \log_a e $$ whereas the derivative of the natural logarithm is simply $$ D[ \log x ] = D[ \log_e x ] = D[ \ln x ] = \frac{1}{x} $$
A Practical Example
Consider the function f(x):
$$ f(x) = \log(2x+3) $$
This is a composite function, consisting of the natural logarithm h(x) and the inner function g(x)=2x+3:
$$ h(g(x)) = \log(g(x)) $$
$$ g(x) = 2x+3 $$
Therefore, we’ll apply the chain rule for differentiating composite functions:
$$ f'(x) = D[h(g(x))] \cdot D[g(x)] $$
$$ f'(x) = D[\log(2x+3)] \cdot D[2x+3] $$
Recall that the derivative of the natural logarithm log(x) is 1/x.
Therefore, the derivative of log(2x+3) becomes:
$$ f'(x) = \frac{1}{2x+3} \cdot D[g(x)] $$
Since the derivative of g(x)=2x+3 is 2, we have:
$$ f'(x) = \frac{1}{2x+3} \cdot 2 $$
Thus, the derivative of f(x) is:
$$ f'(x) = \frac{2}{2x+3} $$
Proof
To derive the derivative of the logarithmic function, start from the definition of the derivative:
\[ f'(x)=\lim_{h\to 0}\frac{f(x+h)-f(x)}{h} \]
Let \( f(x)=\log_a x \). Replacing \( f(x) \) and \( f(x+h) \) in the definition gives
\[ f'(x) = \lim_{h\to 0}\frac{\log_a(x+h)-\log_a x}{h} \]
Using the logarithmic identity
\[ \log_a x-\log_a y=\log_a\frac{x}{y} \]
the expression becomes
\[ f'(x) = \lim_{h\to 0}\frac{\log_a\frac{x+h}{x}}{h} \]
and, after simplifying the fraction inside the logarithm,
\[ f'(x) = \lim_{h\to 0}\frac{\log_a\left(1+\frac{h}{x}\right)}{h} \]
To bring the expression into a more useful form, multiply and divide the denominator by \( x \):
\[ f'(x) = \lim_{h\to 0}\frac{\log_a\left(1+\frac{h}{x}\right)}{h\cdot\frac{x}{x}} \]
\[ f'(x) = \lim_{h\to 0}\frac{\log_a\left(1+\frac{h}{x}\right)}{\frac{h}{x}\cdot x} \]
\[ f'(x) = \lim_{h\to 0}\left(\frac{\log_a\left(1+\frac{h}{x}\right)}{\frac{h}{x}}\right)\cdot\frac{1}{x} \]
Since \( \frac{1}{x} \) does not depend on \( h \), it can be taken outside the limit:
\[ f'(x) = \frac{1}{x}\cdot\lim_{h\to 0}\frac{\log_a\left(1+\frac{h}{x}\right)}{\frac{h}{x}} \]
Now introduce the substitution
\[ t=\frac{h}{x} \]
which transforms the limit into
\[ \lim_{t\to 0}\frac{\log_a(1+t)}{t} \]
This is a standard limit whose value is
\[ \lim_{t\to 0}\frac{\log_a(1+t)}{t}=\log_a e \]
Substituting this result back into the expression for the derivative yields
\[ f'(x)=\frac{1}{x}\cdot\log_a e \]
This proves that the derivative of the logarithm with base \( a \) is
\[ \boxed{f'(x)=\frac{1}{x}\log_a e} \]
Derivative of the Natural Logarithm
The natural logarithm is the special case in which the base is \( e \).
Substituting \( a=e \) into the formula above gives
\[ f'(x)=\frac{1}{x}\cdot\log_e e \]
Since
\[ \log_e e=1 \]
the formula simplifies to
\[ f'(x)=\frac{1}{x} \cdot 1 = \frac{1}{x} \]
Therefore, the derivative of the natural logarithm is
\[ \boxed{D\ln x=\frac{1}{x}} \]
which completes the proof.
Alternative Proof
The derivative of a function can be found using its limit definition:
$$ f'(x) = \lim_{h \rightarrow 0} \frac{f(x+h)-f(x)}{h} $$
In this case, the function is:
$$ f(x) = \log_a x $$
Plugging this into the difference quotient gives:
$$ f'(x) = \lim_{h \rightarrow 0} \frac{\log_a(x+h)-\log_a(x)}{h} $$
We simplify this using properties of logarithms:
$$ f'(x) = \lim_{h \rightarrow 0} \frac{1}{h} \big[ \log_a(x+h)-\log_a x \big] $$
$$ f'(x) = \lim_{h \rightarrow 0} \frac{1}{h} \cdot \log_a \frac{x+h}{x} $$
Explanation. The logarithm of a fraction equals the difference of the logarithms of the numerator and denominator: $$ \log_x \frac{a}{b} = \log_x a - \log_x b $$
$$ f'(x) = \lim_{h \rightarrow 0} \log_a \left( \frac{x+h}{x} \right)^{\frac{1}{h}} $$
Explanation. The logarithm of a power \( b^a \) equals the exponent times the logarithm of the base: $$ \log_x b^a = a \cdot \log_x b $$
$$ f'(x) = \log_a \left[ \lim_{h \rightarrow 0} \left( \frac{x+h}{x} \right)^{\frac{1}{h}} \right] $$
$$ f'(x) = \log_a \left[ \lim_{h \rightarrow 0} \left( 1+\frac{h}{x} \right)^{\frac{1}{h}} \right] $$
We can rewrite the fraction h/x as the ratio (1/x)/(1/h):
$$ f'(x) = \log_a \left[ \lim_{h \rightarrow 0} \left( 1+\frac{\frac{1}{x}}{\frac{1}{h}} \right)^{\frac{1}{h}} \right] $$
As h approaches zero, 1/h tends toward infinity.
This allows us to reduce the limit to the well-known fundamental limit that approaches \( e^{1/x} \):
$$ f'(x) = \log_a \left[ \lim_{h \rightarrow 0} \left( 1+\frac{\frac{1}{x}}{\frac{1}{h}} \right)^{\frac{1}{h}} \right] = \log_a \left( e^{\frac{1}{x}} \right) $$
$$ f'(x) = \log_a e^{\frac{1}{x}} $$
Explanation. The limit $$ \lim_{h \rightarrow 0} \left( 1+\frac{\frac{1}{x}}{\frac{1}{h}} \right)^{\frac{1}{h}} $$ can be rewritten by substituting y = 1/h. As h→0, y→∞. $$ \lim_{y \rightarrow \infty} \left( 1+\frac{\frac{1}{x}}{y} \right)^y $$ This is the classic limit that converges to \( e^{1/x} \): $$ \lim_{y \rightarrow \infty} \left( 1+\frac{\frac{1}{x}}{y} \right)^y = e^{\frac{1}{x}} $$
Applying one last algebraic step gives the differentiation rule:
$$ f'(x) = \frac{1}{x} \cdot \log_a e $$
Explanation. This uses a property of logarithms previously mentioned: the logarithm of a power \( b^a \) equals \( a \) times the logarithm of \( b \): $$ \log_x b^a = a \cdot \log_x b $$ In this case, we’re reversing that rule to extract the exponent from the logarithm.
We’ve thus derived the formula for the derivative of the logarithm with base a:
$$ f'(x) = \frac{1}{x} \cdot \log_a e $$
Case of the Natural Logarithm
For the natural logarithm, the base a is \( e \), Euler’s number:
$$ \ln e = \log_e e $$
Since the argument and the base are the same, the value of the logarithm is 1:
$$ \ln e = \log_e e = \log e = 1 $$
Explanation. When the argument and base of a logarithm are identical, the exponent \( x \) needed to satisfy \( b^x = k \) must be one: $$ x = \log_b k \Rightarrow b^x = k $$
Therefore, the derivative of the natural logarithm simplifies to:
$$ f'(x) = \frac{1}{x} \cdot \log_a e = \frac{1}{x} \cdot 1 = \frac{1}{x} $$
And so on.
