Differentiation of Log x (Logarithmic Function)

Here you will learn differentiation of log x i.e logarithmic function by using first principle and its examples.

Let’s begin –

Differentiation of log x (Logarithmic Function) with base e and a

(1) Differentiation of log x or \(log_e x\):

The differentiation of \(log_e x\), x > 0 with respect to x is \(1\over x\).

i.e. \(d\over dx\) \(log_e x\) = \(1\over x\)

Proof Using first Principle :

Let f(x) = \(log_e x\). Then, f(x + h) = \(log_e (x + h)\)

\(\therefore\) \(d\over dx\)(f(x)) = \(lim_{h\to 0}\) \(f(x + h) – f(x)\over h\)

\(d\over dx\)(f(x)) = \(lim_{h\to 0}\) \(log_e (x + h) – log_e x\over h\)

\(d\over dx\)(f(x)) = \(lim_{h\to 0}\) \(log_e (1 + h/x)\over h\)

Divide and multiply by x in both numerator and denominator,

\(d\over dx\)(f(x)) = \(lim_{h\to 0}\) \(log_e (1 + h/x)\over h/x\) \(1\over x\)

\(\implies\) \(d\over dx\)(f(x)) = \(1\over x\)

because, [\(lim_{h\to 0}\) \(log_e (1 + x)\over x\) = 1]

Hence, \(d\over dx\) (\(log_e x\)) = \(1\over x\)

Example : What is the differentiation of log 5x ?

Solution : Let y = log 5x

\(d\over dx\) (y) = \(d\over dx\) log 5x

By using chain rule,

\(d\over dx\) (y) = \(1\over 5x\) .5 = \(1\over x\)

Hence, \(d\over dx\) (log 5x) = \(1\over x\)

(2) Differentiation of \(log_a x\) :

The differentiation of \(log_a x\), a > 0 , a \(\ne\) 1 with respect to x is \(1\over x log_e a\).

i.e. \(d\over dx\) \(log_a x\) = \(1\over x log_e a\)

Proof Using first Principle :

Let f(x) = \(log_a x\). Then, f(x + h) = \(log_a (x + h)\)

\(\therefore\) \(d\over dx\)(f(x)) = \(lim_{h\to 0}\) \(f(x + h) – f(x)\over h\)

\(d\over dx\)(f(x)) = \(lim_{h\to 0}\) \(log_a (x + h) – log_a x\over h\)

\(d\over dx\)(f(x)) = \(lim_{h\to 0}\) \(log_a (1 + h/x)\over h\)

By base changing theorem,

\(d\over dx\)(f(x)) = \(lim_{h\to 0}\) \(log_e (1 + h/x)\over (log_e a)h\)

\(\implies\) \(d\over dx\)(f(x)) = \(1\over log_e a\) \(lim_{h\to 0}\) \(log_e (1 + h/x)\over x(h/x)\)

\(\implies\) \(d\over dx\)(f(x)) = \(1\over xlog_e a\)

because, [\(lim_{h\to 0}\) \(log_e (1 + h/x)\over h/x\) = 1]

Hence, \(d\over dx\) (\(log_a x\)) = \(1\over xlog_e a\)

Example : What is the differentiation of \(log_3 x\) ?

Solution : Let y = \(log_3 x\)

\(d\over dx\) (y) = \(d\over dx\) \(log_3 x\)

By using above formula,

\(d\over dx\) (y) = \(1\over xlog_e 3\)

Hence, \(d\over dx\) (\(log_3 x\)) = \(1\over xlog_e 3\)