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, \(d\over dx\) (\(log_3 x\)) = \(1\over xlog_e 3\)
Question for Practice
What is the Differentiation of log log x ?
What is the Differentiation of x log x ?
What is the differentiation of log sin x ?
