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

### 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 ?

What is the differentiation of 1/log x ?

What is the differentiation of $$log x^2$$ ?