# Find the derivative of the function logarithmic function (log x) using the first principle of derivative.

The symbol dy/dx is called the derivative of the dependent variable y with respect to the independent variable x. The process of finding the derivatives is called differentiation.

## Answer: The derivative of the logarithmic function (log x) by using the first principle of the derivative is 1/(x ln a).

Let us proceed step by step.

**Explanation:**

We can find the derivative of the function using the first principle of derivative by taking a suitable example.

The first principle of derivatives is \(f'(x) =\lim_{h\rightarrow 0}\dfrac{f(x+h)-f(x)}{h}\)

Let us consider a logarithmic function log \(_a\) x.

Now we will find out the derivative of log x by using the first principle of derivative.

y = f(x) = log\(_a\) x

We are proceeding with the given function by the rule of the first principle of derivatives

y + Δy = log\(_a\)_{ }(x + Δx) [ Δy represents small change in y ]

Δy = log\(_a\)_{ }(x + Δx) – y [ on transposing y]

On substituting the value of function y = log\(_a\) x, in the above equation, we get

Δy = log\(_a\)_{ }(x + Δx) – log\(_a\) x

Δy = log\(_a\)_{ }[(x + Δx) / x] [Using property of logarithm log a - log b = log (a/b)]

Δy = log\(_a\)_{ }[1 + (Δx / x)]

On dividing both sides of the equation by Δx we get,

Δy / Δx = 1 / Δx [ log\(_a\)_{ }{1 + (Δx / x) } ]

Multiplying numerator and denominator of RHS by x, we get

Δy / Δx = x / x Δx [ log\(_a\)_{ }{1 + (Δx / x) } ]

Δy / Δx = 1 / x [ log\(_a\)_{ }{1 + (Δx / x) } ^{x / Δx} ] [Using property of logarithm: a log b = log b^{a}]

Taking limit on both sides of the equation, we get

limΔ𝑥→0 [ Δy / Δx ] = limΔ𝑥→0 (1 / x [ log\(_a\)_{ }{1 + (Δx / x) } ^{x / Δx} ])

limΔ𝑥→0 [ Δy / Δx ] = 1 / x limΔ𝑥→0 [ log\(_a\)_{ }{1 + (Δx / x) } ^{x / Δx} ]

Let us assume Δx / x = u, therefore, x / Δx will become 1 / u

If Δx → 0 then u → 0, we get

dy / dx = 1 / x lim u→0 [ log\(_a\)_{ }{1 + (u) } ^{1 }^{/ u} ] -------(1)

As we know that, lim x→0 (1+x)^{1 / x} = e

dy / dx = 1 / x log\(_a\) e

dy / dx = 1 / (x ln a) [ log\(_a\) e = ln a ]