# What is the derivative of log (x)?

The symbol dy and dx are called differentials. The process of finding the derivatives is called differentiation.

## Answer: The derivative of log x is 1/ (x ln a)

Let us proceed step by step

**Explanation**:

log x is sometimes used for log_{10} x, log_{e} x, and log_{2} x

Let us consider a logarithmic function defined by y = log_{a} x, x > 0

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 ]

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]

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} ]

Taking limit on both sides of the equation, we get

lim_{ }Δ𝑥→0 [ Δy / Δx ] = lim_{ }Δ𝑥→0 1 / x [ log_{a }{1 + (Δx / x) } ^{x / Δx} ]

lim_{ }Δ𝑥→0 [ Δy / Δx ] = 1 / x lim_{ }Δ𝑥→0 [ log_{a }{1 + (Δx / x) } ^{x / Δx} ]

Let us assume Δx / x as 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 ]