What is the derivative of log (x)?

Let us proceed step by step

Solution:

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

log x is sometimes used for log₁₀ x, log_e x, and log₂ 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 ]

Hence, the derivative of log x is  1 / (x ln a)

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What is the derivative of log (x)?

Summary:

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