# Find the derivative of the function using the first principle of derivative.

The symbols dy and dx are called differentials. 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.

Let us consider a logarithmic function log 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 ]

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

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 as u therefore, x / Δx will become 1 / u

If Δx → 0 then u → 0, we get

dy / dx = 1 / x limu→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 ]