Question

find the derivative of logx by abinito method​

Answer 1

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Answer 2

Answer:

Let us suppose that the function is of the form

y=f(x)=logax

y=f(x)=logax

First we take the increment or small change in the function:

y+Δy=loga(x+Δx)Δy=loga(x+Δx)−y

y+Δy=loga(x+Δx)Δy=loga(x+Δx)−y

Putting the value of function y=logaxy=logax in the above equation, we get

Δy=loga(x+Δx)−logaxΔy=loga(x+Δxx)Δy=loga(1+Δxx)

Δy=loga(x+Δx)−logaxΔy=loga(x+Δxx)Δy=loga(1+Δxx)

Dividing both sides by ΔxΔx, we get

ΔyΔx=1Δxloga(1+Δxx)

ΔyΔx=1Δxloga(1+Δxx)

Multiplying and dividing the right hand side by xx, we have

⇒ΔyΔx=1xxΔxloga(1+Δxx)⇒ΔyΔx=1xloga(1+Δxx)xΔx

Explanation: