Question

derivation of logx by abinito method

Answer 1

Let us suppose that the function is of the form

y=f(x)=logaxy=f(x)=logax

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

y+Δy=loga(x+Δx)Δy=loga(x+Δx)−yy+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⇒ΔyΔx=1xxΔxloga(1+Δxx)⇒ΔyΔx=1xloga(1+Δxx)xΔx

Taking the limit of both sides as Δx→0Δx→0, we have

⇒limΔx→0ΔyΔx=limΔx→01xloga(1+Δxx)xΔx⇒limΔx→0ΔyΔx=1xlimΔx→0loga(1+Δxx)xΔx⇒limΔx→0⁡ΔyΔx=limΔx→0⁡1xloga(1+Δxx)xΔx⇒limΔx→0⁡ΔyΔx=1xlimΔx→0⁡loga(1+Δxx)xΔx

Consider Δxx=u⇒xΔx=1uΔxx=u⇒xΔx=1u, as Δx→0Δx→0 then u→0u→0, we get

⇒dydx=1xlimu→0loga(1+u)1u⇒dydx=1xlimu→0⁡loga(1+u)1u

Using the relation from the limit limx→0(1+x)1x=elimx→0⁡(1+x)1x=e, we have

dydx=1xloga(e)⇒dydx=1xlna⇒ddx(logax)=1xlnadydx=1xloga(e)⇒dydx=1xln⁡a⇒ddx(logax)=1xln⁡a

Example: Find the derivative of

y=f(x)=log10x2y=f(x)=log10x2

We have the given function as

y=log10x2y=log10x2

Differentiating with respect to variable xx, we get

dydx=ddxlog10x2dydx=ddxlog10x2

Using the rule, ddx(logax)=1xlnaddx(logax)=1xln⁡a, we get

dydx=1x2ln10ddx(x2)⇒dydx=2xx2ln10⇒dydx