Question

Y=logx? Give it's derivation.

Answer 1

`d/dx[lnx]=1/x`

Explanation:

Let `y=e^x`, then lny=`x`, by definition.

Let `y=lnx`.............`[1]`, therefore `x=e^y`...........`[2]`

Differentiate............`[2]` with respect to `y`, `dx/dy=e^y`,

inverting both sides, `dy/dx=1/e^y=1/x` [Since `x=e^y` from `......[2]]`

Therefore`d/dx[lnx]=1/x`, and so `int1/xdx=lnx+`constant.

Answer 2

Another derivation of `d/dxlogx`

Explanation:

Firstly, let `log-=ln`

So `y=lnx`.

But, by definition, `lnx=int_0^x1/tdt`.

Hence, by the first Fundamental Theorem of Calculus, it follows that `d/dxlnx=1/t=1/x=dy/dx`