Y=logx? Give it's derivation.

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Answer 1 · 2018-03-13T11:16:41

$\frac{d}{d x} [ln x] = \frac{1}{x}$ $d/dx[lnx]=1/x$

Let $y = e^{x}$ $y=e^x$ , then lny= $x$ $x$ , by definition.

Let $y = ln x$ $y=lnx$ ............. $[1]$ $[1]$ , therefore $x = e^{y}$ $x=e^y$ ........... $[2]$ $[2]$

Differentiate............ $[2]$ $[2]$ with respect to $y$ $y$ , $\frac{d x}{d y} = e^{y}$ $dx/dy=e^y$ ,

inverting both sides, $\frac{d y}{d x} = \frac{1}{e^{y}} = \frac{1}{x}$ $dy/dx=1/e^y=1/x$ [Since $x = e^{y}$ $x=e^y$ from $......[2]]$

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

Answer 2 · 2018-03-13T11:50:33

Another derivation of $d/dxlogx$

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 $d/dxlnx=1/t=1/x=dy/dx$