what is derivative of h = (√(ln(x))) by first principle method


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Question Number 82059 by jagoll last updated on 18/Feb/20

$w h a t i s d e r i v a t i v e o f h = \sqrt{l n (x)}$ $b y f i r s t p r i n c i p l e m e t h o d$
	Answered by Henri Boucatchou last updated on 18/Feb/20

	$\frac{dh (x)}{dx} = \frac{d \sqrt{lnx}}{dx}$ $= \frac{\frac{dlnx}{dx}}{2 \sqrt{lnx}}$ $= \frac{1}{2 x \sqrt{lnx}}$
	Answered by mr W last updated on 18/Feb/20
	$h′=lim_(Δx→0) (((√(ln (x+Δx)))−(√(ln x)))/(Δx)) =lim_(Δx→0) (((√(ln x+ln (1+((Δx)/x))))−(√(ln x)))/(Δx)) =(√(ln x))lim_(Δx→0) (((√(1+((ln (1+((Δx)/x)))/(ln x))))−1)/(Δx)) =(√(ln x))lim_(Δx→0) (({1+(1/2)[((ln (1+((Δx)/x)))/(ln x))]+...}−1)/(Δx)) =(√(ln x))lim_(Δx→0) {(1/2)[((ln (1+((Δx)/x)))/(Δx ln x))]+...} =(√(ln x))lim_(Δx→0) {(1/2)[((ln (1+((Δx)/x))^(x/(Δx)) )/(xln x))]+...} =(√(ln x))×(1/2)×((ln e)/(xln x)) =(1/(2x(√(ln x))))$
	$h^{'} = \lim_{Δ x \to 0} \frac{\sqrt{\ln (x + Δ x)} - \sqrt{\ln x}}{Δ x}$ $= \lim_{Δ x \to 0} \frac{\sqrt{\ln x + \ln (1 + \frac{Δ x}{x})} - \sqrt{\ln x}}{Δ x}$ $= \sqrt{\ln x} \lim_{Δ x \to 0} \frac{\sqrt{1 + \frac{\ln (1 + \frac{Δ x}{x})}{\ln x}} - 1}{Δ x}$ $= \sqrt{\ln x} \lim_{Δ x \to 0} \frac{{1 + \frac{1}{2} [\frac{\ln (1 + \frac{Δ x}{x})}{\ln x}] + . . .} - 1}{Δ x}$ $= \sqrt{\ln x} \lim_{Δ x \to 0} {\frac{1}{2} [\frac{\ln (1 + \frac{Δ x}{x})}{Δ x \ln x}] + . . .}$ $= \sqrt{\ln x} \lim_{Δ x \to 0} {\frac{1}{2} [\frac{\ln {(1 + \frac{Δ x}{x})}^{\frac{x}{Δ x}}}{x \ln x}] + . . .}$ $= \sqrt{\ln x} \times \frac{1}{2} \times \frac{\ln e}{x \ln x}$ $= \frac{1}{2 x \sqrt{\ln x}}$
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