from first principle y=xInx find (dy/dx)


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Question Number 188033 by Michaelfaraday last updated on 25/Feb/23

$f r o m f i r s t p r i n c i p l e$ $y = x I n x f i n d \frac{d y}{d x}$
	Answered by a.lgnaoui last updated on 25/Feb/23

	$\frac{d y}{d x} = l n x + 1 = \frac{y}{x} + 1$ $\frac{y}{x} = \frac{d y}{d x} - 1 \Rightarrow$ $\frac{d y}{d x} - \frac{y}{x} - 1 = 0 (e q u a t i o n d i f)$
	Commented by Spillover last updated on 25/Feb/23

	$f r o m t h e f i r s t p r i c i p l e$ $f^{^{'}} (x) = \lim_{h \to 0} \frac{f (x + h) - f (x)}{h}$
	Answered by Mr_X last updated on 25/Feb/23
	$Solution ⇒ y=xln(x) ⇒ y+Δy=(x+Δx)[ln(x+Δx)] ⇒ Δy=(x+Δx)[ln(x+Δx)]−xln(x) ⇒ Δy=(x+Δx){ln[(x)(1+((Δx)/x))]}−xln(x) ⇒ Δy=(x+Δx)[ln(x)+ln(1+((Δx)/x))]−xln(x) ⇒ Δy=xln(x)+Δxln(x)+xln(1+((Δx)/x))+Δxln(1+((Δx)/x))−xln(x) ⇒ ((Δy)/(Δx))=ln(x)+(x/(Δx))ln(1+((Δx)/x))+ln(1+((Δx)/x)) ⇒ ((Δy)/(Δx))=ln(x)+ln[(1+((Δx)/x))]^(x/(Δx)) +ln(1+((Δx)/x)) ⇒ lim_(Δx→0) ((Δy)/(Δx))=ln(x)+lim_(Δx→0) ln[(1+((Δx)/x))]^(x/(Δx)) +lim_(Δx→0) ln(1+((Δx)/x)) ⇒ (dy/dx)=ln(x)+ln[lim_(Δx→0) (1+((Δx)/x))]^(x/(Δx)) +lim_(Δx→0) ln(1+((Δx)/x)) ⇒ (dy/dx)=ln(x)+ln(e)+ln(1) ⇒ (dy/dx)=ln(x)+1+0 or ⇒ (dy/dx)=ln(x)+1$
	$S o l u t i o n \Rightarrow$ $y = x l n (x)$ $\Rightarrow y + Δ y = (x + Δ x) [l n (x + Δ x)]$ $\Rightarrow Δ y = (x + Δ x) [l n (x + Δ x)] - x l n (x)$ $\Rightarrow Δ y = (x + Δ x) {l n [(x) (1 + \frac{Δ x}{x})]} - x l n (x)$ $\Rightarrow Δ y = (x + Δ x) [l n (x) + l n (1 + \frac{Δ x}{x})] - x l n (x)$ $\Rightarrow Δ y = x l n (x) + Δ x l n (x) + x l n (1 + \frac{Δ x}{x}) + Δ x l n (1 + \frac{Δ x}{x}) - x l n (x)$ $\Rightarrow \frac{Δ y}{Δ x} = l n (x) + \frac{x}{Δ x} l n (1 + \frac{Δ x}{x}) + l n (1 + \frac{Δ x}{x})$ $\Rightarrow \frac{Δ y}{Δ x} = l n (x) + l n {[(1 + \frac{Δ x}{x})]}^{\frac{x}{Δ x}} + l n (1 + \frac{Δ x}{x})$ $\Rightarrow \underset{Δ x \to 0}{l i m} \frac{Δ y}{Δ x} = l n (x) + \underset{Δ x \to 0}{l i m l n} {[(1 + \frac{Δ x}{x})]}^{\frac{x}{Δ x}} + \underset{Δ x \to 0}{l i m l n} (1 + \frac{Δ x}{x})$ $\Rightarrow \frac{d y}{d x} = l n (x) + l n {[\underset{Δ x \to 0}{l i m} (1 + \frac{Δ x}{x})]}^{\frac{x}{Δ x}} + \underset{Δ x \to 0}{l i m l n} (1 + \frac{Δ x}{x})$ $\Rightarrow \frac{d y}{d x} = l n (x) + l n (e) + l n (1)$ $\Rightarrow \frac{d y}{d x} = l n (x) + 1 + 0$ $o r$ $\Rightarrow \frac{d y}{d x} = l n (x) + 1$
	Commented by Michaelfaraday last updated on 01/Mar/23

	$t h a n k s s i r$
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