if x^p y^p =(x + y)^(p +q) prove that (dy/dx)=(y/x)


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Question Number 35481 by JOHNMASANJA last updated on 19/May/18

$i f x^{p} y^{p} = (x + y)^{p + q}$ $p r o v e t h a t \frac{d y}{d x} = \frac{y}{x}$
Commented by math1967 last updated on 19/May/18

$p l z c h e c k x^{p} y^{p} {O R x}^{p} y^{q}$
	Answered by math1967 last updated on 19/May/18
	$lnx^p y^q =ln(x+y)^(p+q) plnx+qlny=(p+q)ln(x+y) (p/x)+(q/y)(dy/dx)=((p+q)/(x+y))(1+(dy/dx)) ((q/y)−((p+q)/(x+y)))(dy/dx)=((p+q)/(x+y))−(p/x) {((qx−py)/(y(x+y)))}(dy/dx)=((qx−py)/(x(x+y))) (dy/dx)=(y/x)$
	${l n x}^{p} y^{q} = l n {(x + y)}^{p + q}$ $p l n x + q l n y = (p + q) l n (x + y)$ $\frac{p}{x} + \frac{q}{y} \frac{d y}{d x} = \frac{p + q}{x + y} (1 + \frac{d y}{d x})$ $(\frac{q}{y} - \frac{p + q}{x + y}) \frac{d y}{d x} = \frac{p + q}{x + y} - \frac{p}{x}$ ${\frac{q x - p y}{y (x + y)}} \frac{d y}{d x} = \frac{q x - p y}{x (x + y)}$ $\frac{d y}{d x} = \frac{y}{x}$
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