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Question Number 192545 by peter frank last updated on 20/May/23

Answered by BaliramKumar last updated on 20/May/23

plnx+qlny= (p+q)ln(x+y)  (p/x)−((p+q)/(x+y))= {((p+q)/(x+y))−(q/y)}(dy/dx)  ((py)/x)((−qx)/)= {((py−)/)((qx)/y)}(dy/dx)  (dy/dx) = (y/x)

$$\mathrm{plnx}+\mathrm{qlny}=\:\left(\mathrm{p}+\mathrm{q}\right)\mathrm{ln}\left(\mathrm{x}+\mathrm{y}\right) \\ $$$$\frac{\mathrm{p}}{\mathrm{x}}−\frac{\mathrm{p}+\mathrm{q}}{\mathrm{x}+\mathrm{y}}=\:\left\{\frac{\mathrm{p}+\mathrm{q}}{\mathrm{x}+\mathrm{y}}−\frac{\mathrm{q}}{\mathrm{y}}\right\}\frac{\mathrm{dy}}{\mathrm{dx}} \\ $$$$\frac{\mathrm{py}}{\mathrm{x}}\frac{−\mathrm{qx}}{}=\:\left\{\frac{\mathrm{py}−}{}\frac{\mathrm{qx}}{\mathrm{y}}\right\}\frac{\mathrm{dy}}{\mathrm{dx}} \\ $$$$\frac{\mathrm{dy}}{\mathrm{dx}}\:=\:\frac{\mathrm{y}}{\mathrm{x}} \\ $$$$ \\ $$

Answered by aleks041103 last updated on 21/May/23

x^p y^q =(x+y)^(p+q) =(1+(y/x))^(p+q) x^(p+q)   ⇒((y/x))^p =(1+(y/x))^(p+q)   ⇒k=(1+k)^(1+(q/p))  ⇒ k=const=(y/x)⇒y=kx  ⇒(dy/dx)=k=(y/x)

$${x}^{{p}} {y}^{{q}} =\left({x}+{y}\right)^{{p}+{q}} =\left(\mathrm{1}+\frac{{y}}{{x}}\right)^{{p}+{q}} {x}^{{p}+{q}} \\ $$$$\Rightarrow\left(\frac{{y}}{{x}}\right)^{{p}} =\left(\mathrm{1}+\frac{{y}}{{x}}\right)^{{p}+{q}} \\ $$$$\Rightarrow{k}=\left(\mathrm{1}+{k}\right)^{\mathrm{1}+\frac{{q}}{{p}}} \:\Rightarrow\:{k}={const}=\frac{{y}}{{x}}\Rightarrow{y}={kx} \\ $$$$\Rightarrow\frac{{dy}}{{dx}}={k}=\frac{{y}}{{x}} \\ $$

Answered by Spillover last updated on 21/May/23

x^p y^q =(x+y)^(p+q)    ln (x^p y^q )=ln (x+y)^(p+q)    pln x+qln y=(p+q)ln (x+y)  (p/x)+(q/y)(dy/dx)=(((p+q)(1+(dy/dx)))/(x+y))  ((py+qx(dy/dx))/(xy))=(((p+q)(1+(dy/dx)))/(x+y))  (dy/dx)=(((p+q)xy−(x+y)py)/((x+y)qx−(p+q)xy))  (dy/dx)=((−py^2 +qxy)/(−pxy+qx^2 ))  (dy/dx)=(y/x)

$${x}^{{p}} {y}^{{q}} =\left({x}+{y}\right)^{{p}+{q}} \: \\ $$$$\mathrm{ln}\:\left({x}^{{p}} {y}^{{q}} \right)=\mathrm{ln}\:\left({x}+{y}\right)^{{p}+{q}} \: \\ $$$${p}\mathrm{ln}\:{x}+{q}\mathrm{ln}\:{y}=\left({p}+{q}\right)\mathrm{ln}\:\left({x}+{y}\right) \\ $$$$\frac{{p}}{{x}}+\frac{{q}}{{y}}\frac{{dy}}{{dx}}=\frac{\left({p}+{q}\right)\left(\mathrm{1}+\frac{{dy}}{{dx}}\right)}{{x}+{y}} \\ $$$$\frac{{py}+{qx}\frac{{dy}}{{dx}}}{{xy}}=\frac{\left({p}+{q}\right)\left(\mathrm{1}+\frac{{dy}}{{dx}}\right)}{{x}+{y}} \\ $$$$\frac{{dy}}{{dx}}=\frac{\left({p}+{q}\right){xy}−\left({x}+{y}\right){py}}{\left({x}+{y}\right){qx}−\left({p}+{q}\right){xy}} \\ $$$$\frac{{dy}}{{dx}}=\frac{−{py}^{\mathrm{2}} +{qxy}}{−{pxy}+{qx}^{\mathrm{2}} } \\ $$$$\frac{{dy}}{{dx}}=\frac{{y}}{{x}} \\ $$

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