Question and Answers Forum

All Questions      Topic List

Differentiation Questions

Previous in All Question      Next in All Question      

Previous in Differentiation      Next in Differentiation      

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

Answered by BaliramKumar last updated on 20/May/23

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

Answered by aleks041103 last updated on 21/May/23

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

Answered by Spillover last updated on 21/May/23

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

Terms of Service

Privacy Policy

Contact: info@tinkutara.com