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 84059 by niroj last updated on 09/Mar/20

$$$$$$\mathbf{Find}\frac{\mathbf{dy}}{\mathbf{dx}}\mathbf{of}{\mathbf{x}}^{\mathbf{m}}{\mathbf{x}}^{\mathbf{n}}={(\mathbf{x}+\mathbf{y})}^{\mathbf{m}+\mathbf{n}}.$$

Commented by jagoll last updated on 09/Mar/20

$$x^{m+n}={(x+y)}^{m+n}$$$$(m+n)x^{m+n-1}=(m+n){(x+y)}^{m+n-1}+(m+n){(x+y)}^{m+n-1}y{}^{\prime }$$$$x^{m+n-1}={(x+y)}^{m+n-1}(1+y{}^{\prime })$$$$1+{\mathrm{y}}^{\prime }=\frac{{\mathrm{x}}^{\mathrm{m}+\mathrm{n}-1}}{{(\mathrm{x}+\mathrm{y})}^{\mathrm{m}+\mathrm{n}-1}}$$$${\mathrm{y}}^{\prime }=\frac{{\mathrm{x}}^{\mathrm{m}+\mathrm{n}-1}-{(\mathrm{x}+\mathrm{y})}^{\mathrm{m}+\mathrm{n}-1}}{{(\mathrm{x}+\mathrm{y})}^{\mathrm{m}+\mathrm{n}-1}}$$$$\mathrm{y}{}^{\prime }=\frac{\frac{{\mathrm{x}}^{\mathrm{m}+\mathrm{n}}}{\mathrm{x}}-\frac{{(\mathrm{x}+\mathrm{y})}^{\mathrm{m}+\mathrm{n}}}{\mathrm{x}+\mathrm{y}}}{\frac{{(\mathrm{x}+\mathrm{y})}^{\mathrm{m}+\mathrm{n}}}{\mathrm{x}+\mathrm{y}}}=\frac{\frac{{\mathrm{x}}^{\mathrm{m}+\mathrm{n}}}{\mathrm{x}}-\frac{{\mathrm{x}}^{\mathrm{m}+\mathrm{n}}}{\mathrm{x}+\mathrm{y}}}{\frac{{\mathrm{x}}^{\mathrm{m}+\mathrm{n}}}{\mathrm{x}+\mathrm{y}}}$$$$=\frac{\frac{1}{\mathrm{x}}-\frac{1}{\mathrm{x}+\mathrm{y}}}{\frac{1}{\mathrm{x}+\mathrm{y}}}=(\mathrm{x}+\mathrm{y})\left(\frac{\mathrm{y}}{\mathrm{x}(\mathrm{x}+\mathrm{y})}\right)=\frac{\mathrm{y}}{\mathrm{x}}$$

Commented by MJS last updated on 09/Mar/20

$$\mathrm{let}m+n=q$$$$x^q={(x+y)}^q$$$$\Rightarrow y=0\mathrm{if}\mathrm{we}\mathrm{stay}\mathrm{within}\mathbb{R}$$

Answered by som(math1967) last updated on 09/Mar/20

$$\frac{dy}{dx}=\frac{y}{x}$$

Commented by niroj last updated on 09/Mar/20

$$Expandtheworkingoperation.$$

Commented by som(math1967) last updated on 09/Mar/20

$${lnx}^m{y}^n=ln{(x+y)}^{m+n}$$$$mlnx+nlny=(m+n)ln(x+y)$$$$\frac{m}{x}+\frac{n}{y}\cdot \frac{dy}{dx}=\frac{m+n}{x+y}(1+\frac{dy}{dx})$$$$(\frac{n}{y}-\frac{m+n}{x+y})\frac{dy}{dx}=(\frac{m+n}{x+y}-\frac{m}{x})$$$$\frac{nx-my}{y(x+y)}\cdot \frac{dy}{dx}=\frac{nx-my}{x(x+y)}$$$$\therefore \frac{dy}{dx}=\frac{y}{x}$$

Commented by niroj last updated on 09/Mar/20

$$\mathrm{thank}\mathrm{you}\mathrm{to}\mathrm{all}\mathrm{super}\mathrm{heros}.$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com