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 63301 by Rio Michael last updated on 02/Jul/19

$$find\frac{dy}{dx}ifx(x+y)=y^2$$

Commented by kaivan.ahmadi last updated on 02/Jul/19

$$f(x,y)=x^2+xy-y^2=0$$$$\frac{dy}{dx}=-\frac{f_x^{\prime }}{f_y^{\prime }}=-\frac{2x+y}{x-2y}=\frac{2x+y}{2y-x}$$$$$$

Commented by Rio Michael last updated on 02/Jul/19

$$perfectandgoodbutididntknowthisrule$$$$\frac{dy}{dx}=\frac{f{}_x^{\prime }}{f{}_y^{\prime }}thanksforit$$

Commented by Prithwish sen last updated on 02/Jul/19

$${\mathrm{x}}^2+\mathrm{xy}={\mathrm{y}}^2$$$$\mathrm{diff}.\mathrm{w}.\mathrm{r}.\mathrm{t}.\mathrm{x}$$$$2\mathrm{x}+\mathrm{y}+\mathrm{x}\frac{\mathrm{dy}}{\mathrm{dx}}=2\mathrm{y}\frac{\mathrm{dy}}{\mathrm{dx}}$$$$\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{2\mathrm{x}+\mathrm{y}}{2\mathrm{y}-\mathrm{x}}$$

Commented by Rio Michael last updated on 02/Jul/19

$$yesthatsourstyle..douoffermathematicsmechanics$$

Commented by MJS last updated on 02/Jul/19

$$\mathrm{the}‘‘\mathrm{long}{\mathrm{version}}^{″}$$$$x^2+xy-y^2=0$$$$\mathrm{now}\mathrm{differentiate}\mathrm{each}\mathrm{term}\mathrm{and}\mathrm{multiply}\mathrm{with}$$$$dx\mathrm{or}dy:$$$$x^2\to 2xdx$$$$xy\to ydx+xdy$$$$y^2\to 2ydy$$$$2xdx+ydx+xdy-2ydy=0$$$$\mathrm{now}‘‘{\mathrm{split}}^{″}dx\mathrm{from}dy$$$$(2x+y)dx=(2y-x)dy$$$$\Rightarrow \frac{dy}{dx}=\frac{2x+y}{2y-x}$$

Commented by Rio Michael last updated on 02/Jul/19

$$thatsstillgood$$

Commented by mathmax by abdo last updated on 02/Jul/19

$$\left(e\right)\iff y^2-xy-x^2=0\Rightarrow p\left(y\right)=0$$$$\mathrm{\Delta }=x^2+4x^2=5x^2\Rightarrow y=\frac{1}{2}(x+\sqrt{5}x\xi \left(x\right))ory=\frac{1}{2}(x-\sqrt{5}\xi \left(x\right)x)with\xi \left(x\right)=1ifx>0$$$$and\xi \left(x\right)=-1ifx<0\Rightarrow y=\frac{x}{2}(1+\sqrt{5}\xi \left(x\right))ory=\frac{x}{2}(1-\sqrt{5}\xi \left(x\right))\Rightarrow$$$$\frac{dy}{dx}=\frac{1+\sqrt{5}\xi \left(x\right)}{2}or\frac{dy}{dx}=\frac{1-\sqrt{5}\xi \left(x\right)}{2}$$

Commented by ajfour last updated on 02/Jul/19

$$whats\xi ,pleaselectureelaboratey,$$$$Sir.$$

Commented by mathmax by abdo last updated on 02/Jul/19

$$\mid x\mid =x\xi \left(x\right)so\xi \left(x\right)=\{\begin{array}{l}1ifx>0\\ -1ifx<0\end{array}$$

Answered by Smail last updated on 02/Jul/19

$$x(x+y)=y^2$$$$x^2+xy=y^2\iff y^2-xy=x^2$$$${(y-\frac{x}{2})}^2=x^2+\frac{x^2}{4}$$$$y=\underset{-}{+}x\frac{\sqrt{5}+1}{2}$$$$\frac{dy}{dx}=\underset{-}{+}\frac{\sqrt{5}+1}{2}$$

Commented by Prithwish sen last updated on 02/Jul/19

$$\because \mathrm{the}\mathrm{equation}\mathrm{represents}\mathrm{a}\mathrm{pair}\mathrm{of}\mathrm{straight}$$$$\mathrm{lines}\mathrm{passing}\mathrm{through}\mathrm{origin}.$$$$(\mathrm{y}-\mathrm{x}\frac{\mathrm{l}+\sqrt{5}}{2})(\mathrm{y}-\mathrm{x}\frac{1-\sqrt{5}}{2})=0$$$$\Rightarrow {\mathrm{y}}^2-\mathrm{xy}-{\mathrm{x}}^2=0$$$$\therefore \mathrm{the}\mathrm{equation}\mathrm{has}\mathrm{only}\mathrm{two}\mathrm{pair}\mathrm{of}\mathrm{tangents}$$$$\mathrm{with}\frac{1-\sqrt{5}}{2}\mathrm{and}\frac{1+\sqrt{5}}{2}\mathrm{as}\mathrm{slope}$$

Commented by Rio Michael last updated on 02/Jul/19

$$whatdidudo?$$

Commented by Smail last updated on 02/Jul/19

$$I^{\prime }vewrittenyintermofx.$$

Commented by Rio Michael last updated on 02/Jul/19

$$ahhokaybutyoucanstilldoimplicitdifferentiation$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com