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Question Number 85472 by liki last updated on 22/Mar/20

Commented by liki last updated on 22/Mar/20

$$...pleaseanyonehelpmenow...$$

Commented by jagoll last updated on 22/Mar/20

$$\mathrm{y}=2^{\mathrm{x}}+\frac{1}{\mathrm{y}}\Rightarrow {\mathrm{y}}^2=2^{\mathrm{x}}\mathrm{y}+1$$$${\mathrm{y}}^2-2^{\mathrm{x}}\mathrm{y}-1=0$$$$2\mathrm{y}{\mathrm{y}}^{\prime }-(2^{\mathrm{x}}{\mathrm{y}}^{\prime }+2^{\mathrm{x}}\mathrm{y}\ln 2)=0$$$${2\mathrm{yy}}^{\prime }-2^{\mathrm{x}}{\mathrm{y}}^{\prime }=2^{\mathrm{x}}\mathrm{y}\ln 2$$$${\mathrm{y}}^{\prime }=\frac{2^{\mathrm{x}}\mathrm{y}\ln 2}{2\mathrm{y}-2^{\mathrm{x}}}$$$$$$

Commented by jagoll last updated on 22/Mar/20

$$\mathrm{done}$$

Commented by liki last updated on 22/Mar/20

$$...Thankyousirbeblessed$$

Answered by mind is power last updated on 22/Mar/20

$$y=2^x+\frac{1}{y}\Rightarrow$$$$\frac{dy}{dx}=ln\left(2\right)2^x-\frac{dy}{dx}.\frac{1}{y^2}....$$$$\frac{dy}{dx}=\frac{ln\left(2\right)2^x{y}^2}{(1+y^2)}=\frac{ln\left(2\right)2^x.y}{\frac{1}{y}+y}..E$$$$\frac{1}{y}=y-2^x\Rightarrow E\iff \frac{2^{x}ln\left(2\right)y}{2y-2^x}=\frac{dy}{dx}$$$$$$

Commented by liki last updated on 22/Mar/20

$$...thanksalotsir.$$

Commented by mind is power last updated on 22/Mar/20

$$withePleasur$$

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