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Question Number 11011 by madscientist last updated on 06/Mar/17

if tan(xy)=x then (dy/dx)=

$${if}\:{tan}\left({xy}\right)={x}\:{then}\:\frac{{dy}}{{dx}}= \\ $$

Answered by mrW1 last updated on 06/Mar/17

xy=tan^(−1) x y=((tan^(−1) x)/x) (dy/dx)=−((tan^(−1) x)/x^2 )+(1/(x(1+x^2 ))) way 2: cos^2 (xy)=(1/(1+tan^2 (xy)))=(1/(1+x^2 )) (1/(cos^2 (xy)))(y+x(dy/dx))=1 (dy/dx)=((cos^2 (xy)−y)/x)=(((1/(1+x^2 ))−y)/x)=(1/(x(1+x^2 )))−(y/x)=(1/(x(1+x^2 )))−((tan^(−1) x)/x^2 )

$${xy}=\mathrm{tan}^{−\mathrm{1}} \:{x} \\ $$$${y}=\frac{\mathrm{tan}^{−\mathrm{1}} \:{x}}{{x}} \\ $$$$\frac{{dy}}{{dx}}=−\frac{\mathrm{tan}^{−\mathrm{1}} \:{x}}{{x}^{\mathrm{2}} }+\frac{\mathrm{1}}{{x}\left(\mathrm{1}+{x}^{\mathrm{2}} \right)} \\ $$$$ \\ $$$${way}\:\mathrm{2}: \\ $$$$\mathrm{cos}^{\mathrm{2}} \:\left({xy}\right)=\frac{\mathrm{1}}{\mathrm{1}+\mathrm{tan}^{\mathrm{2}} \:\left({xy}\right)}=\frac{\mathrm{1}}{\mathrm{1}+{x}^{\mathrm{2}} } \\ $$$$ \\ $$$$\frac{\mathrm{1}}{\mathrm{cos}^{\mathrm{2}} \:\left({xy}\right)}\left({y}+{x}\frac{{dy}}{{dx}}\right)=\mathrm{1} \\ $$$$\frac{{dy}}{{dx}}=\frac{\mathrm{cos}^{\mathrm{2}} \:\left({xy}\right)−{y}}{{x}}=\frac{\frac{\mathrm{1}}{\mathrm{1}+{x}^{\mathrm{2}} }−{y}}{{x}}=\frac{\mathrm{1}}{{x}\left(\mathrm{1}+{x}^{\mathrm{2}} \right)}−\frac{{y}}{{x}}=\frac{\mathrm{1}}{{x}\left(\mathrm{1}+{x}^{\mathrm{2}} \right)}−\frac{\mathrm{tan}^{−\mathrm{1}} \:{x}}{{x}^{\mathrm{2}} } \\ $$

Answered by bahmanfeshki last updated on 06/Mar/17

(xy′+y)(1+tan^2 (xy))=1 →y′=(1/x)((1/(1+x^2 ))−y)

$$\left({xy}'+{y}\right)\left(\mathrm{1}+{tan}^{\mathrm{2}} \:\left({xy}\right)\right)=\mathrm{1} \\ $$$$\rightarrow{y}'=\frac{\mathrm{1}}{{x}}\left(\frac{\mathrm{1}}{\mathrm{1}+{x}^{\mathrm{2}} }−{y}\right) \\ $$

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