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Question Number 104542 by mohammad17 last updated on 22/Jul/20

Answered by Dwaipayan Shikari last updated on 22/Jul/20

$$2)$$$${tan}^{-1}\left(\frac{y}{x}\right)=log(x^2+y^2)$$$$\frac{\frac{1}{x}.\frac{dy}{dx}-\frac{y}{x^2}}{\frac{y^2}{x^2}+1}=\frac{2x+2y\frac{dy}{dx}}{x^2+y^2}$$$$x\frac{dy}{dx}-y=2x+2y\frac{dy}{dx}$$$$\frac{dy}{dx}=\frac{2x+y}{x-2y}$$

Commented by mohammad17 last updated on 22/Jul/20

$$thankyousir$$

Answered by Dwaipayan Shikari last updated on 22/Jul/20

$$xy=y^{tanx}$$$$logx+logy=tanxlogy$$$$\frac{1}{x}+\frac{1}{y}\frac{dy}{dx}=tanx.\frac{1}{y}\frac{dy}{dx}+{sec}^{2}xlogy$$$$\frac{1}{y}\frac{dy}{dx}(1-tanx)=\frac{{xsec}^{2}logy-1}{x}$$$$\frac{tanx}{log\left(xy\right)}.\frac{dy}{dx}=\frac{{xsec}^{2}xlogy-1}{x(1-tanx)}$$$$\frac{dy}{dx}=\frac{log\left(xy\right)}{tanx}\left(\frac{{xsec}^{2}xlogy-1}{x(1-tanx)}\right)$$

Commented by mohammad17 last updated on 22/Jul/20

Commented by mohammad17 last updated on 22/Jul/20

$$siristhesolutiontrue?$$

Commented by Dwaipayan Shikari last updated on 22/Jul/20

$$yes$$

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