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Question Number 111216 by mohammad17 last updated on 02/Sep/20

Commented by Dwaipayan Shikari last updated on 02/Sep/20

$$x^y=e^{{tan}^{-1}y}$$$$ylogx={tan}^{-1}y$$$$\frac{dy}{dx}logx+\frac{y}{x}=\frac{1}{1+y^2}.\frac{dy}{dx}$$$$\frac{dy}{dx}(logx-\frac{1}{1+y^2})=-\frac{y}{x}$$$$\frac{dy}{dx}=\frac{y}{x(\frac{1}{1+y^2}-logx)}$$

Commented by mohammad17 last updated on 03/Sep/20

$$thankyousir$$

Answered by Dwaipayan Shikari last updated on 02/Sep/20

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

Commented by mohammad17 last updated on 03/Sep/20

$$thankyousir$$

Answered by Dwaipayan Shikari last updated on 02/Sep/20

$$y={log}_y\sqrt{x}$$$$y^y=\sqrt{x}$$$$ylogy=\frac{1}{2}logx$$$$(logy+1)\frac{dy}{dx}=\frac{1}{2x}$$$$\frac{dy}{dx}=\frac{1}{2x(logy+1)}$$

Commented by mohammad17 last updated on 03/Sep/20

$$thankyousir$$

Answered by Rio Michael last updated on 02/Sep/20

$$\left(1\right)\mathbf{A}=\mathbf{i}-2\mathbf{j}+\mathbf{k},\mathbf{B}=2\mathbf{i}+\mathbf{j}-2\mathbf{k}$$$$\mathbf{A}\mathrm{is}\mathrm{not}\mathrm{a}\mathrm{scalar}\mathrm{multiple}\mathrm{of}\mathbf{B}\mathrm{neigher}\mathrm{is}\mathbf{B}\mathrm{a}\mathrm{scalar}\mathrm{multiple}\mathrm{of}\mathbf{A}$$$$\mathrm{so}\mathrm{the}\mathrm{two}\mathrm{vectors}\mathrm{are}\mathrm{not}\mathrm{parrallel}.$$$$\mathbf{A}.\mathbf{B}=2-2-2=-2\Rightarrow \mathbf{A}\mathrm{and}\mathbf{B}\mathrm{are}\mathrm{nieghter}\mathrm{parrallel}\mathrm{nor}\mathrm{perpendicular}$$

Commented by mohammad17 last updated on 03/Sep/20

$$thankyousir$$

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