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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 (dy/dx)logx+(y/x)=(1/(1+y^2 )).(dy/dx) (dy/dx)(logx−(1/(1+y^2 )))=−(y/x) (dy/dx)=(y/(x((1/(1+y^2 ))−logx)))

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

Commented by mohammad17 last updated on 03/Sep/20

thank you sir

$${thank}\:{you}\:{sir} \\ $$

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

y=x^(tan^(−1) y) logy=(tan^(−1) y)logx (1/y).(dy/dx)=logx(1/(1+y^2 )).(dy/dx)+(1/x)tan^(−1) y (dy/dx)((1/y)−((logx)/(1+y^2 )))=(1/x)tan^(−1) y (dy/dx)=((tan^(−1) y)/(x((1/y)−((logx)/(1+y^2 )))))

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

Commented by mohammad17 last updated on 03/Sep/20

thank you sir

$${thank}\:{you}\:{sir} \\ $$

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

y=log_y (√x) y^y =(√x) ylogy=(1/2)logx (logy+1)(dy/dx)=(1/(2x)) (dy/dx)=(1/(2x(logy+1)))

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

Commented by mohammad17 last updated on 03/Sep/20

thank you sir

$${thank}\:{you}\:{sir} \\ $$

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

(1) A = i−2j+k , B = 2i+j−2k A is not a scalar multiple of B neigher is B a scalar multiple of A so the two vectors are not parrallel. A.B = 2−2−2 = −2 ⇒ A and B are nieghter parrallel nor perpendicular

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

Commented by mohammad17 last updated on 03/Sep/20

thank you sir

$${thank}\:{you}\:{sir} \\ $$

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