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Question Number 107922 by mohammad17 last updated on 13/Aug/20

if: y=(x/(x^2 +1)) then find (d(√y)/d(√x)) ?

$${if}:\:{y}=\frac{{x}}{{x}^{\mathrm{2}} +\mathrm{1}}\:{then}\:{find}\:\frac{{d}\sqrt{{y}}}{{d}\sqrt{{x}}}\:? \\ $$

Answered by 1549442205PVT last updated on 13/Aug/20

Set (√x) =t and (√y)=z⇒y=(t^2 /(t^4 +1))  z^2 =(t^2 /(t^4 +1))⇒2z.z′=((2t(t^4 +1)−4t^3 .t^2 )/((t^4 +1)^2 ))  ⇒2z.z′=((−2t^5 +2)/((t^4 +1)^2 ))  (d(√y)/d(√x))=(dz/dt)=((−2t^5 +2)/((t^4 +1)^2 .2z))=((−2x^2 (√x)+2)/(2(x^2 +1)^2 (√(x/(x^2 +1)))))  ⇒(d(√y)/d(√x))=((−2x^2 (√x)+2)/(2(x^2 +1)(√(x(x^2 +1)))))

$$\mathrm{Set}\:\sqrt{\mathrm{x}}\:=\mathrm{t}\:\mathrm{and}\:\sqrt{\mathrm{y}}=\mathrm{z}\Rightarrow\mathrm{y}=\frac{\mathrm{t}^{\mathrm{2}} }{\mathrm{t}^{\mathrm{4}} +\mathrm{1}} \\ $$$$\mathrm{z}^{\mathrm{2}} =\frac{\mathrm{t}^{\mathrm{2}} }{\mathrm{t}^{\mathrm{4}} +\mathrm{1}}\Rightarrow\mathrm{2z}.\mathrm{z}'=\frac{\mathrm{2t}\left(\mathrm{t}^{\mathrm{4}} +\mathrm{1}\right)−\mathrm{4t}^{\mathrm{3}} .\mathrm{t}^{\mathrm{2}} }{\left(\mathrm{t}^{\mathrm{4}} +\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$\Rightarrow\mathrm{2z}.\mathrm{z}'=\frac{−\mathrm{2t}^{\mathrm{5}} +\mathrm{2}}{\left(\mathrm{t}^{\mathrm{4}} +\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$\frac{\mathrm{d}\sqrt{\mathrm{y}}}{\mathrm{d}\sqrt{\mathrm{x}}}=\frac{\mathrm{dz}}{\mathrm{dt}}=\frac{−\mathrm{2t}^{\mathrm{5}} +\mathrm{2}}{\left(\mathrm{t}^{\mathrm{4}} +\mathrm{1}\right)^{\mathrm{2}} .\mathrm{2z}}=\frac{−\mathrm{2x}^{\mathrm{2}} \sqrt{\mathrm{x}}+\mathrm{2}}{\mathrm{2}\left(\mathrm{x}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} \sqrt{\frac{\mathrm{x}}{\mathrm{x}^{\mathrm{2}} +\mathrm{1}}}} \\ $$$$\Rightarrow\frac{\boldsymbol{\mathrm{d}}\sqrt{\boldsymbol{\mathrm{y}}}}{\boldsymbol{\mathrm{d}}\sqrt{\boldsymbol{\mathrm{x}}}}=\frac{−\mathrm{2}\boldsymbol{\mathrm{x}}^{\mathrm{2}} \sqrt{\boldsymbol{\mathrm{x}}}+\mathrm{2}}{\mathrm{2}\left(\boldsymbol{\mathrm{x}}^{\mathrm{2}} +\mathrm{1}\right)\sqrt{\boldsymbol{\mathrm{x}}\left(\boldsymbol{\mathrm{x}}^{\mathrm{2}} +\mathrm{1}\right)}} \\ $$

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