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

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

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

$$\mathrm{Set}\sqrt{\mathrm{x}}=\mathrm{t}\mathrm{and}\sqrt{\mathrm{y}}=\mathrm{z}\Rightarrow \mathrm{y}=\frac{{\mathrm{t}}^2}{{\mathrm{t}}^4+1}$$$${\mathrm{z}}^2=\frac{{\mathrm{t}}^2}{{\mathrm{t}}^4+1}\Rightarrow 2\mathrm{z}.{\mathrm{z}}^{\prime }=\frac{2\mathrm{t}({\mathrm{t}}^4+1)-{4\mathrm{t}}^3.{\mathrm{t}}^2}{{({\mathrm{t}}^4+1)}^2}$$$$\Rightarrow 2\mathrm{z}.{\mathrm{z}}^{\prime }=\frac{-{2\mathrm{t}}^5+2}{{({\mathrm{t}}^4+1)}^2}$$$$\frac{\mathrm{d}\sqrt{\mathrm{y}}}{\mathrm{d}\sqrt{\mathrm{x}}}=\frac{\mathrm{dz}}{\mathrm{dt}}=\frac{-{2\mathrm{t}}^5+2}{{({\mathrm{t}}^4+1)}^2.2\mathrm{z}}=\frac{-{2\mathrm{x}}^2\sqrt{\mathrm{x}}+2}{2{({\mathrm{x}}^2+1)}^2\sqrt{\frac{\mathrm{x}}{{\mathrm{x}}^2+1}}}$$$$\Rightarrow \frac{\mathbf{d}\sqrt{\mathbf{y}}}{\mathbf{d}\sqrt{\mathbf{x}}}=\frac{-2{\mathbf{x}}^2\sqrt{\mathbf{x}}+2}{2({\mathbf{x}}^2+1)\sqrt{\mathbf{x}({\mathbf{x}}^2+1)}}$$

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