A) \[\frac{y+x}{{{y}^{2}}-2x}\]
B) \[\frac{{{y}^{3}}-x}{2{{y}^{2}}-2xy-1}\]
C) \[\frac{{{y}^{3}}+x}{2{{y}^{2}}-x}\]
D) None of these
Correct Answer: D
Solution :
We have,\[y=\sqrt{x+\sqrt{y+\sqrt{x+\sqrt{y+...\infty }}}}\] \[\Rightarrow \] \[{{y}^{2}}=x+\sqrt{y+\sqrt{x+\sqrt{y+...+\infty }}}\] \[\Rightarrow \]\[{{y}^{2}}=x+\sqrt{y+y}\] \[\Rightarrow \] \[({{y}^{2}}-{{x}^{2}})=2y\] On differentiating both sides w. r. t . \[x\], we get \[2({{y}^{2-x}})\left( 2y\frac{dy}{dx}-1 \right)=2\frac{dy}{dx}\] \[\Rightarrow \] \[2y({{y}^{2}}-x)\frac{dy}{dx}-\frac{dy}{dx}={{y}^{2}}-x\] \[\Rightarrow \] \[\frac{dy}{dx}(2{{y}^{3}}-2xy-1)=({{y}^{2}}-x)\] \[\Rightarrow \] \[\frac{dy}{dx}=\frac{{{y}^{2}}-x}{(2{{y}^{3}}-2xy-1)}\]
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