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 :
\[y=\sqrt{x+\sqrt{y+\sqrt{x+\sqrt{y+......\infty }}}}\] \[\Rightarrow \] \[{{y}^{2}}=x+\sqrt{y+\sqrt{x+\sqrt{y+......\infty }}}\] \[\Rightarrow \] \[{{y}^{2}}=x+\sqrt{y+y}\] \[\Rightarrow \] \[{{y}^{2}}=x+\sqrt{2y}\] \[\Rightarrow \] \[{{({{y}^{2}}-x)}^{2}}=2y\] On differentiating both sides w.r.t. x, we get \[2({{y}^{2}}-x)\left( 2y\frac{dy}{dx}-1 \right)=2\frac{dy}{dx}\] \[\Rightarrow \] \[2({{y}^{3}}-xy)\frac{dy}{dx}-({{y}^{2}}-x)=\frac{dy}{dx}\] \[\Rightarrow \] \[(2{{y}^{3}}-2xy-1)\frac{dy}{dx}={{y}^{2}}-x\] \[\Rightarrow \] \[\frac{dy}{dx}=\frac{{{y}^{2}}-x}{2{{y}^{3}}-2xy-1}\]
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