A) \[\frac{{{y}^{3}}}{2{{y}^{2}}-4{{x}^{2}}}\]
B) \[\frac{2{{y}^{2}}-4{{x}^{2}}}{{{y}^{3}}}\]
C) \[\frac{4y-8x}{{{y}^{3}}}\]
D) 0
Correct Answer: D
Solution :
Given \[3{{y}^{2}}-4{{x}^{2}}=0\] ...(i) On differentiating with respect to\[x,\]we get \[6y\frac{dy}{dx}-8x=0\] \[\Rightarrow \] \[3y\frac{dy}{dx}-4x=0\] ...(ii) Again differentiating with respect to\[x,\] \[3y\frac{{{d}^{2}}y}{d{{x}^{2}}}+3{{\left( \frac{dy}{dx} \right)}^{2}}-4=0\] \[\Rightarrow \] \[3y\frac{{{d}^{2}}y}{d{{x}^{2}}}+3{{\left( \frac{4x}{3y} \right)}^{2}}-4=0\] [using Eq. (i)] \[\Rightarrow \] \[3y\frac{{{d}^{2}}y}{d{{x}^{2}}}+\frac{16{{x}^{2}}}{3{{y}^{2}}}-4=0\] \[\Rightarrow \] \[\frac{{{d}^{2}}y}{d{{x}^{2}}}=\frac{12{{y}^{2}}-16{{x}^{2}}}{9{{y}^{3}}}\] \[\Rightarrow \] \[\frac{{{d}^{2}}y}{d{{x}^{2}}}=\frac{4(3{{y}^{2}}-4{{x}^{2}})}{9{{y}^{3}}}\] [using Eq. (i)] \[\Rightarrow \] \[\frac{{{d}^{2}}y}{d{{x}^{2}}}=0\]
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