A) \[y\]
B) \[\sqrt{1+{{y}^{2}}}\]
C) \[\frac{y}{\sqrt{1+{{y}^{2}}}}\]
D) \[{{y}^{2}}\]
Correct Answer: A
Solution :
\[x=\int_{0}^{y}{\frac{dt}{\sqrt{1+{{t}^{2}}}},}\] by Leibnitz rule, we get \[1=\frac{1}{\sqrt{1+{{y}^{2}}}}.\,\frac{dy}{dx}\] \[\Rightarrow \] \[\frac{dy}{dx}=\sqrt{1+{{y}^{2}}}\] \[\Rightarrow \] \[\frac{{{d}^{2}}y}{d{{x}^{2}}}=\frac{y}{\sqrt{1+{{y}^{2}}}}\cdot \,\frac{dy}{dx}\] \[\therefore \] \[\frac{{{d}^{2}}y}{d{{x}^{2}}}=\,\frac{y}{\sqrt{1+{{y}^{2}}}}\cdot \,\sqrt{1+{{y}^{2}}}=y\]
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