A) \[n{{e}^{nx}}\]
B) \[n{{e}^{-nx}}\]
C) 1
D) \[-n{{e}^{-nx}}\]
Correct Answer: D
Solution :
Given that, \[y={{e}^{nx}}\] Differeniating both sides with respect to x \[\frac{dy}{dx}=n{{e}^{nx}}\]Again differentiating w.r.t 'x', \[\frac{{{d}^{2}}y}{d{{x}^{2}}}=n.n{{e}^{nx}}={{n}^{2}}{{e}^{nx}}\] \[y={{e}^{nx}}\] \[nx={{\log }_{e}}y\] \[x=\frac{1}{n}\log y\] Differentiating x with respect to y\[\frac{dx}{dy}=\frac{1}{n}.\frac{1}{y}\] Again differentiating with respect to y \[\frac{{{d}^{2}}x}{d{{y}^{2}}}=\frac{1}{n}.\left( \frac{-1}{{{y}^{2}}} \right)=\frac{-1}{n{{y}^{2}}}=-\frac{1}{n{{e}^{2nx}}}\] ??.(2) Multiplying equation (1) and (2) \[\left( \frac{{{d}^{2}}y}{d{{x}^{2}}} \right)\left( \frac{{{d}^{2}}x}{d{{y}^{2}}} \right)=({{n}^{2}}enx)\left( \frac{-1}{n}{{e}^{2nx}} \right)=-n{{e}^{-nx}}\]
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