A) \[4{{e}^{2x}}\]
B) \[-\frac{1}{2}\,{{e}^{-4x}}\]
C) \[-\frac{3}{4}\,{{e}^{-5x}}\]
D) \[4{{e}^{x}}\]
Correct Answer: B
Solution :
Given, \[x={{\log }_{e}}t\] \[\Rightarrow \] \[{{e}^{x}}=t\] and \[y+1={{t}^{2}}\] \[\Rightarrow \] \[y={{e}^{2x}}-1\] On differentiating w. r. t. y, we get \[2{{e}^{2x}}\frac{dx}{dy}=1\] \[\Rightarrow \] \[\frac{dx}{dy}=\frac{1}{2{{e}^{2x}}}\] Again differentiating w. r. t. y, we get \[\frac{{{d}^{2}}x}{d{{y}^{2}}}=\frac{1}{2}{{e}^{-2x}}(-2)\frac{dx}{dy}\] \[=-{{e}^{-2x}}.\frac{1}{2{{e}^{2x}}}\] \[=-\frac{1}{2}{{e}^{-4x}}\]
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