A) \[{{y}^{2}}.\log a{{b}^{2}}\]
B) \[y.\log a{{b}^{2}}\]
C) \[{{y}^{2}}\]
D) \[y.{{(\log {{a}^{2}}b)}^{2}}\]
E) \[y.{{(\log a{{b}^{2}})}^{2}}\]
Correct Answer: E
Solution :
\[\because \] \[y={{a}^{x}}{{b}^{2x-1}}\] Taking log on both sides, we get \[log\text{ }y=x\text{ }log\text{ }a+(2x-1)log\text{ }b\] On differentiating w.r.t.\[x,\]we get \[\frac{1}{y}=\frac{dy}{dx}=\log a+\log {{b}^{2}}\] \[\frac{dy}{dx}=y\log a{{b}^{2}}\] Again differentiating, we get \[\frac{{{d}^{2}}y}{d{{x}^{2}}}=\frac{dy}{dx}\log a{{b}^{2}}=y{{(\log \,a{{b}^{2}})}^{2}}\]
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