A) \[{{\left( \frac{{{d}^{2}}y}{d{{x}^{2}}} \right)}^{-1}}\]
B) \[-{{\left( \frac{{{d}^{2}}y}{d{{x}^{2}}} \right)}^{-1}}.{{\left( \frac{dy}{d{{x}^{2}}} \right)}^{-3}}\]
C) \[\left( \frac{{{d}^{2}}y}{d{{x}^{2}}} \right)\times {{\left( \frac{dy}{dx} \right)}^{2}}\]
D) \[-\left( \frac{{{d}^{2}}y}{d{{x}^{2}}} \right).{{\left( \frac{dy}{dx} \right)}^{-3}}\]
Correct Answer: D
Solution :
[d] \[\because \frac{dx}{dy}=\frac{1}{\frac{dy}{dx}}={{\left( \frac{dy}{dx} \right)}^{-1}}\] Now \[\frac{d}{dy}=\left( \frac{dx}{dy} \right)=\frac{d}{dy}\left\{ {{\left( \frac{dy}{dx} \right)}^{-1}} \right\}\] \[=\frac{d}{dx}=\left( \frac{dy}{dx} \right).\frac{dx}{dy}\frac{{{d}^{2}}x}{d{{y}^{2}}}=\frac{-1}{{{\left( \frac{dy}{dx} \right)}^{2}}}\times \frac{d}{dx}\left( \frac{dy}{dx} \right).\frac{dx}{dy}\] \[=-\frac{{{d}^{2}}y}{d{{x}^{2}}}.\frac{-1}{{{\left( \frac{dy}{dx} \right)}^{3}}}=-\frac{{{d}^{2}}y}{d{{x}^{2}}}.{{\left( \frac{dy}{dx} \right)}^{-3}}\] Hence, option [d]is correct.
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