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Hint: Differentiate the given function with respect to \[x\] and with respect to \[y\] twice.

A: Given \[f:[0,\infty )\to [0,\infty ]\]be a function defined by \[y=f\left( x \right)={{x}^{2}}\], then we have to check the value of \[\left( \dfrac{{{d}^{2}}y}{d{{x}^{2}}} \right)\left( \dfrac{{{d}^{2}}x}{d{{y}^{2}}} \right)\].

Now, we take the given function,

\[y={{x}^{2}}....\left( i \right)\]

Now, we differentiate it with respect to \[x\].

\[\left[ \text{Also}\dfrac{d\left( {{x}^{n}} \right)}{dx}=n{{x}^{n-1}} \right]\]

Therefore, \[\dfrac{dy}{dx}=2x.....\left( ii \right)\]

Again differentiating with respect to \[x\],

We get, \[\dfrac{{{d}^{2}}y}{d{{x}^{2}}}=2....\left( iii \right)\]

As we have found that, \[\dfrac{dy}{dx}=2x\]

By taking reciprocal on both sides,

We get, \[\dfrac{dx}{dy}=\dfrac{1}{2x}....\left( iv \right)\]

Now, we differentiate with respect to \[y\].

We get, \[\dfrac{d}{dy}\left( \dfrac{dx}{dy} \right)=\dfrac{d}{dy}\left( \dfrac{1}{2x} \right)\]

\[\Rightarrow \dfrac{{{d}^{2}}x}{d{{y}^{2}}}=\dfrac{-1}{2{{x}^{2}}}\dfrac{dx}{dy}\]

Now, we put the value of \[\dfrac{dx}{dy}\].

We get, \[\dfrac{{{d}^{2}}x}{d{{y}^{2}}}=\dfrac{-1}{2{{x}^{2}}}.\dfrac{1}{2x}\]

Hence, \[\Rightarrow \dfrac{{{d}^{2}}x}{d{{y}^{2}}}=\dfrac{-1}{4{{x}^{3}}}....\left( v \right)\]

Multiplying equation \[\left( iii \right)\]and \[\left( v \right)\],

We get, \[\left( \dfrac{{{d}^{2}}y}{d{{x}^{2}}} \right).\left( \dfrac{{{d}^{2}}x}{d{{y}^{2}}} \right)=2.\left( \dfrac{-1}{4{{x}^{3}}} \right)\]

Therefore, \[\left( \dfrac{{{d}^{2}}y}{d{{x}^{2}}} \right).\left( \dfrac{{{d}^{2}}x}{d{{y}^{2}}}

\right)=\dfrac{-1}{2{{x}^{3}}}\].

Hence, given Assertion (A) is wrong.

R: Here we have to check whether \[\left( \dfrac{dy}{dx} \right).\left( \dfrac{dx}{dy} \right)=1\]or not.

We know that any quantity when multiplied by its reciprocal gives a result as \[1\].

That is \[a\times \dfrac{1}{a}=1\].

Now, we put \[a=\dfrac{dy}{dx}\].

We get, \[\dfrac{dy}{dx}\times \dfrac{1}{\dfrac{dy}{dx}}=1\]

Or, \[\left( \dfrac{dy}{dx} \right).\left( \dfrac{dx}{dy} \right)=1\]

To verify it further, we multiply the equation \[\left( ii \right)\]and \[\left( iv \right)\].

\[\left( \dfrac{dy}{dx} \right).\left( \dfrac{dx}{dy} \right)=2x.\dfrac{1}{2x}\]

Therefore, \[\left( \dfrac{dy}{dx} \right).\left( \dfrac{dx}{dy} \right)=1\] [Hence Proved]

Hence, Reason (R) is correct.

Therefore, option (d) is correct that is A is false and R is true

Note: Some students misunderstand that \[\dfrac{{{d}^{2}}x}{d{{y}^{2}}}\]and \[\dfrac{{{d}^{2}}y}{d{{x}^{2}}}\] are reciprocal of each other like \[\dfrac{dx}{dy}\]and \[\dfrac{dy}{dx}\], but they are not as proved by above result also.

A: Given \[f:[0,\infty )\to [0,\infty ]\]be a function defined by \[y=f\left( x \right)={{x}^{2}}\], then we have to check the value of \[\left( \dfrac{{{d}^{2}}y}{d{{x}^{2}}} \right)\left( \dfrac{{{d}^{2}}x}{d{{y}^{2}}} \right)\].

Now, we take the given function,

\[y={{x}^{2}}....\left( i \right)\]

Now, we differentiate it with respect to \[x\].

\[\left[ \text{Also}\dfrac{d\left( {{x}^{n}} \right)}{dx}=n{{x}^{n-1}} \right]\]

Therefore, \[\dfrac{dy}{dx}=2x.....\left( ii \right)\]

Again differentiating with respect to \[x\],

We get, \[\dfrac{{{d}^{2}}y}{d{{x}^{2}}}=2....\left( iii \right)\]

As we have found that, \[\dfrac{dy}{dx}=2x\]

By taking reciprocal on both sides,

We get, \[\dfrac{dx}{dy}=\dfrac{1}{2x}....\left( iv \right)\]

Now, we differentiate with respect to \[y\].

We get, \[\dfrac{d}{dy}\left( \dfrac{dx}{dy} \right)=\dfrac{d}{dy}\left( \dfrac{1}{2x} \right)\]

\[\Rightarrow \dfrac{{{d}^{2}}x}{d{{y}^{2}}}=\dfrac{-1}{2{{x}^{2}}}\dfrac{dx}{dy}\]

Now, we put the value of \[\dfrac{dx}{dy}\].

We get, \[\dfrac{{{d}^{2}}x}{d{{y}^{2}}}=\dfrac{-1}{2{{x}^{2}}}.\dfrac{1}{2x}\]

Hence, \[\Rightarrow \dfrac{{{d}^{2}}x}{d{{y}^{2}}}=\dfrac{-1}{4{{x}^{3}}}....\left( v \right)\]

Multiplying equation \[\left( iii \right)\]and \[\left( v \right)\],

We get, \[\left( \dfrac{{{d}^{2}}y}{d{{x}^{2}}} \right).\left( \dfrac{{{d}^{2}}x}{d{{y}^{2}}} \right)=2.\left( \dfrac{-1}{4{{x}^{3}}} \right)\]

Therefore, \[\left( \dfrac{{{d}^{2}}y}{d{{x}^{2}}} \right).\left( \dfrac{{{d}^{2}}x}{d{{y}^{2}}}

\right)=\dfrac{-1}{2{{x}^{3}}}\].

Hence, given Assertion (A) is wrong.

R: Here we have to check whether \[\left( \dfrac{dy}{dx} \right).\left( \dfrac{dx}{dy} \right)=1\]or not.

We know that any quantity when multiplied by its reciprocal gives a result as \[1\].

That is \[a\times \dfrac{1}{a}=1\].

Now, we put \[a=\dfrac{dy}{dx}\].

We get, \[\dfrac{dy}{dx}\times \dfrac{1}{\dfrac{dy}{dx}}=1\]

Or, \[\left( \dfrac{dy}{dx} \right).\left( \dfrac{dx}{dy} \right)=1\]

To verify it further, we multiply the equation \[\left( ii \right)\]and \[\left( iv \right)\].

\[\left( \dfrac{dy}{dx} \right).\left( \dfrac{dx}{dy} \right)=2x.\dfrac{1}{2x}\]

Therefore, \[\left( \dfrac{dy}{dx} \right).\left( \dfrac{dx}{dy} \right)=1\] [Hence Proved]

Hence, Reason (R) is correct.

Therefore, option (d) is correct that is A is false and R is true

Note: Some students misunderstand that \[\dfrac{{{d}^{2}}x}{d{{y}^{2}}}\]and \[\dfrac{{{d}^{2}}y}{d{{x}^{2}}}\] are reciprocal of each other like \[\dfrac{dx}{dy}\]and \[\dfrac{dy}{dx}\], but they are not as proved by above result also.

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