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Hint: Directly apply the derivative and apply necessary rules of differentiation. And the given expression should be derived with respect to $x$.

Complete step-by-step answer:

The given expression is

\[{{b}^{2}}{{x}^{2}}+{{a}^{2}}{{y}^{2}}={{a}^{2}}{{b}^{2}}\]

Now we will find the first order derivative of the given expression, so we will differentiate the given

expression with respect to $'x'$, we get

\[\dfrac{d}{dx}\left( {{b}^{2}}{{x}^{2}}+{{a}^{2}}{{y}^{2}} \right)=\dfrac{d}{dx}\left( {{a}^{2}}{{b}^{2}}\right)\]

Now we will apply the the sum rule of differentiation, i.e., differentiation of sum of two functions is same as the sum of individual differentiation of the functions, i.e.,

$\dfrac{d}{dx}(u+v)=\dfrac{d}{x}(u)+\dfrac{d}{x}(v)$

Applying this formula in the above equation, we get

\[\dfrac{d}{dx}\left( {{b}^{2}}{{x}^{2}} \right)+\dfrac{d}{dx}\left( {{a}^{2}}{{y}^{2}}

\right)=\dfrac{d}{dx}\left( {{a}^{2}}{{b}^{2}} \right)\]

Now we know the differentiation of constant term is zero and taking out the constant term on L.H.S., we get

\[{{b}^{2}}\dfrac{d}{dx}\left( {{x}^{2}} \right)+{{a}^{2}}\dfrac{d}{dx}\left( {{y}^{2}} \right)=0\]

Now applying the chain rule and we know $\dfrac{d}{dx}({{x}^{n}})=n{{x}^{n-1}}$ in the above equation, we get

\[2{{b}^{2}}x+2{{a}^{2}}y\dfrac{dy}{dx}=0\]

Dividing throughout by $'2'$ , we get

\[\begin{align}

& {{b}^{2}}x+{{a}^{2}}y\dfrac{dy}{dx}=0 \\

& \Rightarrow {{a}^{2}}y\dfrac{dy}{dx}=-{{b}^{2}}x \\

\end{align}\]

\[\Rightarrow \dfrac{dy}{dx}=-\dfrac{{{b}^{2}}}{{{a}^{2}}}\times \dfrac{x}{y}........(i)\]

Now we will find the second order derivative. For that we will again differentiate the above

expression with respect to $'x'$ , we get

\[\Rightarrow \dfrac{d}{dx}\left( \dfrac{dy}{dx} \right)=\dfrac{d}{dx}\left( -

\dfrac{{{b}^{2}}}{{{a}^{2}}}\times \dfrac{x}{y} \right)\]

Taking out the constant term, we get

\[\Rightarrow \dfrac{{{d}^{2}}y}{d{{x}^{2}}}=-\dfrac{{{b}^{2}}}{{{a}^{2}}}\dfrac{d}{dx}\left( \dfrac{x}{y}\right)\]

Now we know the quotient rule, i.e., \[\dfrac{d}{dx}\left( \dfrac{u}{v} \right)=\dfrac{v\dfrac{d}{dx}u-

u\dfrac{d}{dx}v}{{{v}^{2}}}\], applying this formula in the above equation, we get

\[\begin{align}

& \dfrac{{{d}^{2}}y}{d{{x}^{2}}}=-\dfrac{{{b}^{2}}}{{{a}^{2}}}\times \dfrac{y\dfrac{dx}{dx}-

x\dfrac{dy}{dx}}{{{y}^{2}}} \\

& \Rightarrow \dfrac{{{d}^{2}}y}{d{{x}^{2}}}=-\dfrac{{{b}^{2}}}{{{a}^{2}}}\times \dfrac{y-

x\dfrac{dy}{dx}}{{{y}^{2}}} \\

\end{align}\]

Substitute value from equation (i), we get

\[\Rightarrow \dfrac{{{d}^{2}}y}{d{{x}^{2}}}=-\dfrac{{{b}^{2}}}{{{a}^{2}}}\times \dfrac{y-x\left( -

\dfrac{{{b}^{2}}}{{{a}^{2}}}\times \dfrac{x}{y} \right)}{{{y}^{2}}}\]

Taking the LCM in numerator, we get

\[\Rightarrow \dfrac{{{d}^{2}}y}{d{{x}^{2}}}=-\dfrac{{{b}^{2}}}{{{a}^{2}}}\times \dfrac{\left(

\dfrac{{{a}^{2}}{{y}^{2}}+{{b}^{2}}{{x}^{2}}}{{{a}^{2}}y} \right)}{{{y}^{2}}}\]

Substituting the value from the given equation \[{{b}^{2}}{{x}^{2}}+{{a}^{2}}{{y}^{2}}={{a}^{2}}{{b}^{2}}\], we get

\[\Rightarrow \dfrac{{{d}^{2}}y}{d{{x}^{2}}}=-\dfrac{{{b}^{2}}}{{{a}^{2}}}\times \dfrac{\left(

\dfrac{{{a}^{2}}{{b}^{2}}}{{{a}^{2}}y} \right)}{{{y}^{2}}}\]

Cancelling the like terms, we get

\[\begin{align}

& \Rightarrow \dfrac{{{d}^{2}}y}{d{{x}^{2}}}=-\dfrac{{{b}^{2}}}{{{a}^{2}}}\times \dfrac{\left(

\dfrac{{{b}^{2}}}{y} \right)}{{{y}^{2}}} \\

& \Rightarrow \dfrac{{{d}^{2}}y}{d{{x}^{2}}}=-\dfrac{{{b}^{4}}}{{{a}^{2}}{{y}^{3}}} \\

\end{align}\]

This is the required second order derivative.

Note: Another approach is dividing the given expression by ${{a}^{2}}{{b}^{2}}$, you will get equation

of ellipse.

\[\begin{align}

& {{b}^{2}}{{x}^{2}}+{{a}^{2}}{{y}^{2}}={{a}^{2}}{{b}^{2}} \\

& \Rightarrow

\dfrac{{{b}^{2}}{{x}^{2}}}{{{a}^{2}}{{b}^{2}}}+\dfrac{{{a}^{2}}{{y}^{2}}}{{{a}^{2}}{{b}^{2}}}=\dfrac{{{a}^{2

}}{{b}^{2}}}{{{a}^{2}}{{b}^{2}}} \\

& \Rightarrow \dfrac{{{x}^{2}}}{{{a}^{2}}}+\dfrac{{{y}^{2}}}{{{b}^{2}}}=1 \\

\end{align}\]

Then we can differentiate, you will get the same answer.

Complete step-by-step answer:

The given expression is

\[{{b}^{2}}{{x}^{2}}+{{a}^{2}}{{y}^{2}}={{a}^{2}}{{b}^{2}}\]

Now we will find the first order derivative of the given expression, so we will differentiate the given

expression with respect to $'x'$, we get

\[\dfrac{d}{dx}\left( {{b}^{2}}{{x}^{2}}+{{a}^{2}}{{y}^{2}} \right)=\dfrac{d}{dx}\left( {{a}^{2}}{{b}^{2}}\right)\]

Now we will apply the the sum rule of differentiation, i.e., differentiation of sum of two functions is same as the sum of individual differentiation of the functions, i.e.,

$\dfrac{d}{dx}(u+v)=\dfrac{d}{x}(u)+\dfrac{d}{x}(v)$

Applying this formula in the above equation, we get

\[\dfrac{d}{dx}\left( {{b}^{2}}{{x}^{2}} \right)+\dfrac{d}{dx}\left( {{a}^{2}}{{y}^{2}}

\right)=\dfrac{d}{dx}\left( {{a}^{2}}{{b}^{2}} \right)\]

Now we know the differentiation of constant term is zero and taking out the constant term on L.H.S., we get

\[{{b}^{2}}\dfrac{d}{dx}\left( {{x}^{2}} \right)+{{a}^{2}}\dfrac{d}{dx}\left( {{y}^{2}} \right)=0\]

Now applying the chain rule and we know $\dfrac{d}{dx}({{x}^{n}})=n{{x}^{n-1}}$ in the above equation, we get

\[2{{b}^{2}}x+2{{a}^{2}}y\dfrac{dy}{dx}=0\]

Dividing throughout by $'2'$ , we get

\[\begin{align}

& {{b}^{2}}x+{{a}^{2}}y\dfrac{dy}{dx}=0 \\

& \Rightarrow {{a}^{2}}y\dfrac{dy}{dx}=-{{b}^{2}}x \\

\end{align}\]

\[\Rightarrow \dfrac{dy}{dx}=-\dfrac{{{b}^{2}}}{{{a}^{2}}}\times \dfrac{x}{y}........(i)\]

Now we will find the second order derivative. For that we will again differentiate the above

expression with respect to $'x'$ , we get

\[\Rightarrow \dfrac{d}{dx}\left( \dfrac{dy}{dx} \right)=\dfrac{d}{dx}\left( -

\dfrac{{{b}^{2}}}{{{a}^{2}}}\times \dfrac{x}{y} \right)\]

Taking out the constant term, we get

\[\Rightarrow \dfrac{{{d}^{2}}y}{d{{x}^{2}}}=-\dfrac{{{b}^{2}}}{{{a}^{2}}}\dfrac{d}{dx}\left( \dfrac{x}{y}\right)\]

Now we know the quotient rule, i.e., \[\dfrac{d}{dx}\left( \dfrac{u}{v} \right)=\dfrac{v\dfrac{d}{dx}u-

u\dfrac{d}{dx}v}{{{v}^{2}}}\], applying this formula in the above equation, we get

\[\begin{align}

& \dfrac{{{d}^{2}}y}{d{{x}^{2}}}=-\dfrac{{{b}^{2}}}{{{a}^{2}}}\times \dfrac{y\dfrac{dx}{dx}-

x\dfrac{dy}{dx}}{{{y}^{2}}} \\

& \Rightarrow \dfrac{{{d}^{2}}y}{d{{x}^{2}}}=-\dfrac{{{b}^{2}}}{{{a}^{2}}}\times \dfrac{y-

x\dfrac{dy}{dx}}{{{y}^{2}}} \\

\end{align}\]

Substitute value from equation (i), we get

\[\Rightarrow \dfrac{{{d}^{2}}y}{d{{x}^{2}}}=-\dfrac{{{b}^{2}}}{{{a}^{2}}}\times \dfrac{y-x\left( -

\dfrac{{{b}^{2}}}{{{a}^{2}}}\times \dfrac{x}{y} \right)}{{{y}^{2}}}\]

Taking the LCM in numerator, we get

\[\Rightarrow \dfrac{{{d}^{2}}y}{d{{x}^{2}}}=-\dfrac{{{b}^{2}}}{{{a}^{2}}}\times \dfrac{\left(

\dfrac{{{a}^{2}}{{y}^{2}}+{{b}^{2}}{{x}^{2}}}{{{a}^{2}}y} \right)}{{{y}^{2}}}\]

Substituting the value from the given equation \[{{b}^{2}}{{x}^{2}}+{{a}^{2}}{{y}^{2}}={{a}^{2}}{{b}^{2}}\], we get

\[\Rightarrow \dfrac{{{d}^{2}}y}{d{{x}^{2}}}=-\dfrac{{{b}^{2}}}{{{a}^{2}}}\times \dfrac{\left(

\dfrac{{{a}^{2}}{{b}^{2}}}{{{a}^{2}}y} \right)}{{{y}^{2}}}\]

Cancelling the like terms, we get

\[\begin{align}

& \Rightarrow \dfrac{{{d}^{2}}y}{d{{x}^{2}}}=-\dfrac{{{b}^{2}}}{{{a}^{2}}}\times \dfrac{\left(

\dfrac{{{b}^{2}}}{y} \right)}{{{y}^{2}}} \\

& \Rightarrow \dfrac{{{d}^{2}}y}{d{{x}^{2}}}=-\dfrac{{{b}^{4}}}{{{a}^{2}}{{y}^{3}}} \\

\end{align}\]

This is the required second order derivative.

Note: Another approach is dividing the given expression by ${{a}^{2}}{{b}^{2}}$, you will get equation

of ellipse.

\[\begin{align}

& {{b}^{2}}{{x}^{2}}+{{a}^{2}}{{y}^{2}}={{a}^{2}}{{b}^{2}} \\

& \Rightarrow

\dfrac{{{b}^{2}}{{x}^{2}}}{{{a}^{2}}{{b}^{2}}}+\dfrac{{{a}^{2}}{{y}^{2}}}{{{a}^{2}}{{b}^{2}}}=\dfrac{{{a}^{2

}}{{b}^{2}}}{{{a}^{2}}{{b}^{2}}} \\

& \Rightarrow \dfrac{{{x}^{2}}}{{{a}^{2}}}+\dfrac{{{y}^{2}}}{{{b}^{2}}}=1 \\

\end{align}\]

Then we can differentiate, you will get the same answer.

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