Question

Find the second derivative i.e. \[\dfrac{{{d}^{2}}y}{d{{x}^{2}}}\] of \[{{b}^{2}}{{x}^{2}}+{{a}^{2}}{{y}^{2}}={{a}^{2}}{{b}^{2}}\]

Answer 1

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.