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

\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{aligned} & {{b}^{2}}x+{{a}^{2}}y\dfrac{dy}{dx}=0 \\\ & \Rightarrow {{a}^{2}}y\dfrac{dy}{dx}=-{{b}^{2}}x \\\ \end{aligned}$$ $$\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{aligned} & \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{aligned}$$ Substitute value from equation (i), we get $$\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{aligned} & \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{aligned}$$ This is the required second order derivative. Note: Another approach is dividing the given expression by ${{a}^{2}}{{b}^{2}}$, you will get equation of ellipse. $$\begin{aligned} & {{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{aligned}$$ Then we can differentiate, you will get the same answer.