Question

If ${{x}^{3}}+{{y}^{3}}-3axy=0$, then prove that \[\dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{2{{a}^{2}}xy}{{{\left( ax-{{y}^{2}} \right)}^{3}}}\].

Answer 1

Hint: We have an implicit function $\left\{ \text{function in the form }\phi \left( x,y \right)=0 \right\}$. To find $\dfrac{dy}{dx}$ here in the case of implicit function, we differentiate each term with the respect to x regarding y as a function of x and then collect terms in $\dfrac{dy}{dx}$ together on one side to finally get $\dfrac{dy}{dx}$.
After getting $\dfrac{dy}{dx}$, we will differentiate the equation with respect to x, regarding y and $\dfrac{dy}{dx}$ as a function of x and then collect terms in \[\dfrac{{{d}^{2}}y}{d{{x}^{2}}}\] together on one side to finally get \[\dfrac{{{d}^{2}}y}{d{{x}^{2}}}\].

Complete step-by-step answer:
We have given,

${{x}^{3}}+{{y}^{3}}-3axy=0$

We will differentiate the above expression with respect to x.

On differentiating with respect to x, we get,

\[\begin{align}

  & \dfrac{d}{dx}\left\{ {{x}^{3}}+{{y}^{3}}-3axy \right\}=\dfrac{d\left( 0 \right)}{dx} \\

 & \Rightarrow \dfrac{d\left( {{x}^{3}} \right)}{dx}+\dfrac{d\left( {{y}^{3}} \right)}{dx}-\dfrac{d\left( 3axy \right)}{dx}=0 \\

 & \left( \text{since, derivative of constant term is always zero} \right) \\

 & \Rightarrow 3{{x}^{2}}+3{{y}^{2}}.\dfrac{dy}{dx}-3a\dfrac{d\left( xy \right)}{dx}=0 \\

 & \left( \text{since, }\dfrac{d\left( {{x}^{n}} \right)}{dx}=n{{x}^{n-1}} \right) \\

\end{align}\]

$\Rightarrow $and to find $\dfrac{d\left( {{y}^{3}} \right)}{dx}$, we are applying chain rule of differentiation here.

$\begin{align}

  & \Rightarrow 3{{x}^{2}}+3{{y}^{2}}.\dfrac{dy}{dx}-3a\left[ \dfrac{d\left( x \right)}{dx}.y+x.\dfrac{d\left( y \right)}{dx} \right]=0 \\

 & \left( \text{Applying product rule of differentiation i}\text{.e}\text{. }\dfrac{d\left( AB \right)}{dx}=\dfrac{dA}{dx}.B+\dfrac{dB}{dx}.A \right) \\

 & \Rightarrow 3{{x}^{2}}+3{{y}^{2}}.\dfrac{dy}{dx}-3a\left[ 1\times y+x.\dfrac{dy}{dx} \right]=0 \\

 & \Rightarrow 3{{x}^{2}}+3{{y}^{2}}.\dfrac{dy}{dx}-3ay-3ax.\dfrac{dy}{dx}=0 \\

\end{align}$

(On expanding the terms of bracket)

Now, collecting terms in $\dfrac{dy}{dx}$ together on one side, we get,

$\begin{align}

  & \Rightarrow 3{{x}^{2}}-3ay+3{{y}^{2}}\dfrac{dy}{dx}-3ax\dfrac{dy}{dx}=0 \\

 & \Rightarrow 3{{x}^{2}}-3ay+\dfrac{dy}{dx}\left( 3{{y}^{2}}-3ax \right)=0 \\

 & \Rightarrow \dfrac{dy}{dx}=\dfrac{\left( 3{{x}^{2}}-3ay \right)}{\left( 3{{y}^{2}}-3ax \right)} \\
 & \Rightarrow \dfrac{dy}{dx}=\dfrac{3ay-3{{x}^{2}}}{3{{y}^{2}}-3ax}=\dfrac{ay-{{x}^{2}}}{{{y}^{2}}-ax}.............\left( 1 \right) \\

\end{align}$

(Taking 3 common from numerator and denominator and canceling it.)

Again we differentiate equation (1), with respect to x.

On differentiating equation (1) both sides,

\[\dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{d}{dx}\left\{ \dfrac{ay-{{x}^{2}}}{{{y}^{2}}-ax} \right\}\]
$\Rightarrow $To differentiate right hand side, we will apply the quotient rule of differentiation,

i.e. $\dfrac{d}{dx}\left( \dfrac{f}{g} \right)=\dfrac{g\left( \dfrac{df}{dx} \right)-f\left( \dfrac{dg}{dx} \right)}{{{g}^{2}}}$

Where $g\ne 0$

Therefore,

\[\begin{align}

  & \dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{\left( {{y}^{2}}-ax \right)\dfrac{d}{dx}\left( ay-{{x}^{2}} \right)-\left( ay-{{x}^{2}} \right)\dfrac{d}{dx}\left\{ {{y}^{2}}-ax \right\}}{{{\left( {{y}^{2}}-ax \right)}^{2}}} \\

 & \Rightarrow \dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{\left( {{y}^{2}}-ax \right)\left\{ a.\left( \dfrac{dy}{dx} \right).\dfrac{d\left( {{x}^{2}} \right)}{dx} \right\}-\left( ay-{{x}^{2}} \right)\left\{ \dfrac{d\left( {{y}^{2}} \right)}{dx}-a\dfrac{d\left( x \right)}{dx} \right\}}{{{\left( {{y}^{2}}-ax \right)}^{2}}} \\

 & \Rightarrow \dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{\left( {{y}^{2}}-ax \right)\left\{ \dfrac{ady}{dx}-2x \right\}-\left( ay-{{x}^{2}} \right)\left\{ 2y.\dfrac{dy}{dx}-a \right\}}{{{\left( {{y}^{2}}-ax \right)}^{2}}}..........\left( 2 \right) \\

\end{align}\]

Taking a (constant) outside from differentiation and using $\dfrac{d\left( {{x}^{n}} \right)}{dx}=n{{x}^{n-1}}$.

To differentiate terms containing y, we will differentiate the terms using chain rule of differentiation.

From equation (1), we have,

$\dfrac{dy}{dx}=\dfrac{ay-{{x}^{2}}}{{{y}^{2}}-ax}$

Putting this value of $\dfrac{dy}{dx}$ in equation (2), we will get,

\[\begin{align}

  & \dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{\left( {{y}^{2}}-ax \right)\left\{ a\left( \dfrac{ay-{{x}^{2}}}{{{y}^{2}}-ax} \right)-2x \right\}-\left( ay-{{x}^{2}} \right)\left\{ 2y\left( \dfrac{ay-{{x}^{2}}}{{{y}^{2}}-ax} \right)-a \right\}}{{{\left( {{y}^{2}}-ax \right)}^{2}}} \\

 & \dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{\left( {{y}^{2}}-ax \right)\left\{ \dfrac{{{a}^{2}}y-a{{x}^{2}}-2x\left( {{y}^{2}}-ax \right)}{\left( {{y}^{2}}-ax \right)} \right\}-\left( ay-{{x}^{2}} \right)\left\{ \dfrac{2a{{y}^{2}}-2y{{x}^{2}}-a\left( {{y}^{2}}-ax \right)}{\left( {{y}^{2}}-ax \right)} \right\}}{{{\left( {{y}^{2}}-ax \right)}^{2}}} \\

 & \dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{\left( {{a}^{2}}y-a{{x}^{2}}-2x{{y}^{2}}+2a{{x}^{2}} \right)-\left( ay-{{x}^{2}} \right)\left( \dfrac{2a{{y}^{2}}-2y{{x}^{2}}-a{{y}^{2}}+{{a}^{2}}x}{{{y}^{2}}-ax} \right)}{{{\left( {{y}^{2}}-ax \right)}^{2}}} \\

 & \Rightarrow \dfrac{\left( {{a}^{2}}y+a{{x}^{2}}-2x{{y}^{2}} \right)\left( {{y}^{2}}-ax \right)-\left( ay-{{x}^{2}} \right)\left( a{{y}^{2}}-2{{x}^{2}}y+{{a}^{2}}x \right)}{{{\left( {{y}^{2}}-ax \right)}^{3}}} \\

 & \Rightarrow \dfrac{{{a}^{2}}{{y}^{3}}-{{a}^{3}}xy+a{{x}^{2}}{{y}^{2}}-{{a}^{2}}{{x}^{3}}-2x{{y}^{4}}+2a{{x}^{2}}{{y}^{2}}-\left( a{{y}^{2}}-2{{x}^{2}}y+{{a}^{2}}x \right)\left( ay-{{x}^{2}} \right)}{{{\left( {{y}^{2}}-ax \right)}^{3}}} \\

 & \Rightarrow \dfrac{{{a}^{2}}{{y}^{3}}-{{a}^{3}}yx+3a{{x}^{2}}{{y}^{2}}-{{a}^{2}}{{x}^{3}}-2x{{y}^{4}}-{{a}^{2}}{{y}^{3}}+a{{x}^{2}}{{y}^{2}}+2a{{x}^{2}}{{y}^{2}}-2{{x}^{4}}y-{{a}^{3}}xy+{{a}^{2}}{{x}^{3}}}{{{\left( {{y}^{2}}-ax \right)}^{3}}} \\

\end{align}\]

$\Rightarrow $Cancelling the terms having equal positive and negative values.

\[\begin{align}

  & \Rightarrow \dfrac{6a{{x}^{2}}{{y}^{2}}-2x{{y}^{4}}-2{{x}^{4}}y-2{{a}^{3}}xy}{{{\left( {{y}^{2}}-ax \right)}^{3}}} \\

 & \Rightarrow \dfrac{-2xy\left( {{x}^{3}}+{{y}^{3}}+{{a}^{3}}-3axy \right)}{{{\left( {{y}^{2}}-ax \right)}^{3}}} \\

 & \Rightarrow \dfrac{-2xy\left\{ \left( {{x}^{3}}+{{y}^{3}}-3axy \right)+{{a}^{3}} \right\}}{{{\left( {{y}^{2}}-ax \right)}^{3}}} \\

 & \Rightarrow \dfrac{-2xy\left( 0+{{a}^{3}} \right)}{{{\left( {{y}^{2}}-ax \right)}^{3}}}\text{, since it is given that }{{x}^{3}}+{{y}^{3}}-3axy=0 \\

 & \Rightarrow \dfrac{-2xy{{a}^{3}}}{{{\left( {{y}^{2}}-ax \right)}^{3}}}=\dfrac{2xy{{a}^{3}}}{{{\left( ax-{{y}^{2}} \right)}^{3}}}=\text{Right Hand Side} \\
\end{align}\]


Note: In this question, we have given an implicit function $\left\{ \text{function in the form }\phi \left( x,y \right)=0 \right\}$. Here students can make mistakes by starting with solving for y and then finding $\dfrac{dy}{dx}$.
But, the correct method to find $\dfrac{dy}{dx}$ in this question is by differentiating each term with respect to x, regarding y as a function of x and the collecting terms in $\dfrac{dy}{dx}$ together on one side.