Question

If we have an expression as ${{x}^{2}}+6xy+{{y}^{2}}=10$ then show that $\dfrac{{d^{2}y}}{d{{x}^{2}}}=\dfrac{80}{{{\left( 3x+y \right)}^{3}}}$. $$$$

Answer 1

We begin by differentiating the given equation ${{x}^{2}}+6xy+{{y}^{2}}=10$ implicitly with respect to $x$ and find the expression for $\dfrac{dy}{dx}$. We differentiate $\dfrac{dy}{dx}$ with respect to $x$ to find $\dfrac{{{d}^{2}}y}{d{{x}^{2}}}$. We simplify until we get an expression of ${{x}^{2}}+6xy+{{y}^{2}}$ in the numerator where we put the given values to conclude the proof. We use the sum rule$\dfrac{d}{dx}\left( f+g \right)=\dfrac{d}{dx}f+\dfrac{d}{dx}g$, the product rule $\dfrac{d}{dx}\left( fg \right)=g\dfrac{d}{dx}f+f\dfrac{d}{dx}g$, the quotient rule $\dfrac{d}{dx}\left( \dfrac{f}{g} \right)=\dfrac{g\dfrac{d}{dx}f-f\dfrac{d}{dx}g}{{{\left( g \right)}^{2}}}$ and the chain rule $\dfrac{d}{dx}f\left( g\left( x \right) \right)=\left( \dfrac{d}{dx}f\left( g\left( x \right) \right) \right)\times \left( \dfrac{d}{dx}g\left( x \right) \right)$ to find derivatives.

Complete step-by-step solution
We know that an explicit equation can be expressed $y$ in terms of $x$ but for an implicit equation $y$cannot be expressed in terms of $x$. The given equation is an implicit equation in $x$ and $y$which is
${{x}^{2}}+6xy+{{y}^{2}}=10$
We are asked to prove$\dfrac{d{{y}^{2}}}{d{{x}^{2}}}=\dfrac{80}{{{\left( 3x+y \right)}^{3}}}$. We know that when we differentiate implicitly either with respect to $x$ or with respect to $y$; we use the chain rule to treat $x$ or $y$ as a function of $y$ or $x$ respectively. Since we have to find $\dfrac{{{d}^{2}}y}{d{{x}^{2}}}$; let us differentiate the given equation by $x$.We have;
$\Rightarrow \dfrac{d}{dx}\left( {{x}^{2}}+6xy+{{y}^{2}} \right)=0$
We use sum rule of derivative and have;
$\Rightarrow \dfrac{d}{dx}{{x}^{2}}+\dfrac{d}{dx}6xy+\dfrac{d}{dx}{{y}^{2}}=0$
We use product rule of differentiation for $\dfrac{d}{dx}6xy$ and chain rule for $\dfrac{d}{dx}{{y}^{2}}$ to have;

& \Rightarrow \dfrac{d}{dx}{{x}^{2}}+6\left( y\dfrac{d}{dx}x+x\dfrac{d}{dx}y \right)+\dfrac{d}{dx}{{y}^{2}}=0 \\\ & \Rightarrow 2x+6\left( y+x\dfrac{dy}{dx} \right)+2y\dfrac{dy}{dx}=0 \\\ & \Rightarrow 2x+6y+6x\dfrac{dy}{dx}+2y\dfrac{dy}{dx}=0 \\\ & \Rightarrow \dfrac{dy}{dx}=\dfrac{-\left( 2x+6y \right)}{\left( 6x+2y \right)}=-\dfrac{x+3y}{3x+y} \\\ \end{aligned}$$ We differentiate the above once again with respect to $x$ and represent the derivatives as ${{y}^{'}},{{y}^{''}}$. We have $$\Rightarrow \dfrac{d}{dx}\left( \dfrac{dy}{dx} \right)=\dfrac{d}{dx}\dfrac{-\left( x+3y \right)}{3x+y}$$ We use the quotient rule of differentiation in the above step to differentiate the expression at the right hand side to have; $$\begin{aligned} & \Rightarrow \dfrac{d}{dx}\left( \dfrac{dy}{dx} \right)=\dfrac{\left( 3x+y \right)\dfrac{d}{dx}-\left( x+3y \right)-\left( -\left( x+3y \right) \right)\dfrac{d}{dx}\left( 3x+y \right)}{{{\left( 3x+y \right)}^{2}}} \\\ & \Rightarrow \dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{-\left( 3x+y \right)\left( 1+3\dfrac{dy}{dx} \right)+\left( x+3y \right)\left( 3+\dfrac{dy}{dx} \right)}{{{\left( 3x+y \right)}^{2}}} \\\ & \Rightarrow \dfrac{{{d}^{2}}y}{d{{x}^{2}}}=-\dfrac{\left( 3x+y \right)\left( 1+3\dfrac{dy}{dx} \right)-\left( x+3y \right)\left( 3+\dfrac{dy}{dx} \right)}{{{\left( 3x+y \right)}^{2}}} \\\ \end{aligned}$$ We put the obtained expression of $\dfrac{dy}{dx}$ in the above step to have; $$\begin{aligned} & \Rightarrow \dfrac{{{d}^{2}}y}{d{{x}^{2}}}=-\dfrac{\left( 3x+y \right)\left( 1-3\times \dfrac{x+3y}{3x+y} \right)-\left( x+3y \right)\left( 3-\dfrac{x+3y}{3x+y} \right)}{{{\left( 3x+y \right)}^{2}}} \\\ & \Rightarrow \dfrac{{{d}^{2}}y}{d{{x}^{2}}}=-\dfrac{\left( 3x+y \right)\left( \dfrac{3x+y-3x-9y}{3x+y} \right)-\left( x+3y \right)\left( \dfrac{9x+3y-x-3y}{3x+y} \right)}{{{\left( 3x+y \right)}^{2}}} \\\ & \Rightarrow \dfrac{{{d}^{2}}y}{d{{x}^{2}}}=-\dfrac{\left( 3x+y \right)\left( \dfrac{-8y}{3x+y} \right)-\left( x+3y \right)\left( \dfrac{8x}{3x+y} \right)}{{{\left( 3x+y \right)}^{2}}} \\\ \end{aligned}$$ Let us take $\dfrac{-8}{3x+y}$ common in the numerator and have, $$\begin{aligned} & \Rightarrow \dfrac{{{d}^{2}}y}{d{{x}^{2}}}=-\dfrac{\dfrac{-8}{3x+y}\left\\{ \left( 3x+y \right)y+\left( x+3y \right)x \right\\}}{{{\left( 3x+y \right)}^{2}}} \\\ & \Rightarrow \dfrac{{{d}^{2}}y}{d{{x}^{2}}}=8\dfrac{3xy+{{y}^{2}}+{{x}^{2}}+3xy}{{{\left( 3x+y \right)}^{3}}} \\\ & \Rightarrow \dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{8\left( {{x}^{2}}+6xy+{{y}^{2}} \right)}{{{\left( 3x+y \right)}^{3}}} \\\ \end{aligned}$$ We put back the given value ${{x}^{2}}+6xy+{{y}^{2}}=10$ in the above step to have; $$\begin{aligned} & \Rightarrow \dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{8\times 10}{{{\left( 3x+y \right)}^{3}}} \\\ & \therefore \dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{80}{{{\left( 3x+y \right)}^{3}}} \\\ \end{aligned}$$ Hence the given statement is proved. $$$$ **Note:** We note that the statement of proof is a representation of the given equation as a differential equation which has order (highest order derivative) 2 and degree 3. We note the given equation is a quadratic equation of two variable whose general form is given by ${{x}^{2}}+2hxy+b{{y}^{2}}+2gx+2fy+c=0$. Since ${{h}^{2}}-ab>0$and it is a curve the given equation is an equation of a hyperbola.