Question

If\[a{{x}^{2}}+2hxy+b{{y}^{2}}=1\], then find the value of \[\dfrac{{{d}^{2}}y}{d{{x}^{2}}}\]

Answer 1

Hint: Here we Directly apply the derivative and apply necessary rules of differentiation which describes the relation x and y.

Complete step-by-step solution -
 Given that \[a{{x}^{2}}+2hxy+b{{y}^{2}}=1\]
This can be re-written as,
\[\Rightarrow a{{x}^{2}}+2hxy+b{{y}^{2}}-1=0\]
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
\[\Rightarrow \dfrac{d}{dx}\left( a{{x}^{2}}+2hxy+b{{y}^{2}}-1 \right)=0\]
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
\[\Rightarrow \dfrac{d}{dx}\left( a{{x}^{2}} \right)+\dfrac{d}{dx}\left( 2hxy \right)+\dfrac{d}{dx}\left( b{{y}^{2}} \right)-\dfrac{d}{dx}\left( 1 \right)=0\]
Taking out the constant terms and we know differentiation of constant term is zero, so the above equation can be written as,
\[\Rightarrow a\dfrac{d}{dx}\left( {{x}^{2}} \right)+2h\dfrac{d}{dx}\left( xy \right)+b\dfrac{d}{dx}\left( {{y}^{2}} \right)-0=0\]
Now we know $\dfrac{d}{dx}({{x}^{n}})=n{{x}^{n-1}}$ and derivative of a constant is always zero, applying this formula, the above equation becomes,
\[\Rightarrow a(2x)+2h\dfrac{d}{dx}\left( xy \right)+b(2y)\dfrac{d}{dx}\left( y \right)-0=0\]
We know the product rule as, \[\dfrac{d}{dx}\left( u\cdot v \right)=u\dfrac{d}{dx}v+v\dfrac{d}{dx}u\], applying this formula in the above equation, we get
\[\begin{align}
  & \Rightarrow a(2x)+2h\left[ x\dfrac{d}{dx}\left( y \right)+y\dfrac{d}{dx}\left( x \right) \right]+b(2y)\dfrac{dy}{dx}=0 \\
 & \Rightarrow 2ax+2hx\dfrac{dy}{dx}+2hy+2by\dfrac{dy}{dx}=0 \\
\end{align}\]
Bring similar terms on one side, we get
\[\Rightarrow \dfrac{dy}{dx}(2hx+2by)=-2ax-2hy\]
\[\Rightarrow \dfrac{dy}{dx}=\dfrac{-(2ax+2hy)}{2hx+2by}\]
Taking out $'2'$ common, we get
\[\Rightarrow \dfrac{dy}{dx}=\dfrac{-(ax+hy)}{hx+by}\]
Now we will find the second order derivative. For that we will again differentiate the above expression with respect to $'x'$ , we get
\[\Rightarrow \dfrac{d}{dx}\left( \dfrac{dy}{dx} \right)=-\dfrac{d}{dx}\left( \dfrac{(ax+hy)}{hx+by} \right)\]
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
\[\Rightarrow \dfrac{d}{dx}\left( \dfrac{dy}{dx} \right)=-\left( \dfrac{(hx+by)\dfrac{d}{dx}(ax+hy)-(ax+hy)\dfrac{d}{dx}(hx+by)}{{{\left( hx+by \right)}^{2}}} \right)\]
Now we will apply the the sum rule of differentiation,$\dfrac{d}{dx}(u+v)=\dfrac{d}{x}(u)+\dfrac{d}{x}(v)$ in the above equation, we get
\[\Rightarrow \dfrac{d}{dx}\left( \dfrac{dy}{dx} \right)=-\left( \dfrac{(hx+by)\left( a(1)+h\dfrac{dy}{dx} \right)-(ax+hy)\left( h+b\dfrac{dy}{dx} \right)}{{{\left( hx+by \right)}^{2}}} \right)\]
\[\Rightarrow \dfrac{{{d}^{2}}y}{d{{x}^{2}}}=-\left( \dfrac{\left( a(hx+by)+h(hx+by)\dfrac{dy}{dx} \right)-\left( h(ax+hy)+b(ax+hy)\dfrac{dy}{dx} \right)}{{{\left( hx+by \right)}^{2}}} \right)\]
\[\Rightarrow \dfrac{{{d}^{2}}y}{d{{x}^{2}}}=-\left( \dfrac{hax+aby+{{h}^{2}}x\dfrac{dy}{dx}+hby\dfrac{dy}{dx}-hax-{{h}^{2}}y-bax\dfrac{dy}{dx}-bhy\dfrac{dy}{dx}}{{{\left( hx+by \right)}^{2}}} \right)\]
Cancelling the like terms, we get
\[\Rightarrow \dfrac{{{d}^{2}}y}{d{{x}^{2}}}=-\left( \dfrac{aby+{{h}^{2}}x\dfrac{dy}{dx}-{{h}^{2}}y-bax\dfrac{dy}{dx}}{{{\left( hx+by \right)}^{2}}} \right)\]
Taking out common terms, we get
\[\begin{align}
  & \Rightarrow \dfrac{{{d}^{2}}y}{d{{x}^{2}}}=-\left( \dfrac{y(ab-{{h}^{2}})+x\dfrac{dy}{dx}({{h}^{2}}-ab)}{{{\left( hx+by \right)}^{2}}} \right) \\
 & \Rightarrow \dfrac{{{d}^{2}}y}{d{{x}^{2}}}=-\left( \dfrac{y(ab-{{h}^{2}})-x\dfrac{dy}{dx}(ab-{{h}^{2}})}{{{\left( hx+by \right)}^{2}}} \right) \\
\end{align}\]
Taking out the common terms, we get
\[\Rightarrow \dfrac{{{d}^{2}}y}{d{{x}^{2}}}=-\left( \dfrac{(ab-{{h}^{2}})\left( y-x\dfrac{dy}{dx} \right)}{{{\left( hx+by \right)}^{2}}} \right)\]
Let us denote $\dfrac{dy}{dx}=y'$ , for simplicity, then the above equation becomes,
\[\therefore \dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{(ab-{{h}^{2}})(y-x{y}')}{{{(by+hx)}^{2}}}\]
Note: Alternative approach is applying the below formula to find out first order differentiate of function in $x,y$
\[\dfrac{dy}{dx}=-\dfrac{\dfrac{\partial \phi (x,y)}{\partial x}}{\dfrac{\partial \phi (x,y)}{\partial y}}\]
This also will lead us to the same result.