Question

If the value of ${{x}^{2}}+{{y}^{2}}={{r}^{2}}$, find the value of ${{\left( \dfrac{{{d}^{3}}y}{d{{x}^{3}}} \right)}_{x=0}}$.

Answer 1

Hint: Differentiate the given function thrice to find the third derivative of the given function. This is an implicit function which is written in terms of both dependent and independent variables. Substitute the value $x=0$ in the derivatives of the given function to find the value of ${{\left( \dfrac{{{d}^{3}}y}{d{{x}^{3}}} \right)}_{x=0}}$.

Complete step-by-step answer:
We have the function ${{x}^{2}}+{{y}^{2}}={{r}^{2}}$. We have to evaluate its third derivative.
We will begin by differentiating the given function with respect to the variable $x$.
Thus, we have $\dfrac{d}{dx}\left( {{x}^{2}}+{{y}^{2}} \right)=\dfrac{d}{dx}\left( {{r}^{2}} \right)$.
We will use the sum rule of differentiation. Thus, we have $\dfrac{d}{dx}\left( {{x}^{2}}+{{y}^{2}} \right)=\dfrac{d}{dx}\left( {{x}^{2}} \right)+\dfrac{d}{dx}\left( {{y}^{2}} \right)=\dfrac{d}{dx}\left( {{r}^{2}} \right).....\left( 1 \right)$.
We know that differentiation of a constant is zero with respect to any variable. Thus, we have $\dfrac{d}{dx}\left( {{r}^{2}} \right)=0.....\left( 2 \right)$.
We know that differentiation of any function of the form $y=a{{x}^{n}}$ is $\dfrac{dy}{dx}=an{{x}^{n-1}}$.
Substituting $a=1,n=2$ in the above equation, we have $\dfrac{d}{dx}\left( {{x}^{2}} \right)=2x.....\left( 3 \right)$.
To find the value of $\dfrac{d}{dx}\left( {{y}^{2}} \right)$, we will multiply and divide the given equation by $dy$.
Thus, we have $\dfrac{d}{dx}\left( {{y}^{2}} \right)=\dfrac{d}{dy}\left( {{y}^{2}} \right)\times \dfrac{dy}{dx}$.
We know that $\dfrac{d}{dy}\left( {{y}^{2}} \right)=2y$.
Thus, we have $\dfrac{d}{dx}\left( {{y}^{2}} \right)=\dfrac{d}{dy}\left( {{y}^{2}} \right)\times \dfrac{dy}{dx}=2y\dfrac{dy}{dx}.....\left( 4 \right)$.
Substituting the value of equation $\left( 2 \right),\left( 3 \right),\left( 4 \right)$ in equation $\left( 1 \right)$, we get $\dfrac{d}{dx}\left( {{x}^{2}}+{{y}^{2}} \right)=\dfrac{d}{dx}\left( {{x}^{2}} \right)+\dfrac{d}{dx}\left( {{y}^{2}} \right)=\dfrac{d}{dx}\left( {{r}^{2}} \right)\Rightarrow 2x+2y\dfrac{dy}{dx}=0$.
Thus, we have $2x+2y\dfrac{dy}{dx}=0$.
We can rewrite this equation as $\dfrac{dy}{dx}=\dfrac{-x}{y}.....\left( 5 \right)$.
We will now find the second derivative of the given function. So, we will differentiate the above function again.
We know that $\dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{d}{dx}\left( \dfrac{dy}{dx} \right)$.
Thus, we have $\dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{d}{dx}\left( \dfrac{dy}{dx} \right)=\dfrac{d}{dx}\left( \dfrac{-x}{y} \right)$.
We will use quotient rule for differentiation of two functions of the form $z=\dfrac{f\left( y \right)}{g\left( y \right)}$ such that $\dfrac{dz}{dy}=\dfrac{g\left( y \right)f'\left( y \right)-f\left( y \right)g'\left( y \right)}{{{g}^{2}}\left( y \right)}$.
Thus, we have $\dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{d}{dx}\left( \dfrac{dy}{dx} \right)=\dfrac{d}{dx}\left( \dfrac{-x}{y} \right)=\dfrac{y\dfrac{d}{dx}\left( -x \right)-\left( -x \right)\dfrac{d}{dx}\left( y \right)}{{{y}^{2}}}.....\left( 6 \right)$.
We know that differentiation of any function of the form $y=a{{x}^{n}}$ is $\dfrac{dy}{dx}=an{{x}^{n-1}}$.
Thus, we have $\dfrac{d}{dx}\left( -x \right)=-1.....\left( 7 \right)$.
Substituting equation $\left( 5 \right),\left( 7 \right)$ in equation $\left( 6 \right)$, we have $\dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{y\dfrac{d}{dx}\left( -x \right)-\left( -x \right)\dfrac{d}{dx}\left( y \right)}{{{y}^{2}}}=\dfrac{y\left( -1 \right)+x\left( \dfrac{-x}{y} \right)}{{{y}^{2}}}=-\dfrac{1}{y}+\dfrac{x}{{{y}^{2}}}\left( -\dfrac{x}{y} \right)=-\dfrac{1}{y}-\dfrac{{{x}^{2}}}{{{y}^{3}}}$.
Hence, we have $\dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{-1}{y}-\dfrac{{{x}^{2}}}{{{y}^{3}}}$.
We will differentiate the above equation again to find the third derivative of the given function.
Thus, we have $\dfrac{{{d}^{3}}y}{d{{x}^{3}}}=\dfrac{d}{dx}\left( \dfrac{{{d}^{2}}y}{d{{x}^{2}}} \right)=\dfrac{d}{dx}\left( \dfrac{-1}{y}-\dfrac{{{x}^{2}}}{{{y}^{3}}} \right)$.
We can rewrite this as $\dfrac{{{d}^{3}}y}{d{{x}^{3}}}=\dfrac{d}{dx}\left( \dfrac{-1}{y}-\dfrac{{{x}^{2}}}{{{y}^{3}}} \right)=\dfrac{d}{dx}\left( \dfrac{-1}{y} \right)+\dfrac{d}{dx}\left( \dfrac{-{{x}^{2}}}{{{y}^{3}}} \right).....\left( 8 \right)$.
To find the value of $\dfrac{d}{dx}\left( \dfrac{-1}{y} \right)$, we will multiply and divide by $dy$.
Thus, we have $\dfrac{d}{dx}\left( \dfrac{-1}{y} \right)=\dfrac{d}{dy}\left( \dfrac{-1}{y} \right)\times \dfrac{dy}{dx}$.
We know that $\dfrac{d}{dy}\left( \dfrac{-1}{y} \right)=\dfrac{1}{{{y}^{2}}}$ as if $y=a{{x}^{n}}$ then $\dfrac{dy}{dx}=an{{x}^{n-1}}$.
Thus, we have $\dfrac{d}{dx}\left( \dfrac{-1}{y} \right)=\dfrac{d}{dy}\left( \dfrac{-1}{y} \right)\times \dfrac{dy}{dx}=\dfrac{1}{{{y}^{2}}}\left( \dfrac{-x}{y} \right)=\dfrac{-x}{{{y}^{3}}}.....\left( 9 \right)$.
To find the value of $\dfrac{d}{dx}\left( \dfrac{-{{x}^{2}}}{{{y}^{3}}} \right)$, we will use quotient rule.
Thus, we have $\dfrac{d}{dx}\left( \dfrac{-{{x}^{2}}}{{{y}^{3}}} \right)=\dfrac{{{y}^{3}}\times \dfrac{d}{dx}\left( -{{x}^{2}} \right)-\left( -{{x}^{2}} \right)\dfrac{d}{dx}\left( {{y}^{3}} \right)}{{{y}^{6}}}.....\left( 10 \right)$.
We know that $\dfrac{d}{dx}\left( -{{x}^{2}} \right)=-2x.....\left( 11 \right)$ as if $y=a{{x}^{n}}$ then $\dfrac{dy}{dx}=an{{x}^{n-1}}$.
To find the value of $\dfrac{d}{dx}\left( {{y}^{3}} \right)$, multiply and divide the equation by $dy$.
Thus, we have $\dfrac{d}{dx}\left( {{y}^{3}} \right)=\dfrac{d}{dy}\left( {{y}^{3}} \right)\dfrac{dy}{dx}$
We know that $\dfrac{d}{dy}\left( {{y}^{3}} \right)=3{{y}^{2}}$.
Thus, we get $\dfrac{d}{dx}\left( {{y}^{3}} \right)=\dfrac{d}{dy}\left( {{y}^{3}} \right)\dfrac{dy}{dx}=3{{y}^{2}}\left( \dfrac{-x}{y} \right)=-3xy.....\left( 12 \right)$
Substituting equation $\left( 11 \right),\left( 12 \right)$ in equation $\left( 10 \right)$, we have $\dfrac{d}{dx}\left( \dfrac{-{{x}^{2}}}{{{y}^{3}}} \right)=\dfrac{{{y}^{3}}\times \dfrac{d}{dx}\left( -{{x}^{2}} \right)-\left( -{{x}^{2}} \right)\dfrac{d}{dx}\left( {{y}^{3}} \right)}{{{y}^{6}}}=\dfrac{-2x{{y}^{3}}-3{{x}^{3}}y}{{{y}^{6}}}.....\left( 13 \right)$
Substituting equation $\left( 9 \right),\left( 13 \right)$ in equation $\left( 8 \right)$, we have $\dfrac{{{d}^{3}}y}{d{{x}^{3}}}=\dfrac{d}{dx}\left( \dfrac{-1}{y} \right)+\dfrac{d}{dx}\left( \dfrac{-{{x}^{2}}}{{{y}^{3}}} \right)=\dfrac{-x}{{{y}^{3}}}-\dfrac{2x{{y}^{3}}+3{{x}^{3}}y}{{{y}^{6}}}$.
Thus, we have $\dfrac{{{d}^{3}}y}{d{{x}^{3}}}=\dfrac{-x}{{{y}^{3}}}-\dfrac{2x{{y}^{3}}+3{{x}^{3}}y}{{{y}^{6}}}$.
We will now evaluate the value of $\dfrac{{{d}^{3}}y}{d{{x}^{3}}}$ at $x=0$.
Firstly, we will substitute $x=0$ in the equation ${{x}^{2}}+{{y}^{2}}={{r}^{2}}$. Thus, we have ${{y}^{2}}={{r}^{2}}\Rightarrow y=\pm r$.
We will now substitute this value in the equation $\dfrac{{{d}^{3}}y}{d{{x}^{3}}}=\dfrac{-x}{{{y}^{3}}}-\dfrac{2x{{y}^{3}}+3{{x}^{3}}y}{{{y}^{6}}}$.
Thus, we have ${{\left( \dfrac{{{d}^{3}}y}{d{{x}^{3}}} \right)}_{x=0}}=\dfrac{0}{{{\left( \pm r \right)}^{3}}}-\dfrac{2\left( 0 \right){{\left( \pm r \right)}^{3}}+3{{\left( 0 \right)}^{3}}\left( \pm r \right)}{{{\left( \pm r \right)}^{6}}}=0+0=0$.
Hence the value of $\dfrac{{{d}^{3}}y}{d{{x}^{3}}}$ at $x=0$ is $0$.

Note: One needs to use quotient rule and product rule to find the third derivative of the given function. To differentiate an implicit function, we differentiate both sides of an equation by treating one of the variables as a function of the other one.