Question

If ${{y}^{2}}=a{{x}^{2}}+bx+c$, where a, b and c are constant, then ${{y}^{3}}\dfrac{{{d}^{2}}y}{d{{x}^{2}}}$ is?
(a) A constant
(b) A function of x
(c) A function of y
(d) A function of x and y both

Answer 1

Hint:Differentiate both the sides with respect to x. In the L.H.S use the chain rule of differentiation given as $\dfrac{d\left( {{y}^{2}} \right)}{dx}=\dfrac{d\left( {{y}^{2}} \right)}{dy}\times \dfrac{dy}{dx}$ to evaluate. In the R.H.S use the formula $\dfrac{d\left[ {{\left( x \right)}^{n}} \right]}{d\left[ x \right]}=n{{\left( x \right)}^{n-1}}$ to simplify. Again differentiate the function both the sides with respect to x and use the product rule in the L.H.S given as $\dfrac{d\left( u\times v \right)}{dx}=u\dfrac{dv}{dx}+v\dfrac{du}{dx}$ to simplify. Substitute the value of y and $\dfrac{dy}{dx}$ in the second derivative and simplify the relation to get the correct option.

Complete step-by-step solution:
Here we have been provided with the function ${{y}^{2}}=a{{x}^{2}}+bx+c$ and we are asked to find the value of ${{y}^{3}}\dfrac{{{d}^{2}}y}{d{{x}^{2}}}$ and choose the correct option regarding the obtained relation.
On differentiating both the sides with respect to x we get,
\[\begin{align}
  & \Rightarrow \dfrac{d\left( {{y}^{2}} \right)}{dx}=\dfrac{d\left( a{{x}^{2}}+bx+c \right)}{dx} \\
 & \Rightarrow \dfrac{d\left( {{y}^{2}} \right)}{dx}=\dfrac{d\left( a{{x}^{2}} \right)}{dx}+\dfrac{d\left( bx \right)}{dx}+\dfrac{d\left( c \right)}{dx} \\
\end{align}\]
In the L.H.S applying the chain rule of derivative given as $\dfrac{d\left( {{y}^{2}} \right)}{dx}=\dfrac{d\left( {{y}^{2}} \right)}{dy}\times \dfrac{dy}{dx}$ and in the R.H.S using the fact that the derivative of the constant function is 0, using the formula $\dfrac{d\left[ {{\left( x \right)}^{n}} \right]}{d\left[ x \right]}=n{{\left( x \right)}^{n-1}}$ we get,
\[\Rightarrow 2y\dfrac{dy}{dx}=2ax+b\] ……. (1)
Again differentiating the function with respect to x both the sides we get,
\[\begin{align}
  & \Rightarrow \dfrac{d\left( 2y\dfrac{dy}{dx} \right)}{dx}=\dfrac{d\left( 2ax+b \right)}{dx} \\
 & \Rightarrow 2\dfrac{d\left( y\dfrac{dy}{dx} \right)}{dx}=\dfrac{d\left( 2ax \right)}{dx}+\dfrac{d\left( b \right)}{dx} \\
 & \Rightarrow 2\dfrac{d\left( y\dfrac{dy}{dx} \right)}{dx}=2a \\
 & \Rightarrow \dfrac{d\left( y\dfrac{dy}{dx} \right)}{dx}=a \\
\end{align}\]
In the L.H.S using the product rule of the derivative given as $\dfrac{d\left( u\times v \right)}{dx}=u\dfrac{dv}{dx}+v\dfrac{du}{dx}$ we get,
\[\begin{align}
  & \Rightarrow y\left[ \dfrac{d}{dx}\left( \dfrac{dy}{dx} \right) \right]+\dfrac{dy}{dx}\left[ \dfrac{dy}{dx} \right]=a \\
 & \Rightarrow y\dfrac{{{d}^{2}}y}{d{{x}^{2}}}+{{\left( \dfrac{dy}{dx} \right)}^{2}}=a \\
\end{align}\]
Substituting the value of $\dfrac{dy}{dx}$ from equation (1) we get,
\[\begin{align}
  & \Rightarrow y\dfrac{{{d}^{2}}y}{d{{x}^{2}}}+\dfrac{{{\left( 2ax+b \right)}^{2}}}{4{{y}^{2}}}=a \\
 & \Rightarrow 4{{y}^{3}}\dfrac{{{d}^{2}}y}{d{{x}^{2}}}+{{\left( 2ax+b \right)}^{2}}=4{{y}^{2}}a \\
\end{align}\]
In the R.H.S substituting the value of ${{y}^{2}}$ from the given function we get,
\[\begin{align}
  & \Rightarrow 4{{y}^{3}}\dfrac{{{d}^{2}}y}{d{{x}^{2}}}+{{\left( 2ax+b \right)}^{2}}=4\left( a{{x}^{2}}+bx+c \right)a \\
 & \Rightarrow 4{{y}^{3}}\dfrac{{{d}^{2}}y}{d{{x}^{2}}}=4{{a}^{2}}{{x}^{2}}+4abx+4ac-{{\left( 2ax+b \right)}^{2}} \\
\end{align}\]
Using the algebraic identity ${{\left( m+n \right)}^{2}}={{m}^{2}}+{{n}^{2}}+2mn$ we get,
\[\begin{align}
  & \Rightarrow 4{{y}^{3}}\dfrac{{{d}^{2}}y}{d{{x}^{2}}}=4{{a}^{2}}{{x}^{2}}+4abx+4ac-4{{a}^{2}}{{x}^{2}}-4abx-{{b}^{2}} \\
 & \Rightarrow 4{{y}^{3}}\dfrac{{{d}^{2}}y}{d{{x}^{2}}}=4ac-{{b}^{2}} \\
 & \therefore {{y}^{3}}\dfrac{{{d}^{2}}y}{d{{x}^{2}}}=ac-\dfrac{{{b}^{2}}}{4} \\
\end{align}\]
Clearly we can see that the expression of ${{y}^{3}}\dfrac{{{d}^{2}}y}{d{{x}^{2}}}$ is independent of both x and y and is a constant because a, b and c are constants provided in the question.
Hence, option (a) is the correct answer.

Note:You must remember all the basic rules and formulas of differentiation like: - the Product rule, Chain rule, \[\dfrac{u}{v}\] rule etc. Substitute the value of ${{y}^{2}}$ only after multiplying with both the sides because in the L.H.S we have to keep the variable y and only simplify the R.H.S. Conclude the result only after getting the simplest form otherwise you may choose the wrong option.