# If \[xy=a{{x}^{2}}+\left( \dfrac{b}{x} \right)\], then find the value of \[x\dfrac{{{d}^{2}}y}{d{{x}^{2}}}+2\dfrac{dy}{dx}=\]

(A) \[\dfrac{y}{x}\] (B) \[\dfrac{-y}{x}\]

(C) \[\dfrac{2y}{x}\] (D) \[\dfrac{-2y}{x}\]

Answer

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Hint: Carefully examine the given equation and try to convert it such that LHS just has $y$ term. In this way it will be easy for us to find the first and second derivative and substitute it in the final equation to calculate.

The given expression is \[xy=a{{x}^{2}}+\left( \dfrac{b}{x} \right)\]

We need to find, \[x\dfrac{{{d}^{2}}y}{d{{x}^{2}}}+2\dfrac{dy}{dx}.\]

For this problem first let us find \[\dfrac{dy}{dx}\]and \[\dfrac{{{d}^{2}}y}{d{{x}^{2}}}\].

So, consider the given expression,\[xy=a{{x}^{2}}+\left( \dfrac{b}{x} \right)\]

Dividing with $'x'$ on both sides, we get

\[\dfrac{xy}{x}=\dfrac{a{{x}^{2}}}{x}+\dfrac{\dfrac{b}{x}}{x}\]

Cancelling the like terms, we get

\[\Rightarrow y=ax+\dfrac{b}{{{x}^{2}}}=ax+b{{x}^{-2}}\]

Now, by differentiating both sides with respect to\[x\], we get

\[\dfrac{dy}{dx}=\dfrac{d}{dx}(ax)+\dfrac{d}{dx}(b{{x}^{-2}})\]

Taking out the constant terms, we get

\[\dfrac{dy}{dx}=a\dfrac{d}{dx}(x)+b\dfrac{d}{dx}({{x}^{-2}})\]

Now we know, \[\dfrac{d\left( {{x}^{n}} \right)}{dx}=n\left( {{x}^{n-1}} \right)\] , so the above equation becomes

\[\begin{align}

& \dfrac{dy}{dx}=a(1)+b\left( -2{{x}^{-2-1}} \right) \\

& \dfrac{dy}{dx}=a+b(-2{{x}^{-3}}) \\

\end{align}\]

\[\dfrac{dy}{dx}=a-2b{{x}^{-3}}............(i)\]

Now we find the second derivative.

Let us differentiate both sides of equation (i) with respect to $'x'$, we get

\[\dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{d}{dx}\left( a-2b{{x}^{-3}} \right)\]

This can be written as,

\[\dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{d}{dx}\left( a \right)-\dfrac{d}{dx}\left( 2b{{x}^{-3}} \right)\]

We know differentiation of constant term is zero and taking out the constant terms, we get\[\dfrac{{{d}^{2}}y}{d{{x}^{2}}}=0-2b\dfrac{d}{dx}\left( {{x}^{-3}} \right)\]

Now we know, \[\dfrac{d\left( {{x}^{n}} \right)}{dx}=n\left( {{x}^{n-1}} \right)\] , so the above equation becomes

\[\dfrac{{{d}^{2}}y}{d{{x}^{2}}}=-2b\left( -3{{x}^{-3-1}} \right)\]

\[\dfrac{{{d}^{2}}y}{d{{x}^{2}}}=6b{{x}^{-4}}.............(ii)\]

Now consider the other equation,

\[x\dfrac{{{d}^{2}}y}{d{{x}^{2}}}+2\dfrac{dy}{dx}=\]

Now by substituting values from equation (i) and (ii), we get

\[x\dfrac{{{d}^{2}}y}{d{{x}^{2}}}+2\dfrac{dy}{dx}=x\left( 6b{{x}^{-4}} \right)+2\left( a-2b{{x}^{-3}} \right)\]

\[\begin{align}

& x\dfrac{{{d}^{2}}y}{d{{x}^{2}}}+2\dfrac{dy}{dx}=6b{{x}^{-3}}+2a-4b{{x}^{-3}} \\

& x\dfrac{{{d}^{2}}y}{d{{x}^{2}}}+2\dfrac{dy}{dx}=2a+2b{{x}^{-3}} \\

& x\dfrac{{{d}^{2}}y}{d{{x}^{2}}}+2\dfrac{dy}{dx}=2\left( a+\dfrac{b}{{{x}^{3}}} \right) \\

\end{align}\]

Now by multiplying and dividing with \[\left( {{x}^{2}} \right)\] on RHS we get,

\[\begin{align}

& x\dfrac{{{d}^{2}}y}{d{{x}^{2}}}+2\dfrac{dy}{dx}=2\left[ a+\dfrac{b}{{{x}^{3}}} \right]\times \dfrac{{{x}^{2}}}{{{x}^{2}}} \\

& x\dfrac{{{d}^{2}}y}{d{{x}^{2}}}+2\dfrac{dy}{dx}=2\left[ ax{}^{2}+\dfrac{b{{x}^{2}}}{{{x}^{3}}} \right]\times \dfrac{1}{{{x}^{2}}} \\

& x\dfrac{{{d}^{2}}y}{d{{x}^{2}}}+2\dfrac{dy}{dx}=2\left[ a{{x}^{2}}+\dfrac{b}{x} \right]\dfrac{1}{{{x}^{2}}} \\

\end{align}\]

But given, \[xy=a{{x}^{2}}+\left( \dfrac{b}{x} \right)\], so the above equation becomes,

\[x\dfrac{{{d}^{2}}y}{d{{x}^{2}}}+2\dfrac{dy}{dx}=2\left( xy \right)\dfrac{1}{{{x}^{2}}}\]

Cancelling the like terms, we get

\[x\dfrac{{{d}^{2}}y}{d{{x}^{2}}}+2\dfrac{dy}{dx}=2\dfrac{y}{x}\]

Hence the correct answer is option (C).

Note: Another way to solve this problem is directly differentiating the given expression with respect to $x$ instead of dividing the expression by $x$ .

\[\dfrac{d(xy)}{dx}=\dfrac{d}{dx}\left( a{{x}^{2}}+\left( \dfrac{b}{x} \right) \right)\]

Now we will apply product rule, so we get

\[x\dfrac{d(y)}{dx}+y\dfrac{d(x)}{dy}=\dfrac{d}{dx}\left( a{{x}^{2}} \right)+\dfrac{d}{dx}\left( \dfrac{b}{x} \right)\]

This gets a little complicated. But you will get the same answer.

The given expression is \[xy=a{{x}^{2}}+\left( \dfrac{b}{x} \right)\]

We need to find, \[x\dfrac{{{d}^{2}}y}{d{{x}^{2}}}+2\dfrac{dy}{dx}.\]

For this problem first let us find \[\dfrac{dy}{dx}\]and \[\dfrac{{{d}^{2}}y}{d{{x}^{2}}}\].

So, consider the given expression,\[xy=a{{x}^{2}}+\left( \dfrac{b}{x} \right)\]

Dividing with $'x'$ on both sides, we get

\[\dfrac{xy}{x}=\dfrac{a{{x}^{2}}}{x}+\dfrac{\dfrac{b}{x}}{x}\]

Cancelling the like terms, we get

\[\Rightarrow y=ax+\dfrac{b}{{{x}^{2}}}=ax+b{{x}^{-2}}\]

Now, by differentiating both sides with respect to\[x\], we get

\[\dfrac{dy}{dx}=\dfrac{d}{dx}(ax)+\dfrac{d}{dx}(b{{x}^{-2}})\]

Taking out the constant terms, we get

\[\dfrac{dy}{dx}=a\dfrac{d}{dx}(x)+b\dfrac{d}{dx}({{x}^{-2}})\]

Now we know, \[\dfrac{d\left( {{x}^{n}} \right)}{dx}=n\left( {{x}^{n-1}} \right)\] , so the above equation becomes

\[\begin{align}

& \dfrac{dy}{dx}=a(1)+b\left( -2{{x}^{-2-1}} \right) \\

& \dfrac{dy}{dx}=a+b(-2{{x}^{-3}}) \\

\end{align}\]

\[\dfrac{dy}{dx}=a-2b{{x}^{-3}}............(i)\]

Now we find the second derivative.

Let us differentiate both sides of equation (i) with respect to $'x'$, we get

\[\dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{d}{dx}\left( a-2b{{x}^{-3}} \right)\]

This can be written as,

\[\dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{d}{dx}\left( a \right)-\dfrac{d}{dx}\left( 2b{{x}^{-3}} \right)\]

We know differentiation of constant term is zero and taking out the constant terms, we get\[\dfrac{{{d}^{2}}y}{d{{x}^{2}}}=0-2b\dfrac{d}{dx}\left( {{x}^{-3}} \right)\]

Now we know, \[\dfrac{d\left( {{x}^{n}} \right)}{dx}=n\left( {{x}^{n-1}} \right)\] , so the above equation becomes

\[\dfrac{{{d}^{2}}y}{d{{x}^{2}}}=-2b\left( -3{{x}^{-3-1}} \right)\]

\[\dfrac{{{d}^{2}}y}{d{{x}^{2}}}=6b{{x}^{-4}}.............(ii)\]

Now consider the other equation,

\[x\dfrac{{{d}^{2}}y}{d{{x}^{2}}}+2\dfrac{dy}{dx}=\]

Now by substituting values from equation (i) and (ii), we get

\[x\dfrac{{{d}^{2}}y}{d{{x}^{2}}}+2\dfrac{dy}{dx}=x\left( 6b{{x}^{-4}} \right)+2\left( a-2b{{x}^{-3}} \right)\]

\[\begin{align}

& x\dfrac{{{d}^{2}}y}{d{{x}^{2}}}+2\dfrac{dy}{dx}=6b{{x}^{-3}}+2a-4b{{x}^{-3}} \\

& x\dfrac{{{d}^{2}}y}{d{{x}^{2}}}+2\dfrac{dy}{dx}=2a+2b{{x}^{-3}} \\

& x\dfrac{{{d}^{2}}y}{d{{x}^{2}}}+2\dfrac{dy}{dx}=2\left( a+\dfrac{b}{{{x}^{3}}} \right) \\

\end{align}\]

Now by multiplying and dividing with \[\left( {{x}^{2}} \right)\] on RHS we get,

\[\begin{align}

& x\dfrac{{{d}^{2}}y}{d{{x}^{2}}}+2\dfrac{dy}{dx}=2\left[ a+\dfrac{b}{{{x}^{3}}} \right]\times \dfrac{{{x}^{2}}}{{{x}^{2}}} \\

& x\dfrac{{{d}^{2}}y}{d{{x}^{2}}}+2\dfrac{dy}{dx}=2\left[ ax{}^{2}+\dfrac{b{{x}^{2}}}{{{x}^{3}}} \right]\times \dfrac{1}{{{x}^{2}}} \\

& x\dfrac{{{d}^{2}}y}{d{{x}^{2}}}+2\dfrac{dy}{dx}=2\left[ a{{x}^{2}}+\dfrac{b}{x} \right]\dfrac{1}{{{x}^{2}}} \\

\end{align}\]

But given, \[xy=a{{x}^{2}}+\left( \dfrac{b}{x} \right)\], so the above equation becomes,

\[x\dfrac{{{d}^{2}}y}{d{{x}^{2}}}+2\dfrac{dy}{dx}=2\left( xy \right)\dfrac{1}{{{x}^{2}}}\]

Cancelling the like terms, we get

\[x\dfrac{{{d}^{2}}y}{d{{x}^{2}}}+2\dfrac{dy}{dx}=2\dfrac{y}{x}\]

Hence the correct answer is option (C).

Note: Another way to solve this problem is directly differentiating the given expression with respect to $x$ instead of dividing the expression by $x$ .

\[\dfrac{d(xy)}{dx}=\dfrac{d}{dx}\left( a{{x}^{2}}+\left( \dfrac{b}{x} \right) \right)\]

Now we will apply product rule, so we get

\[x\dfrac{d(y)}{dx}+y\dfrac{d(x)}{dy}=\dfrac{d}{dx}\left( a{{x}^{2}} \right)+\dfrac{d}{dx}\left( \dfrac{b}{x} \right)\]

This gets a little complicated. But you will get the same answer.

Last updated date: 01st Oct 2023

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