# 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) \[\dfrac{-2y}{x}\]

Last updated date: 22nd Mar 2023

•

Total views: 309.3k

•

Views today: 6.91k

Answer

Verified

309.3k+ views

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.

Recently Updated Pages

Calculate the entropy change involved in the conversion class 11 chemistry JEE_Main

The law formulated by Dr Nernst is A First law of thermodynamics class 11 chemistry JEE_Main

For the reaction at rm0rm0rmC and normal pressure A class 11 chemistry JEE_Main

An engine operating between rm15rm0rm0rmCand rm2rm5rm0rmC class 11 chemistry JEE_Main

For the reaction rm2Clg to rmCrmlrm2rmg the signs of class 11 chemistry JEE_Main

The enthalpy change for the transition of liquid water class 11 chemistry JEE_Main

Trending doubts

Name the Largest and the Smallest Cell in the Human Body ?

Fill in the blanks A 1 lakh ten thousand B 1 million class 9 maths CBSE

Epipetalous and syngenesious stamens occur in aSolanaceae class 11 biology CBSE

Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE

A ball impinges directly on a similar ball at rest class 11 physics CBSE

Lysosomes are known as suicidal bags of cell why class 11 biology CBSE

Two balls are dropped from different heights at different class 11 physics CBSE

A 30 solution of H2O2 is marketed as 100 volume hydrogen class 11 chemistry JEE_Main

A sample of an ideal gas is expanded from 1dm3 to 3dm3 class 11 chemistry CBSE