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Question: Find the value of \(\dfrac{{{d^2}x}}{{d{y^2}}}\) A. \({\left( {\dfrac{{{d^2}y}}{{d{x^2}}}} \right)...

Find the value of d2xdy2\dfrac{{{d^2}x}}{{d{y^2}}}
A. (d2ydx2)−1{\left( {\dfrac{{{d^2}y}}{{d{x^2}}}} \right)^{ - 1}}
B. −(d2ydx2)−1(dydx)−3 - {\left( {\dfrac{{{d^2}y}}{{d{x^2}}}} \right)^{ - 1}}{\left( {\dfrac{{dy}}{{dx}}} \right)^{ - 3}}
C. (d2ydx2)(dydx)−2\left( {\dfrac{{{d^2}y}}{{d{x^2}}}} \right){\left( {\dfrac{{dy}}{{dx}}} \right)^{ - 2}}
D. −(d2ydx2)(dydx)−3 - \left( {\dfrac{{{d^2}y}}{{d{x^2}}}} \right){\left( {\dfrac{{dy}}{{dx}}} \right)^{ - 3}}

Explanation

Solution

First of all we will write the given by breaking it into two parts and then write the second part as something’s inverse. After that multiply and divide and keep on doing the modifications until we get the required result.

Complete step-by-step solution:
We are required to find the equivalent operator of d2xdy2\dfrac{{{d^2}x}}{{d{y^2}}}.
We can write this as:-
⇒d2xdy2=ddy(dxdy)\Rightarrow \dfrac{{{d^2}x}}{{d{y^2}}} = \dfrac{d}{{dy}}\left( {\dfrac{{dx}}{{dy}}} \right)
We see that the fraction inside the bracket can be written as inverse as follows:-
\Rightarrow \dfrac{{{d^2}x}}{{d{y^2}}} = \dfrac{d}{{dy}}\left\\{ {{{\left( {\dfrac{{dy}}{{dx}}} \right)}^{ - 1}}} \right\\}
Now, we will multiply and divide the right hand side of the above expression by dx to obtain the following expression:-
\Rightarrow \dfrac{{{d^2}x}}{{d{y^2}}} = \dfrac{d}{{dy}}\left\\{ {{{\left( {\dfrac{{dy}}{{dx}}} \right)}^{ - 1}}} \right\\} \times \dfrac{{dx}}{{dx}}
Re – arranging the terms on the right hand side to obtain the following expression:-
\Rightarrow \dfrac{{{d^2}x}}{{d{y^2}}} = \dfrac{d}{{dx}}\left\\{ {{{\left( {\dfrac{{dy}}{{dx}}} \right)}^{ - 1}}} \right\\} \times \dfrac{{dx}}{{dy}}
Now, we will just find the derivative of the function inside the parenthesis with respect to x to obtain:-
⇒d2xdy2=−(dydx)−2d2ydx2dxdy\Rightarrow \dfrac{{{d^2}x}}{{d{y^2}}} = - {\left( {\dfrac{{dy}}{{dx}}} \right)^{ - 2}}\dfrac{{{d^2}y}}{{d{x^2}}}\dfrac{{dx}}{{dy}}
[Because of the chain rule which states that \dfrac{d}{{dx}}\left\\{ {f(g(x))} \right\\} = f'(g(x)).g'(x)]
Now, we can rewrite the expression as follows:-
⇒d2xdy2=−(dydx)−2d2ydx2(dydx)−1\Rightarrow \dfrac{{{d^2}x}}{{d{y^2}}} = - {\left( {\dfrac{{dy}}{{dx}}} \right)^{ - 2}}\dfrac{{{d^2}y}}{{d{x^2}}}{\left( {\dfrac{{dy}}{{dx}}} \right)^{ - 1}}
Now, we also know that ax.ay=ax+y{a^x}.{a^y} = {a^{x + y}}. Using this in the above derived expression, we will club the like terms to get the following expression:-
⇒d2xdy2=−(dydx)−2−1d2ydx2\Rightarrow \dfrac{{{d^2}x}}{{d{y^2}}} = - {\left( {\dfrac{{dy}}{{dx}}} \right)^{ - 2 - 1}}\dfrac{{{d^2}y}}{{d{x^2}}}
Simplifying the calculations in the power, we will thus obtain:-
⇒d2xdy2=−(dydx)−3d2ydx2\Rightarrow \dfrac{{{d^2}x}}{{d{y^2}}} = - {\left( {\dfrac{{dy}}{{dx}}} \right)^{ - 3}}\dfrac{{{d^2}y}}{{d{x^2}}}

Hence, the correct option is (D).

Note: The students must note that they may make the mistake of crossing over denominator from the numerator and the denominator but we cannot do that because d is not a variable or anything but it represents a differential and it is attached to something.
The students must also know that the chain rule was used in one step of the solution. Let us understand it in a bit detail how did we use that so that there is no confusion:-
Chain rule suggests that \dfrac{d}{{dx}}\left\\{ {f(g(x))} \right\\} = f'(g(x)).g'(x)
And, we did \dfrac{d}{{dx}}\left\\{ {{{\left( {\dfrac{{dy}}{{dx}}} \right)}^{ - 1}}} \right\\} = - {\left( {\dfrac{{dy}}{{dx}}} \right)^{ - 2}}\dfrac{{{d^2}y}}{{d{x^2}}}
If we compare both of them, we get f(x)=x−1f(x) = {x^{ - 1}} and g(x)=dydxg(x) = \dfrac{{dy}}{{dx}}.
So, we get: f′(x)=−x−2f'(x) = - {x^{ - 2}} and g′(x)=d2ydx2g'(x) = \dfrac{{{d^2}y}}{{d{x^2}}} and thus we have the required result which we used in the solution.