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# Starting with $\dfrac{dx}{dy}=\dfrac{1}{\dfrac{dy}{dx}}$. Prove that $\dfrac{{{d}^{2}}y}{d{{y}^{2}}}=-\dfrac{\dfrac{{{d}^{2}}y}{d{{x}^{2}}}}{{{\left( \dfrac{dy}{dx} \right)}^{3}}}$ and deduce that for the parabola ${{y}^{2}}=4ax$, $\dfrac{{{d}^{2}}y}{d{{x}^{2}}}.\dfrac{{{d}^{2}}x}{d{{y}^{2}}}=-\dfrac{2a}{{{y}^{3}}}$..

Last updated date: 27th Mar 2023
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Hint: Use the quotient rule of differentiation in the given question to get the desired result.

Here, starting with $\dfrac{dx}{dy}=\dfrac{1}{\dfrac{dy}{dx}}$, we have to prove that $\dfrac{{{d}^{2}}y}{d{{y}^{2}}}=-\dfrac{\dfrac{{{d}^{2}}y}{d{{x}^{2}}}}{{{\left( \dfrac{dy}{dx} \right)}^{3}}}$.
Also, we have to prove $\dfrac{{{d}^{2}}y}{d{{x}^{2}}}.\dfrac{{{d}^{2}}x}{d{{y}^{2}}}=-\dfrac{2a}{{{y}^{3}}}$for parabola ${{y}^{2}}=4ax$.
Taking, $\dfrac{dx}{dy}=\dfrac{1}{\dfrac{dy}{dx}}$
Now, we will differentiate both sides with respect to $y$.
Also, we know that quotient rule says that,
$\dfrac{d}{dy}\left( \dfrac{f}{g} \right)=\dfrac{g\left( \dfrac{dt}{dy} \right)-f\left( \dfrac{dg}{dy} \right)}{{{g}^{2}}}$
In $\dfrac{dx}{dy}=\dfrac{1}{\dfrac{dy}{dx}}$
$f=1$and $g=\dfrac{dy}{dx}$
Now, differentiating both sides with respect to $y$.
We get, $\dfrac{d}{dy}\left( \dfrac{dx}{dy} \right)=\dfrac{\left( \dfrac{dy}{dx} \right)\dfrac{d}{dy}\left( 1 \right)-\left( 1 \right)\left[ \dfrac{d}{dy}\left( \dfrac{dy}{dx} \right) \right]}{{{\left( \dfrac{dy}{dx} \right)}^{2}}}$
As we know that $\dfrac{d}{dy}\left( \text{constant} \right)=0$
Therefore, we get
$\dfrac{{{d}^{2}}x}{d{{y}^{2}}}=\dfrac{0-\dfrac{d}{dy}\left( \dfrac{dy}{dx} \right)}{{{\left( \dfrac{dy}{dx} \right)}^{2}}}$
Now we will multiply by $\dfrac{dy}{dx}$ on both the numerator and denominator of $RHS$.
We get $\dfrac{{{d}^{2}}x}{d{{y}^{2}}}=\dfrac{-\dfrac{d}{dy}\left( \dfrac{dy}{dx} \right).\dfrac{dy}{dx}}{{{\left( \dfrac{dy}{dx} \right)}^{2}}.\dfrac{dy}{dx}}$
We know that $\dfrac{d}{dy}\left( \dfrac{dy}{dx} \right).\dfrac{dy}{dx}=\dfrac{{{d}^{2}}y}{d{{x}^{2}}}$
Therefore, we get $\dfrac{{{d}^{2}}x}{d{{y}^{2}}}=\dfrac{-\dfrac{{{d}^{2}}y}{d{{x}^{2}}}}{{{\left( \dfrac{dy}{dx} \right)}^{2}}}$[Hence Proved]
which is our required result.
Now, we have to prove that $\dfrac{{{d}^{2}}y}{d{{x}^{2}}}.\dfrac{{{d}^{2}}x}{d{{y}^{2}}}=-\dfrac{2a}{{{y}^{3}}}$ for parabola ${{y}^{2}}=4ax$.
Now we take parabola, ${{y}^{2}}=4ax$.
So, we differentiate the above equation with respect to $x$.
Also, we know that $\dfrac{d}{dx}\left( {{x}^{n}} \right)=n{{x}^{n-1}}$
Therefore, $2y\dfrac{dy}{dx}=4a$
$\dfrac{dy}{dx}=\dfrac{2a}{y}....\left( i \right)$
Again differentiating both sides with respect to $x$,
We get, $\dfrac{{{d}^{2}}y}{d{{x}^{2}}}=-\left( 2a \right){{y}^{-1-1}}.\dfrac{dy}{dx}$
Now, we put the value of $\dfrac{dy}{dx}$from equation $\left( i \right)$.
We get $\dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{-2a}{{{y}^{2}}}.\dfrac{2a}{y}$
Therefore, $\dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{-4{{a}^{2}}}{{{y}^{3}}}....\left( ii \right)$
Now, from previous results, we know that
$\dfrac{{{d}^{2}}x}{d{{y}^{2}}}=\dfrac{\dfrac{-{{d}^{2}}y}{d{{x}^{2}}}}{{{\left( \dfrac{dy}{dx} \right)}^{2}}}$
By putting values from equation $\left( i \right)$and $\left( ii \right)$
We get, $\dfrac{{{d}^{2}}x}{d{{y}^{2}}}=\dfrac{-\left[ \dfrac{-4{{a}^{2}}}{{{y}^{3}}} \right]}{{{\left[ \dfrac{2a}{y} \right]}^{3}}}$
$\Rightarrow \dfrac{{{d}^{2}}x}{d{{y}^{2}}}=\dfrac{\dfrac{+4{{a}^{2}}}{{{y}^{3}}}}{\dfrac{8{{a}^{3}}}{{{y}^{3}}}}$
By cancelling the like terms,
We get, $\dfrac{{{d}^{2}}x}{d{{y}^{2}}}=\dfrac{+1}{2a}....\left( iii \right)$
Now we will multiply the equation $\left( ii \right)$and $\left( iii \right)$.
We get, $\left( \dfrac{{{d}^{2}}y}{d{{x}^{2}}} \right)\left( \dfrac{{{d}^{2}}x}{d{{y}^{2}}} \right)=\left( \dfrac{-4{{a}^{2}}}{{{y}^{3}}} \right).\dfrac{1}{2a}$
By cancelling the like terms,
We get, $\dfrac{{{d}^{2}}y}{d{{x}^{2}}}.\dfrac{{{d}^{2}}x}{d{{y}^{2}}}=\dfrac{-2a}{{{y}^{3}}}$which is the required result.

Note: In the term $\dfrac{d}{dy}\left( \dfrac{dy}{dx} \right)$, some students cancel $dy$ from numerator and denominator considering them to be like terms but that is wrong and $\dfrac{d}{dy}\left( \dfrac{dy}{dx} \right)=\dfrac{\dfrac{{{d}^{2}}y}{d{{x}^{2}}}}{\dfrac{dy}{dx}}$.