Question

If $x = \log p$ and $y = \dfrac{1}{p}$, then:
(A) $\dfrac{{{d^2}y}}{{d{x^2}}} - 2p = 0$
(B) $\dfrac{{{d^2}y}}{{d{x^2}}} + y = 0$
(C) $\dfrac{{{d^2}y}}{{d{x^2}}} + \dfrac{{dy}}{{dx}} = 0$
(D) $\dfrac{{{d^2}y}}{{d{x^2}}} - \dfrac{{dy}}{{dx}} = 0$

Answer 1

Hint: In the given problem, we are required to differentiate a parametric function with respect to x.
For differentiating parametric functions, we first find derivatives of x and y with respect to p separately and then divide both of them to find the differential $\dfrac{{dy}}{{dx}}$. Also, we will use the chain rule of differentiation in order to solve the problem. We must remember the derivatives of logarithmic function and power rule of differentiation in order to get to the final answer.

Complete step-by-step answer:
Now, we have, $x = \log p$ and $y = \dfrac{1}{p}$. So, to find the derivative of the parametric function, we first differentiate x and y separately with respect to the parameter, p.
Hence, differentiating x with respect to p, we get,
$ \Rightarrow \dfrac{{dx}}{{dp}} = \dfrac{d}{{dp}}\left( {\log p} \right)$
We know that the derivative of logarithmic function $\log x$ with respect to x is $\left( {\dfrac{1}{x}} \right)$. So, we get,
$ \Rightarrow \dfrac{{dx}}{{dp}} = \dfrac{1}{p} - - - \left( 1 \right)$
Now, we differentiate y with respect to p. So, we get,
\[ \Rightarrow \dfrac{{dy}}{{dp}} = \dfrac{d}{{dp}}\left( {\dfrac{1}{p}} \right)\]
Now, we know the power rule of differentiation \[\dfrac{{d\left( {{x^n}} \right)}}{{dx}} = n{x^{n - 1}}\]. So, we get,
\[ \Rightarrow \dfrac{{dy}}{{dp}} = \left( { - \dfrac{1}{{{p^2}}}} \right) - - - \left( 2 \right)\]
Now, we divide the equation $\left( 2 \right)$ by the equation $\left( 1 \right)$. So, we get,
\[ \Rightarrow \dfrac{{\left( {\dfrac{{dy}}{{dp}}} \right)}}{{\left( {\dfrac{{dx}}{{dp}}} \right)}} = \dfrac{{\left( { - \dfrac{1}{{{p^2}}}} \right)}}{{\left( {\dfrac{1}{p}} \right)}}\]
Cancelling the common factors in numerator and denominator, we get,
\[ \Rightarrow \dfrac{{dy}}{{dx}} = - \dfrac{1}{p}\]
Now, differentiating both sides of the above equation with respect to x, we get,
\[ \Rightarrow \dfrac{d}{{dx}}\left( {\dfrac{{dy}}{{dx}}} \right) = \dfrac{d}{{dx}}\left( { - \dfrac{1}{p}} \right)\]
Using the chain rule of differentiation $\dfrac{{dy}}{{dx}} = \dfrac{{dy}}{{dt}} \times \dfrac{{dt}}{{dx}}$, we get,
\[ \Rightarrow \dfrac{{{d^2}y}}{{d{x^2}}} = \dfrac{d}{{dp}}\left( { - \dfrac{1}{p}} \right) \times \dfrac{{dp}}{{dx}}\]
Again, using the power rule of differentiation, we get,
\[ \Rightarrow \dfrac{{{d^2}y}}{{d{x^2}}} = - \left( { - \dfrac{1}{{{p^2}}}} \right) \times \dfrac{1}{{\left( {\dfrac{{dx}}{{dp}}} \right)}}\]
Now, substituting the value of \[\left( {\dfrac{{dx}}{{dp}}} \right)\] from equation $\left( 1 \right)$, we get,
\[ \Rightarrow \dfrac{{{d^2}y}}{{d{x^2}}} = \dfrac{1}{{{p^2}}} \times \dfrac{1}{{\left( {\dfrac{1}{p}} \right)}}\]
Simplifying the expression further, we get,
\[ \Rightarrow \dfrac{{{d^2}y}}{{d{x^2}}} = \dfrac{1}{{{p^2}}} \times p\]
\[ \Rightarrow \dfrac{{{d^2}y}}{{d{x^2}}} = \dfrac{1}{p}\]
Now, we can clearly notice that the expressions of $\dfrac{{dy}}{{dx}}$ and \[\dfrac{{{d^2}y}}{{d{x^2}}}\] are negative of each other. So, we get,
$\dfrac{{{d^2}y}}{{d{x^2}}} + \dfrac{{dy}}{{dx}} = 0$ since $\dfrac{{{d^2}y}}{{d{x^2}}} = \dfrac{1}{p}$ and \[\dfrac{{dy}}{{dx}} = \dfrac{-1}{p}\].
Hence, the correct answer is the option (C).
So, the correct answer is “Option C”.

Note: We must remember this method to find the derivatives of the parametric function. Finding the second derivative of a parametric function is a bit complex as the process involves chain rule of differentiation. The chain rule of differentiation involves differentiating a composite by introducing new unknowns to ease the process and examine the behaviour of function layer by layer.