Question

If \[x=\varphi \left( t \right)\], \[y=\psi \left( t \right)\], then \[\dfrac{{{d}^{2}}y}{d{{x}^{2}}}\] is equal to
(1) \[\dfrac{\left( \varphi' \psi'' -\text{ }\psi' \varphi'' \right)}{{{\left( \varphi' \right)}^{2}}}\]
(2) \[\dfrac{\left( \varphi' \psi'' -\text{ }\psi' \varphi'' \right)}{{{\left( \varphi' \right)}^{3}}}\]
(3) \[\dfrac{\varphi'' }{\psi'' }\]
(4) \[\dfrac{\psi'' }{\varphi'' }\]

Answer 1

Hint: Here, in this given question we need to find the value of \[\dfrac{{{d}^{2}}y}{d{{x}^{2}}}\], first we will differentiate \[x\] and \[~y\] with respect to \[t\]and after that we will divide \[\dfrac{dy}{dx}\] and then again differentiate it with respect to \[x\]at last simplify it to get the requires answer.

Formula Used:
A formula for obtaining the derivative of a quotient of two functions is the quotient rule. It makes remembering all the terms a little bit simpler.
$\dfrac{d}{dx}\left( \dfrac{u}{v} \right)=\dfrac{vd\left( u \right)-ud\left( v \right)}{{{v}^{2}}}$
According to the formula, you must: in order to determine the derivative of \[u\] divided by \[v\]:
Multiply the derivative of \[u\] by \[v\].
The product of \[u\] times the derivative of \[v\] must then be subtracted from that result.
The final step is to multiply those terms by \[v\] squared.

Complete step by step Solution:
Given \[x=\varphi \left( t \right)\] and \[y=\psi \left( t \right).\]
Differentiate \[x\] and \[~y\] with respect to \[t\]
\[\dfrac{dx}{dt}=\varphi' \left( t \right)\] $...\left( i \right)$
\[\dfrac{dy}{dt}=\psi' \left( t \right)\] $...\left( ii \right)$
Divide equation $\left( ii \right)$by equation $\left( i \right)$
\[\dfrac{dy}{dx}=\dfrac{\psi' \left( t \right)}{\varphi' \left( t \right)}\]
Double differentiate \[\dfrac{dy}{dx}\] with respect to \[x\]
\[\dfrac{{{d}^{2}}y}{d{{x}^{2}}}~\]
\[=\left( \dfrac{d}{dx} \right)\left( \dfrac{dy}{dx} \right)\]
Multiply and divide R.H.S by $dt$
\[=\left( \dfrac{d}{dx} \right)\left( \dfrac{dy}{dx} \right)\left( \dfrac{dt}{dt} \right)\]
Now rewrite the equation as
\[=\left( \dfrac{d}{dt} \right)\left( \dfrac{dy}{dx} \right)\left( \dfrac{dt}{dx} \right)\]
Here, we can write \[\dfrac{dy}{dx}\] as \[\dfrac{\psi' \left( t \right)}{\varphi' \left( t \right)}\]
\[=\text{ }\left( \dfrac{d}{dt} \right)\text{ }\left\{ \dfrac{\psi' \left( t \right)}{\varphi' \left( t \right)} \right\}\times \left( \dfrac{dt}{dx} \right)\]
Again, we can write \[\dfrac{dt}{dx}\]as \[\dfrac{1}{\varphi' \left( t \right)}\]
\[=\text{ }\left( \dfrac{d}{dt} \right)\text{ }\left\{ \dfrac{\psi' \left( t \right)}{\varphi' \left( t \right)} \right\}\times \text{ }\dfrac{1}{\varphi' \left( t \right)}\]
Here we will use the quotient rule formula of differentiation
$\dfrac{d}{dx}\left( \dfrac{u}{v} \right)=\dfrac{vd\left( u \right)-ud\left( v \right)}{{{v}^{2}}}$
Here, \[\dfrac{\psi' \left( t \right)}{\varphi' \left( t \right)}\] are in $\dfrac{u}{v}$ form
\[=\text{ }\dfrac{\left[ \varphi' \left( t \right)\left( \dfrac{d}{dt} \right)\text{ }\psi' \left( t \right)\text{ }\text{ }\psi' \left( t \right)\left( \dfrac{d}{dt} \right)\varphi' \left( t \right) \right]}{{{\left( \varphi' \left( t \right) \right)}^{2}}}\times \dfrac{1}{\varphi' \left( t \right)}\]
\[=\text{ }\dfrac{\left[ \varphi' \left( t \right)\left( \dfrac{d}{dt} \right)\text{ }\psi' \left( t \right)\text{ }\text{ }\psi' \left( t \right)\left( \dfrac{d}{dt} \right)\varphi' \left( t \right) \right]}{{{\left( \varphi' \left( t \right) \right)}^{2}}\varphi' \left( t \right)}\]
\[=\text{ }\dfrac{\left[ \varphi' \left( t \right)\text{ }\psi'' \left( t \right)\text{ }\text{ }\psi' \left( t \right)\varphi'' \left( t \right) \right]}{{{\left( \varphi' \left( t \right) \right)}^{3}}}\]

Hence, the correct option is 2.

Note:To solve this type of questions, one must remember the rules of differentiation
Here to solve this question we have used differentiation, double differentiation
(For example, $\dfrac{d}{dx}\left( \dfrac{dy}{dx} \right)=\dfrac{{{d}^{2}}y}{d{{x}^{2}}}~$ ) and a formula of quotient rule $\dfrac{d}{dx}\left( \dfrac{u}{v} \right)=\dfrac{vd\left( u \right)-ud\left( v \right)}{{{v}^{2}}}$
We can use this formula when we have to differentiate a numerator and denominator with respect to \[x\]
We can understand this from an example
$\dfrac{d}{dx}\left( \dfrac{u}{v} \right)=\dfrac{{{x}^{2}}+x}{x}$
$=\dfrac{xd\left( {{x}^{2}}+x \right)-\left( {{x}^{2}}+x \right)d\left( x \right)}{{{x}^{2}}}$
$=\dfrac{x\left( 2x+1 \right)-\left( {{x}^{2}}+x \right)\cdot 1}{{{x}^{2}}}$
$=\dfrac{2{{x}^{2}}+x-{{x}^{2}}-x}{{{x}^{2}}}$
$=\dfrac{{{x}^{2}}}{{{x}^{2}}}$
$=1$
Hence, $\dfrac{d}{dx}\left( \dfrac{u}{v} \right)=\dfrac{{{x}^{2}}+x}{x}=1$