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Question
If x = f(t) and y = g(t), then \[\frac{d^2 y}{d x^2}\] is equal to
Options
\[\frac{f' g'' - g'f''}{\left( f' \right)^3}\]
\[\frac{f' g'' - g'f''}{\left( f' \right)^2}\]
\[\frac{g''}{f''}\]
\[\frac{f'' g' - g'' f'}{\left( g' \right)^3}\]
Solution
(a) \[\frac{f' g'' - g'f''}{\left( f' \right)^3}\]
Here,
x = f(t) and y = g(t)
\[\Rightarrow \frac{dx}{dt} = f'\left( t \right) \text { and } \frac{dy}{dt} = g'\left( t \right)\]
\[ \therefore \frac{dy}{dx} = \frac{g'\left( t \right)}{f'\left( t \right)}\]
\[\Rightarrow \frac{d^2 y}{d x^2} = \frac{d}{dt}\left\{ \frac{g^{'} \left( t \right)}{f^{'} \left( t \right)} \right\} \times \frac{dt}{dx}\]
\[ = \frac{f^{'} \left( t \right) g^{''} \left( t \right) - g^{'} \left( t \right) f^{''} \left( t \right)}{\left[ f^{'} \left( t \right) \right]^2} \times \frac{1}{f'\left( t \right)}\]
\[ = \frac{f^{'} \left( t \right) g^{''} \left( t \right) - g^{'} \left( t \right) f^{''} \left( t \right)}{\left[ f^{'} \left( t \right) \right]^3}\]
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