If x=f (t ) and y=g (t ) are differentiable functions of t, then d^2ydx^2 is

Correct answer: A. (a): We have, x=f (t ) ,y=g (t ) ⇒ dx dt=f^ ' (t ) and dy dt=g^ ' (t ) Now, dy dx= dy dt dx dt= g^ ' (t )f^ ' (t ), d^2ydx^2 = f^ ' (t ) g^ ' ' (t )-g^ ' (t ) f^ ' ' (t ) (f^ ' (t ) )^2 dt dx = f^ ' (t ) g^ ' ' (t )-g^ ' (t ) f^ ' ' (t ) (f^ ' (t ) )^2 1f^ ' (t ) = f^ ' (t ) g^ ' ' (t )-g^ ' (t )f^ ' ' (t ) (f^ ' (t ) )^3

JEE Previous Year Question

$If x = f (t) and y = g (t) are differentiable functions of t, then \frac{d ^{2} y}{d x ^{2}} is$

Topic: JEE PYQ

Answer: A. Full step-by-step solution below.

A $\frac{f ^{’} ( t ) \cdot g ^{’ ’} ( t ) - g ^{’} ( t ) \cdot f ^{’ ’} ( t )}{( f ^{’} ( t ) ) ^{3}}$
B $\frac{f ^{’} ( t ) \cdot g ^{’ ’} ( t ) - g ^{’} ( t ) \cdot f ^{’ ’} ( t )}{( f ^{’} ( t ) ) ^{2}}$
C $\frac{g ^{’} ( t ) \cdot f ^{’ ’} ( t ) - f ^{’} ( t ) \cdot g ^{’ ’} ( t )}{( f ^{’} ( t ) ) ^{3}}$
D $\frac{g ^{’} ( t ) \cdot f ^{’ ’} ( t ) + f ^{’} ( t ) \cdot g ^{’ ’} ( t )}{( f ^{’} ( t ) ) ^{3}}$

Correct answer: A

Solution

(a): We have, x = f (t), y = g (t) \Rightarrow \frac{dx}{dt} = f^{’} (t) and \frac{dy}{dt} = g^{’} (t) Now, \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{g ^{’} ( t )}{f ^{’} ( t )}, \frac{d ^{2} y}{d x ^{2}} = \frac{f ^{’} ( t ) \cdot g ^{’ ’} ( t ) - g ^{’} ( t ) \cdot f ^{’ ’} ( t )}{( f ^{’} ( t ) ) ^{2}} \cdot \frac{dt}{dx} = \frac{f ^{’} ( t ) \cdot g ^{’ ’} ( t ) - g ^{’} ( t ) \cdot f ^{’ ’} ( t )}{( f ^{’} ( t ) ) ^{2}} \cdot \frac{1}{f ^{’} ( t )} = \frac{f ^{’} ( t ) \cdot g ^{’ ’} ( t ) - g ^{’} ( t ) f ^{’ ’} ( t )}{( f ^{’} ( t ) ) ^{3}}

Solution

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