If $x = \phi \left( t \right),y = {\rm{\psi }}\left( t \right),$ then $\frac{{{d^2}y}}{{d{x^2}}}$ is equal to

$\frac{{\phi '{\rm{\psi ''}} - {\rm{\psi '}}\phi ''}}{{{{\left( {\phi '} \right)}^2}}}$

$\frac{{\phi ''}}{{{\rm{\psi ''}}}}$

JEE MATHEMATICS MCQ: DIFFERENTIATION

1 Easy

1 of 1

Application +4 | -1

x = \phi \left( t \right),y = \mathrm{\psi }\left( t \right),

then

\frac{{{d^2}y}}{{d{x^2}}}

is equal to

View answer and explanation

Correct option: B

\frac{{\phi '\psi '' - \psi '\phi ''}}{{{{\left( {\phi '} \right)}^3}}}

Option B (

\frac{{\phi '\psi '' - \psi '\phi ''}}{{{{\left( {\phi '} \right)}^3}}})

is correct because it represents the second derivative of y with respect to x when both x and y are parameterized by t.