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  • Stuck here, help me understand: - Limit , continuity and differentiability - JEE Main-5

\frac{d^{2}x}{dy^{2}}  equals  to

  • Option 1)

    \left ( \frac{d^{2}y}{dx^{2}} \right )\left ( \frac{dy}{dx} \right )^{-2}\;

  • Option 2)

    \; \; -\left ( \frac{d^{2}y}{dx^{2}} \right )\left ( \frac{dy}{dx} \right )^{-3}\;

  • Option 3)

    \; \; \left ( \frac{d^{2}y}{dx^{2}} \right )^{-1}\;

  • Option 4)

    \; -\left ( \frac{d^{2}y}{dx^{2}} \right )^{-1}\left ( \frac{dy}{dx} \right )^{-3}

 

Answers (1)

best_answer

As we learnt in

Second order derivative for parametric function -

When we find 

\frac{dy}{dx}=F(t)\:then\:\frac{d^{2}y}{dx^{2}}=\frac{\frac{d}{dt}\:F(t)}{\frac{dx}{dt}}

-

 

 \frac{d^{2}x}{d^{2}y}=\frac{d}{dy}(\frac{dx}{dy})

=\:\frac{d}{dx}(\frac{dx}{dy}).\frac{dx}{dy}

=\:\frac{d}{dx}=\frac{1}{(\frac{dy}{dx})}.\frac{1}{(\frac{dy}{dx})}

=\:-\frac{1}{(\frac{dy}{dx})^{2}}\times\frac{d^{2}y}{dx^{2}}\times\frac{1}{(\frac{dy}{dx})}

=\:-\frac{1}{(\frac{dy}{dx})^{3}}\times\frac{d^{2}y}{dx^{2}}

=\:-(\frac{dy}{dx})^{-3}.\frac{d^{2}y}{dx^{2}}


Option 1)

\left ( \frac{d^{2}y}{dx^{2}} \right )\left ( \frac{dy}{dx} \right )^{-2}\;

This option is incorrect.

Option 2)

\; \; -\left ( \frac{d^{2}y}{dx^{2}} \right )\left ( \frac{dy}{dx} \right )^{-3}\;

This option is correct.

Option 3)

\; \; \left ( \frac{d^{2}y}{dx^{2}} \right )^{-1}\;

This option is incorrect.

Option 4)

\; -\left ( \frac{d^{2}y}{dx^{2}} \right )^{-1}\left ( \frac{dy}{dx} \right )^{-3}

This option is incorrect.

Posted by

divya.saini

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