Question

If \[ x=f(t), \qquad y=g(t), \] then \[ \frac{d^2y}{dx^2} \] equals

• A. \[ \frac{ \frac{dx}{dt}\frac{d^2y}{dt^2} - \frac{d^2x}{dt^2}\frac{dy}{dt} } {\left(\frac{dx}{dt}\right)^3} \]
• B. \[ \frac{ \frac{dx}{dt}y-x\frac{dy}{dt} } {\left(\frac{dx}{dt}\right)^2} \]
• C. \[ \frac{ x\frac{dy}{dt}-y\frac{dx}{dt} } {x^2} \]
• D. \[ \frac{ \frac{dx}{dt}\frac{d^2y}{dt^2} - \frac{d^2x}{dt^2}\frac{dy}{dt} } {\left(\frac{dx}{dt}\right)^2} \]

Hint:

For parametric curves: \[ \frac{dy}{dx} = \frac{\frac{dy}{dt}} {\frac{dx}{dt}} \] and \[ \frac{d^2y}{dx^2} = \frac{ \frac{dx}{dt}\frac{d^2y}{dt^2} - \frac{d^2x}{dt^2}\frac{dy}{dt} } {\left(\frac{dx}{dt}\right)^3}. \] This formula is frequently used in JEE and entrance examinations.

Answer 1

The Correct Option is A

Step 1: Write the first derivative parametrically.
With $x=f(t)$ and $y=g(t)$, the chain rule gives $\dfrac{dy}{dx}=\dfrac{y'}{x'}$, where $x'=\dfrac{dx}{dt}$ and $y'=\dfrac{dy}{dt}$.

Step 2: Differentiate again with respect to $t$.
$\dfrac{d}{dt}\!\left(\dfrac{y'}{x'}\right)=\dfrac{x'y''-y'x''}{(x')^2}$ by the quotient rule, where $x''=\dfrac{d^2x}{dt^2}$ and $y''=\dfrac{d^2y}{dt^2}$.

Step 3: Convert back to a derivative in $x$.
Since $\dfrac{d^2y}{dx^2}=\dfrac{d}{dx}\!\left(\dfrac{dy}{dx}\right)$, and $\dfrac{d}{dx}=\dfrac{1}{x'}\dfrac{d}{dt}$, we divide the previous result by $x'$.

Step 4: Carry out the division.
$\dfrac{d^2y}{dx^2}=\dfrac{1}{x'}\cdot\dfrac{x'y''-y'x''}{(x')^2}=\dfrac{x'y''-y'x''}{(x')^3}$.

Step 5: Restore the full notation.
Writing the derivatives in full, the numerator is $\dfrac{dx}{dt}\dfrac{d^2y}{dt^2}-\dfrac{d^2x}{dt^2}\dfrac{dy}{dt}$ and the denominator is $\left(\dfrac{dx}{dt}\right)^3$.

Step 6: State the answer.
So $\dfrac{d^2y}{dx^2}=\dfrac{\frac{dx}{dt}\frac{d^2y}{dt^2}-\frac{d^2x}{dt^2}\frac{dy}{dt}}{\left(\frac{dx}{dt}\right)^3}$. \[ \boxed{\dfrac{\frac{dx}{dt}\frac{d^2y}{dt^2}-\frac{d^2x}{dt^2}\frac{dy}{dt}}{\left(\frac{dx}{dt}\right)^3}} \]