Q82 / 150 · MHT CET 2026 (15 April Shift 1) · Mathematics

If $y = \left(\dfrac{ax+b}{cx+d}\right)$, then $2\dfrac{dy}{dx} \cdot \dfrac{d^3 y}{dx^3}$ is equal to

  • A
    $\left(\dfrac{d^2 y}{dx^2}\right)^2$
  • B
    $3\dfrac{d^2 y}{dx^2}$
  • Correct answer
    C
    $3\left(\dfrac{d^2 y}{dx^2}\right)^2$
  • D
    $3\dfrac{d^2 x}{dy^2}$

Solution

Given $y = \dfrac{ax+b}{cx+d}$

Differentiating with respect to $x$, we get

$\dfrac{dy}{dx} = \dfrac{a(cx+d) - c(ax+b)}{(cx+d)^2} = \dfrac{ad-bc}{(cx+d)^2}$

Let $ad-bc = k$. Then $\dfrac{dy}{dx} = k(cx+d)^{-2}$

Differentiating again with respect to $x$, we get

$\dfrac{d^2 y}{dx^2} = -2kc(cx+d)^{-3}$

Differentiating once more with respect to $x$, we get

$\dfrac{d^3 y}{dx^3} = 6kc^2(cx+d)^{-4}$

Evaluating $2\dfrac{dy}{dx} \cdot \dfrac{d^3 y}{dx^3}$:

$2\dfrac{dy}{dx} \cdot \dfrac{d^3 y}{dx^3} = 2 \left( k(cx+d)^{-2} \right) \left( 6kc^2(cx+d)^{-4} \right) = 12k^2 c^2 (cx+d)^{-6}$

Evaluating $3\left(\dfrac{d^2 y}{dx^2}\right)^2$:

$3\left(\dfrac{d^2 y}{dx^2}\right)^2 = 3 \left( -2kc(cx+d)^{-3} \right)^2 = 12k^2 c^2 (cx+d)^{-6}$

Therefore, $2\dfrac{dy}{dx} \cdot \dfrac{d^3 y}{dx^3} = 3\left(\dfrac{d^2 y}{dx^2}\right)^2$

Answer: $3\left(\dfrac{d^2 y}{dx^2}\right)^2$

Hence, the correct answer is option: 'C'.

Relevant Questions