Q83 / 150 · MHT CET 2026 (17 April Shift 2) · Mathematics
If $y = \dfrac{1}{3x + 5}$, then the value of $\dfrac{d^9 y}{dx^9}$ is..
- A$\dfrac{9! \times 3^9}{(3x + 5)^9}$
- B$\dfrac{(-1)^8 \times 9! \times 3^9}{(3x + 5)^9}$
- Correct answerC$\dfrac{(-1)^9 \times 9! \times 3^9}{(3x + 5)^{10}}$
- D$\dfrac{(-1)^8 \times 8! \times 3^9}{(3x + 5)^{10}}$
Solution
Given $y = (3x + 5)^{-1}$
Differentiating with respect to $x$, we get:
$\dfrac{dy}{dx} = -1(3x + 5)^{-2} \times 3$
$\dfrac{d^2y}{dx^2} = (-1)(-2)(3x + 5)^{-3} \times 3^2$
$\dfrac{d^3y}{dx^3} = (-1)(-2)(-3)(3x + 5)^{-4} \times 3^3$
Generalizing for the $n$-th derivative:
$\dfrac{d^ny}{dx^n} = (-1)^n n! (3x + 5)^{-(n+1)} \times 3^n$
Substituting $n = 9$:
$\dfrac{d^9y}{dx^9} = (-1)^9 9! (3x + 5)^{-10} \times 3^9 = \dfrac{(-1)^9 \times 9! \times 3^9}{(3x + 5)^{10}}$
Answer: $\dfrac{(-1)^9 \times 9! \times 3^9}{(3x + 5)^{10}}$
Differentiating with respect to $x$, we get:
$\dfrac{dy}{dx} = -1(3x + 5)^{-2} \times 3$
$\dfrac{d^2y}{dx^2} = (-1)(-2)(3x + 5)^{-3} \times 3^2$
$\dfrac{d^3y}{dx^3} = (-1)(-2)(-3)(3x + 5)^{-4} \times 3^3$
Generalizing for the $n$-th derivative:
$\dfrac{d^ny}{dx^n} = (-1)^n n! (3x + 5)^{-(n+1)} \times 3^n$
Substituting $n = 9$:
$\dfrac{d^9y}{dx^9} = (-1)^9 9! (3x + 5)^{-10} \times 3^9 = \dfrac{(-1)^9 \times 9! \times 3^9}{(3x + 5)^{10}}$
Answer: $\dfrac{(-1)^9 \times 9! \times 3^9}{(3x + 5)^{10}}$
Hence, the correct answer is option: 'C'.