(C) Given $y = (3x + 5)^{-1}$.Using the general formula for the $n^{th}$ derivative of $(ax + b)^{-1}$, which is $\frac{d^n}{dx^n}(ax + b)^{-1} = \fra

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$\frac{(-1)^9 \times 9! \times 3^9}{(3x + 5)^{10}}$

(C) Given $y = (3x + 5)^{-1}$.Using the general formula for the $n^{th}$ derivative of $(ax + b)^{-1}$, which is $\frac{d^n}{dx^n}(ax + b)^{-1} = \frac{(-1)^n \cdot n! \cdot a^n}{(ax + b)^{n+1}}$.Here, $a = 3$, $b = 5$, and $n = 9$.Substituting these values, we get $\frac{d^9y}{dx^9} = \frac{(-1)^9 \cdot 9! \cdot 3^9}{(3x + 5)^{9+1}}$.Thus, $\frac{d^9y}{dx^9} = \frac{(-1)^9 \cdot 9! \cdot 3^9}{(3x + 5)^{10}}$.

Class 12 Mathematics Continuity and Differentiability Higher order derivatives

Difficult

If $y = \frac{1}{3x + 5}$, then the value of $\frac{d^9y}{dx^9}$ is:

MHT CET 2026 All MHT CET

A
$\frac{9! \times 3^9}{(3x + 5)^9}$
B
$\frac{(-1)^8 \times 9! \times 3^9}{(3x + 5)^9}$
C
$\frac{(-1)^9 \times 9! \times 3^9}{(3x + 5)^{10}}$
D
$\frac{(-1)^8 \times 8! \times 3^9}{(3x + 5)^{10}}$

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If $y = \frac{1}{3x + 5}$, then the value of $\frac{d^9y}{dx^9}$ is:

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