If y = frac{1}{3x + 5}, then the value of fra | vedclass

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(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

Vedclass · Accepted Answer

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

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(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, $\frac{d^9y}{dx^9} = \frac{(-1)^9 \cdot 9! \cdot 3^9}{(3x + 5)^{10}}$.

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

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