Discussion
The nth derivative of the function \(y = \dfrac{1}{x + 3}\) is
Answer: A
The nth derivative is \(\dfrac{(-1)^{n}n!}{(x + 3)^{n+1}}\).
Write \(y = (x + 3)^{-1}\) and differentiate repeatedly:
\[ \begin{aligned} y' &= -(x + 3)^{-2} \\ y'' &= 2(x + 3)^{-3} \\ y''' &= -6(x + 3)^{-4} \end{aligned} \]
Each step multiplies by the current power (giving \(1, 2, 6 = n!\)), flips the sign and raises the power of \((x+3)\) in the denominator by one, so \(y^{(n)} = (-1)^{n}n!/(x + 3)^{n+1}\).
The form with \((-1)^{n+1}\) has the wrong sign (it gives \(+1\) for the first derivative), and the forms with \((x + 3)^{n}\) or \((n + 1)!\) fail at \(n = 1\).
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