For n in mathbb{N},find the n-th derivative o | vedclass

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(D) Let $y = \log x$. Then,the first derivative is $y_1 = \frac{d}{dx}(\log x) = x^{-1}$. The second derivative is $y_2 = \frac{d}{dx}(x^{-1}) = -1 \c

Vedclass · Accepted Answer

$(-1)^{n-1} \frac{(n-1)!}{x^{n}}$

Vedclass · Answer

(D) Let $y = \log x$. Then,the first derivative is $y_1 = \frac{d}{dx}(\log x) = x^{-1}$. The second derivative is $y_2 = \frac{d}{dx}(x^{-1}) = -1 \cdot x^{-2}$. The third derivative is $y_3 = \frac{d}{dx}(-1 \cdot x^{-2}) = (-1)(-2) \cdot x^{-3} = 2 \cdot x^{-3}$. The fourth derivative is $y_4 = \frac{d}{dx}(2 \cdot x^{-3}) = 2(-3) \cdot x^{-4} = -6 \cdot x^{-4} = -3! \cdot x^{-4}$. By observing the pattern,the $n$-th derivative is given by $y_n = (-1)^{n-1} \cdot (n-1)! \cdot x^{-n}$. Thus,$\frac{d^{n}}{d x^{n}}(\log x) = (-1)^{n-1} \frac{(n-1)!}{x^{n}}$.

For $n \in \mathbb{N}$,find the $n$-th derivative of $\log x$,i.e.,$\frac{d^{n}}{d x^{n}}(\log x) = $

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