frac{d^n}{dx^n}(log x) = | vedclass

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Question

(D) Let $y = \log x$. First derivative: $y_1 = \frac{d}{dx}(\log x) = \frac{1}{x} = x^{-1}$. Second derivative: $y_2 = \frac{d}{dx}(x^{-1}) = -1 \cdot

Vedclass · Accepted Answer

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

Vedclass · Answer

(D) Let $y = \log x$. First derivative: $y_1 = \frac{d}{dx}(\log x) = \frac{1}{x} = x^{-1}$. Second derivative: $y_2 = \frac{d}{dx}(x^{-1}) = -1 \cdot x^{-2} = \frac{-1}{x^2}$. Third derivative: $y_3 = \frac{d}{dx}(-x^{-2}) = -1 \cdot (-2) \cdot x^{-3} = \frac{2}{x^3} = \frac{2!}{x^3}$. Fourth derivative: $y_4 = \frac{d}{dx}(2x^{-3}) = 2 \cdot (-3) \cdot x^{-4} = \frac{-6}{x^4} = \frac{-3!}{x^4}$. By observing the pattern,the $n^{th}$ derivative is given by: $y_n = \frac{(-1)^{n-1}(n-1)!}{x^n}$.

$\frac{d^n}{dx^n}(\log x) =$

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