## Calculus with Applications (10th Edition)

$$f(x)=\ln{x},$$ (a) To find the second derivative of find the first derivative, and then take its derivative. $$f^{\prime}(x)=\frac{1}{x}=x^{-1}=\frac{(-1)^{(0)}. (0)! }{x},$$ $$f^{\prime\prime}(x)=(-1)x^{-2}=-x^{-2} =\frac{(-1)^{(1)}. (1)! }{x^{2}},$$ $$f^{\prime\prime\prime}(x)=-(-2)x^{-3}=2x^{-3}=\frac{(-1)^{(2)}. (2)! }{x^{3}},$$ $$f^{\prime\prime\prime\prime}(x)=2(-3)x^{-4}=-6x^{-4}=\frac{(-1)^{(3)}. (3)! }{x^{4}},,$$ $$f^{(5)}(x)=-6(-4)x^{-5}=24x^{-5}=\frac{(-1)^{(4)}. (4)! }{x^{5}},$$ (b) The $nth$ derivative of $f$ with respect to $x$ is given by $$f^{(n)}(x)=\frac{(-1)^{(n-1)}. (n-1)! }{x^{n}},$$
$$f(x)=\ln{x},$$ (a) To find the second derivative of find the first derivative, and then take its derivative. $$f^{\prime}(x)=\frac{1}{x}=x^{-1}=\frac{(-1)^{(0)}. (0)! }{x},$$ $$f^{\prime\prime}(x)=(-1)x^{-2}=-x^{-2} =\frac{(-1)^{(1)}. (1)! }{x^{2}},$$ $$f^{\prime\prime\prime}(x)=-(-2)x^{-3}=2x^{-3}=\frac{(-1)^{(2)}. (2)! }{x^{3}},$$ $$f^{\prime\prime\prime\prime}(x)=2(-3)x^{-4}=-6x^{-4}=\frac{(-1)^{(3)}. (3)! }{x^{4}},,$$ $$f^{(5)}(x)=-6(-4)x^{-5}=24x^{-5}=\frac{(-1)^{(4)}. (4)! }{x^{5}},$$ (b) The $nth$ derivative of $f$ with respect to $x$ is given by $$f^{(n)}(x)=\frac{(-1)^{(n-1)}. (n-1)! }{x^{n}},$$