Answer
$$
f(x)=\ln{x},
$$
(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
$$
f^{(n)}(x)=\frac{(-1)^{(n-1)}. (n-1)! }{x^{n}},
$$
Work Step by Step
$$
f(x)=\ln{x},
$$
(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
$$
f^{(n)}(x)=\frac{(-1)^{(n-1)}. (n-1)! }{x^{n}},
$$