# nth Derivative in Proof by Induction

• Feb 13th 2011, 02:50 AM
BG5965
nth Derivative in Proof by Induction
Hey, I've been working through some proof by induction problems, and most of them are manageable, but then I saw one which involved differentiation which I have no idea what to do. Could someone please lead me in the right direction?

$\displaystyle \displaystyle \mbox{Prove that the nth derivative of } y=ln(x), x>0 \mbox{ is: } \frac{d^ny}{dx^n} = \frac{(-1)^{n+1}(n-1)!}{x^n}$

I have the basic steps done:

$\displaystyle ===================$

Step 1: Prove for n = 1

$\displaystyle \dfrac{dy}{dx} [ln(x)] = \dfrac{1}{x}$

$\displaystyle --------$

$\displaystyle \dfrac{d^1y}{dx^1} = \dfrac{(-1)^2(1-1)!}{x^1}$

$\displaystyle \dfrac{d^1y}{dx^1} = \dfrac{1}{x} = \dfrac{dy}{dx} [ln(x)] \mbox{ \therefore True }$

$\displaystyle ===================$

Step 2: Assume for n = k

$\displaystyle \dfrac{d^ky}{dx^k} = \dfrac{(-1)^{k+1}(k-1)!}{x^k}$

$\displaystyle ===================$

Step 3: Prove for n = k+1

I know how to start this, but how do I continue?

Thanks in advance for help. BG
• Feb 13th 2011, 03:14 AM
zzzoak
For

$\displaystyle n=k+1$

$\displaystyle \dfrac{d}{dx} \; \dfrac{d^ky}{dx^k}$