Proving a conjecture in differentiation

How do I prove the conjecture that

d^(n)y

______ = k^(n) y

dx^(n)

The n are in the same position like 2 in the second derivative, d2y/dx2. And the n on the RHS is an exponent, but the y is not.

I know to use induction but I'm stuck at proving P(k+1),

d^(k+1)y

______

dx^(k+1)

Thank you very much!

Re: Proving a conjecture in differentiation

for n = 1

$\displaystyle \frac{dy}{dx} = ky$

assume true for $\displaystyle n$ ...

$\displaystyle \frac{d^ny}{dx^n} = k^n y$

$\displaystyle \frac{d}{dx} \left[\frac{d^ny}{dx^n} = k^n y\right] = \frac{d^{n+1}y}{dx^{n+1}} = k^n \cdot \frac{dy}{dx} = k^n \cdot ky = k^{n+1} y$

Re: Proving a conjecture in differentiation

Quote:

Originally Posted by

**Tutu** How do I prove the conjecture that

d^(n)y

______ = k^(n) y

dx^(n)

skeeter does this using induction on n and **assuming** that dy/dx= ky is that true? You do not mention it in your post but without that conclusion is not true.

Quote:

The n are in the same position like 2 in the second derivative, d2y/dx2. And the n on the RHS is an exponent, but the y is not.

I know to use induction but I'm stuck at proving P(k+1),

d^(k+1)y

______

dx^(k+1)

Thank you very much!