# Proving a conjecture in differentiation

• Nov 27th 2012, 04:56 AM
Tutu
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!
• Nov 27th 2012, 05:29 AM
skeeter
Re: Proving a conjecture in differentiation
for n = 1

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

assume true for $n$ ...

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

$\frac{d}{dx} \left[\frac{d^ny}{dx^n} = k^n y\right] = \frac{d^{n+1}y}{dx^{n+1}} = k^n \cdot \frac{dy}{dx} = k^n \cdot ky = k^{n+1} y$
• Nov 27th 2012, 06:24 AM
HallsofIvy
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!