# Simple Question

• Jan 17th 2007, 09:07 AM
Simple Question
Is e a constant?
$\displaystyle D_x (e) = e^1(D_x 1)$
Derivative of e ---> will it equal to zero?

THanks :)
• Jan 17th 2007, 09:19 AM
ThePerfectHacker
Quote:

Is e a constant?
$\displaystyle D_x (e) = e^1(D_x 1)$
Derivative of e ---> will it equal to zero?

THanks :)

The derivative of $\displaystyle y=e$ is $\displaystyle y'=0$.
Since it is a constant function its derivative is zero.
Ah!
But what is the derivative of $\displaystyle y=\sqrt{e}$.
• Jan 17th 2007, 09:30 AM
CaptainBlack
Quote:

Originally Posted by ThePerfectHacker
But what is the derivative of $\displaystyle y=\sqrt{e}$.

Use the chain rule:

$\displaystyle \frac{d}{dx}\sqrt{e}=\frac{1}{2}\, \frac{1}{\sqrt{e}}\frac{de}{dx}=\frac{1}{2}\, \frac{1}{\sqrt{e}}\, 0=0$:D

RonL
• Jan 17th 2007, 09:35 AM
ThePerfectHacker
Quote:

Originally Posted by CaptainBlack
Use the chain rule:

$\displaystyle \frac{d}{dx}\sqrt{e}=\frac{1}{2}\, \frac{1}{\sqrt{e}}\frac{de}{dx}=\frac{1}{2}\, \frac{1}{\sqrt{e}}\, 0=0$:D

RonL

I am just saying that, is because I once was teaching my classmates Calculus. Just the moment I explain the derivative of 'e' is zero and give them this problem they start doing this.