# Basic derivative, understanding a question.

• Mar 30th 2010, 01:06 AM
Kakariki
Basic derivative, understanding a question.
Hey,

The derivative of $\displaystyle t e^3$ is $\displaystyle e^3$. I do not understand why. The answer in my text says that the t is a constant. However, I do not understand how to tell when something is a constant, and when something is not.

Originally in this problem I used the chain rule, and found the solution to be
$\displaystyle (1)(e^3) + (t)(e^3)$
$\displaystyle = e^3 + t e^3$

I am looking for an explanation, in plain english please.

Thank you!
• Mar 30th 2010, 01:36 AM
Prove It
Quote:

Originally Posted by Kakariki
Hey,

The derivative of $\displaystyle t e^3$ is $\displaystyle e^3$. I do not understand why. The answer in my text says that the t is a constant. However, I do not understand how to tell when something is a constant, and when something is not.

Originally in this problem I used the chain rule, and found the solution to be
$\displaystyle (1)(e^3) + (t)(e^3)$
$\displaystyle = e^3 + t e^3$

I am looking for an explanation, in plain english please.

Thank you!

Actually, $\displaystyle e^3$ is a constant. $\displaystyle t$ is a variable.

So $\displaystyle \frac{d}{dx}(e^3 t) = \frac{d}{dx}(e^3 t^1)$

$\displaystyle = 1e^3 t^0$

$\displaystyle = e^3$.
• Mar 30th 2010, 01:39 AM
mathemagister
Quote:

Originally Posted by Kakariki
Hey,

The derivative of $\displaystyle t e^3$ is $\displaystyle e^3$. I do not understand why. The answer in my text says that the t is a constant. However, I do not understand how to tell when something is a constant, and when something is not.

Originally in this problem I used the chain rule, and found the solution to be
$\displaystyle (1)(e^3) + (t)(e^3)$
$\displaystyle = e^3 + t e^3$

I am looking for an explanation, in plain english please.

Thank you!

$\displaystyle t$ is not the constant. $\displaystyle e^3$ is the constant. Remember, $\displaystyle e$ is just a number (constant, not a variable). Any constant cubed will also be a constant.

Just as the derivative of $\displaystyle 4t$ will be $\displaystyle 4$: the derivative of $\displaystyle e^3t$ will be $\displaystyle e^3$.

Does that help? :)

Mathemagister
• Mar 30th 2010, 02:01 AM
Kakariki
OH!!!!!!!!!! That makes complete sense. I was under the impression that t would be a constant, and it just didn't make any sense to me at all. e^3 being a constant does make sense though! The graph of te^3 would be a line, would it not? Then the derivative would be e^3 (a horizontal line).

Wow, I was just looking at it wrong.

THANK YOU!