Discussion about math, puzzles, games and fun.   Useful symbols: ÷ × ½ √ ∞ ≠ ≤ ≥ ≈ ⇒ ± ∈ Δ θ ∴ ∑ ∫ • π ƒ -¹ ² ³ °

You are not logged in.

|
Options

mathsyperson
2005-12-03 03:34:31

Or we could say that we were finding d/de.

mikau
2005-12-02 13:40:06

[Sh]e's right! Technically we should say the slope of e^x when x equals two.

I think I see where the confusion is now, irspow, thanks.

2005-12-02 12:51:42

#### irspow wrote:

The derivative of e^2 would be 2e because the exponent was a constant and you could use the x^n = nx^(n-1) rule.

eh?

e²  is a constant, and so d/dx would just be zero...

irspow
2005-12-02 12:23:39

I think that you are just a little bit confused on this one.  The derivative of e^x is e^x, but that is because you are differentiating a variable exponent.  There are a lot of derivatives associated with logarithmic and exponential functions.

The derivative of e^2 would be 2e because the exponent was a constant and you could use the x^n = nx^(n-1) rule.  This rule is used when a variable is raised to a constant power. Play close attention to what the variable is in the equation.  Look around the internet for rules for differentiation of logarithmic functions and you will see where you went wrong.

mikau
2005-12-02 11:57:07

one thing thats bothering me. Sometimes it seems the rules for differentiating are not always true. For instance:

e^2

we know the differential of e^x = e^x dx so the slope of the line at 2 is e^2

But what if we used the fact that the derivative of x^n = nx^(n-1) ?

by this definition the derivative of e^2 should be ne^1 or 2e^1 but this is different then e^2!

So what if you took the derivitive of n^2 and n just happens to equal e or close to it? Would the derivative be wrong?

This is disturbing.