derivaive (rule of sums)

ajone216 · #1 Nov 1, 2009, 02:20 PM

find the derivative of f(x)= x^3+2e^3

I'm not sure if I'm right so that's why I'm putting this up.

I believe that the answer is 3x^2+6e^2

Nhatkiem · Nov 1, 2009, 05:38 PM

A very good attempt ajone216.

lets look at this equation. We want to find

$\frac{d}{dx}[x^3+2e^3]$
Equivalently this is

$\frac{d}{dx}[x^3]+\frac{d}{dx}[2e^3]$

$\frac{d}{dx}[x^3]=3x^2$ This is correct.

However!

$\frac{d}{dx}[2e^3]=0$ not $6e^2$

This is because you are taking the derivative with respect to x. 2e^3 is a constant. Regardless of the value of x, 2e^3 remains the same.

And the derivative of any constant (C) $\frac{d}{dx}[C]=0$ So you should have...

$3x^2$

Unknown008 · Nov 1, 2009, 11:01 PM

Yup, e is a constant, which is approximately 2.718. Having to it to the power of three, is is more or less equivalent to 20.1. And, as Nhatkiem said, the derivative of any constant is 0.

However, if you had e^x, then, that's something completely different, but which also applies to your question.

$\frac{d(e^{f(x)})}{dx} = f'(x)e^{f(x)}$

In your question, f'(x) is zero, and anything times zero becomes 0.