# Derivative of a power, algebra problem

1. Feb 6, 2009

I am unclear about an example of calculating a derivative in a book I am reading. I think the problem is my algebra.

\begin{flalign*} \mathrm{if} & & y & = & & x^n \\ \mathrm{then} & & \frac{dy}{dx} & = & & nx^{n-1} \\ \mathrm{so if} & & y & = & & x^2 - x - 2 \\ \mathrm{then} & & \frac{dy}{dx} & = & & 2x - 1 \\ \end{flalign*}

I can't find how it ended up as $$2x - 1$$. Here's is my attempt:

\begin{flalign*} y & = & & x^2 - x - 2 \\ \frac{dy}{dx} & = & & nx^{n-1} \\ & = & & 2x^{2-1} - 1x^{1-1} - 1(2)^{1-1} \\ & = & & 2x - 1 - 1 \\ & = & & 2x - 2 \\ \end{flalign*}

How do I get to the same result of $$2x - 1$$ ?

2. Feb 6, 2009

### Hurkyl

Staff Emeritus
That's not right. (I think) I know what you meant, but you'll do much better if you learn how to express yourself accurately, rather than sloppily.

Why do you think

$$\frac{d}{dx} 2 = 1 (2)^{1-1}$$

? I imagine you were trying to use

$$\frac{d}{dx} x^n = n x^{n-1}$$

but the left hand sides don't match....

(Understand why this doesn't match. That's actually more important than getting the right answer on this problem!)

3. Feb 6, 2009

### symbolipoint

What is x1-1 ? At one time, 1 - 1 = 0. I hope this is still true. If it is, then
x1-1=x0 = 1.

As for your term on the end, realize intuitively that the rate of change of a constant function is 0 .... (zero).

4. Feb 7, 2009

### AtaraxiaAle

The derivative is simply this:

\begin{flalign*} y & = & & x^2 - x - 2 \\ \frac{dy}{dx} & = & & 2x^1 - 1x^0 - 0 \\ & = & & 2x - 1 \\ \end{flalign*}

The derivative of any constant is 0. This is an important concept.

5. Feb 8, 2009

What do you mean by sloppily?

Thanks, that's actually what I am trying to understand.

Thanks, that's the type of information I was after.

6. Feb 9, 2009

### Hurkyl

Staff Emeritus
In the problem you're trying to solve, the very first thing you did was to assert that the derivative of $x^2 - x - 2$ with respect to x is equal to $n x^{n-1}$. I'm pretty sure that's not what you meant to do. Can you state your intentions more accurately?

The expression

$$\frac{d}{dx} x^n$$

has only one free variable: n. It also has a bound variable x that you could rename, if desired. By claiming that this expression matches

$$\frac{d}{dx} 2$$

that means you assert that you can choose a value for n (and possibly rename x) that makes the first expression the same as the second one -- not just in value, but in terms of the actual symbols used! (Although, we can gloss over some trivial intermediate steps, if we are confident everyone understands them)

It seems clear that you intended to choose the value "1" for n... but that doesn't work, because that just leaves you with

$$\frac{d}{dx} x^1$$

or, if we do a simplification,

$$\frac{d}{dx} x$$

which is not the same as

$$\frac{d}{dx} 2$$

....

7. Feb 9, 2009

### MrSparky

make the equation as x^2 - x^1 - 2 x^0
Now dy/dx = 2x - (1 . x^1) - (0 . 2x^-1)
now after multiplying everything you get 2x - 1 - 0= 2x - 1
PS . The dots are multiplication signs as well if you didnt know that

8. Feb 21, 2009

Intention was to find the derivative of $y = x^2 - x - 2$ following the equation in the book and understand how they came at $2x - 1$. In my original attempt I was trying to use a value of -1 for x in $nx^{n-1}$ which is why I ended up with $y = 2x - 2$. Now that I know the derivative of is constant is zero, I can understand how they arrived at the answer.

9. Feb 21, 2009

### HallsofIvy

Yes, that was your intent. That was not what you wrote.
You wrote:
Surely you can see that is not what you meant nor is it true.
Mathematics is hard enough without expecting people whom you are asking for help to guess what you really mean!

10. Feb 21, 2009