Derivative of a power, algebra problem

In summary: It is greatly appreciated.In summary, the conversation was about finding the derivative of a function and understanding how to use the equation from the book to arrive at the correct answer. The confusion stemmed from using incorrect values and not fully understanding the concept of a constant having a derivative of 0.
  • #1
username12345
48
0
I am unclear about an example of calculating a derivative in a book I am reading. I think the problem is my algebra.

[tex]
\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*}
[/tex]

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

[tex]
\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*}
[/tex]

How do I get to the same result of [tex]2x - 1[/tex] ?
 
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  • #2
[tex]
\begin{flalign*}
y & = & & x^2 - x - 2 \\
\frac{dy}{dx} & = & & nx^{n-1}
\end{flalign*}
[/tex]
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.


[tex]
\begin{flalign*}
& = & & 2x^{2-1} - 1x^{1-1} - 1(2)^{1-1} \\
\end{flalign*}
[/tex]
Why do you think

[tex]\frac{d}{dx} 2 = 1 (2)^{1-1}[/tex]

? I imagine you were trying to use

[tex]\frac{d}{dx} x^n = n x^{n-1}[/tex]

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
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
The derivative is simply this:

[tex]

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

[/tex]

The derivative of any constant is 0. This is an important concept.
 
  • #5
Hurkyl said:
but you'll do much better if you learn how to express yourself accurately, rather than sloppily.

What do you mean by sloppily?


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

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

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

Thanks, that's the type of information I was after.
 
  • #6
username12345 said:
What do you mean by sloppily?
In the problem you're trying to solve, the very first thing you did was to assert that the derivative of [itex]x^2 - x - 2[/itex] with respect to x is equal to [itex]n x^{n-1}[/itex]. I'm pretty sure that's not what you meant to do. Can you state your intentions more accurately?


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

[tex]\frac{d}{dx} x^n[/tex]

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

[tex]\frac{d}{dx} 2[/tex]

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

[tex]\frac{d}{dx} x^1[/tex]

or, if we do a simplification,

[tex]\frac{d}{dx} x[/tex]

which is not the same as

[tex]\frac{d}{dx} 2[/tex]

...
 
  • #7
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
Hurkyl said:
In the problem you're trying to solve, the very first thing you did was to assert that the derivative of [itex]x^2 - x - 2[/itex] with respect to x is equal to [itex]n x^{n-1}[/itex]. I'm pretty sure that's not what you meant to do. Can you state your intentions more accurately?

Intention was to find the derivative of [itex] y = x^2 - x - 2[/itex] following the equation in the book and understand how they came at [itex] 2x - 1[/itex]. In my original attempt I was trying to use a value of -1 for x in [itex]nx^{n-1}[/itex] which is why I ended up with [itex] y = 2x - 2[/itex]. Now that I know the derivative of is constant is zero, I can understand how they arrived at the answer.
 
  • #9
Yes, that was your intent. That was not what you wrote.
You wrote:
[tex]y = x^2 - x - 2[/tex]
[tex]\frac{dy}{dx}= nx^{n-1}[/tex]
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!:rolleyes:
 
  • #10
Thankyou to everyone who provided help or feedback.
 

Related to Derivative of a power, algebra problem

1. What is the general formula for finding the derivative of a power function?

The general formula for finding the derivative of a power function is:
f'(x) = n * x^(n-1), where n is the exponent of the power function.

2. How do I apply the power rule when finding the derivative of a power function?

To apply the power rule, you simply multiply the exponent by the coefficient and then subtract 1 from the exponent. For example, if the function is f(x) = 2x^3, the derivative would be f'(x) = 6x^2.

3. Can the power rule be used for any type of power function?

Yes, the power rule can be used for any type of power function, including fractional and negative exponents. For fractional exponents, the power rule can be rewritten as: f'(x) = n * x^(n-1) * (fractional exponent). For negative exponents, the power rule can be rewritten as: f'(x) = n * x^(n-1) / (negative exponent).

4. Is there a shortcut for finding the derivative of a power function?

Yes, there is a shortcut called the logarithmic differentiation method. This method involves taking the natural logarithm of both sides of the function, using logarithm properties to simplify, and then finding the derivative. It is especially useful for more complex power functions.

5. How does the derivative of a power function relate to the slope of the original function?

The derivative of a power function represents the slope of the tangent line to the original function at a given point. This means that the derivative can be used to find the rate of change or slope of a power function at any point on the graph.

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