Find the derivative, but using the definition thereof

In summary, the proof for the derivative of x(-2/3) using the definition of a derivative is done by directly substituting f(x) into the limit expression and expanding it using the binomial expansion. This results in the final expression of (-2/3)x(-5/3).
  • #1
stripes
266
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1. Prove that the derivative x(-2/3) is equal to (-2/3)x(-5/3) using the definition of a derivative "limit as h approaches zero of [(f(x+h) - f(x))]/h".



2. Homework Equations

f(x) = x(-2/3)

limit as h approaches zero of [(f(x+h) - f(x))]/h

sorry I couldn't type out the full equation, I hope that everyone understands what this is...




3. So first I substituted everything in

lim as h approaches zero of [f(x+h) - f(x)]/h


then I multiplied the numerator by the conjugate root. After doing so, I then expanded (x + h)4 = x4+4hx3+6h2x2+4h3x+h4

then i plugged it in and got 1 over the cubed root of x4+4hx3+6h2x2+4h3x+h4 MINUS 1 over the cubed root of x4.

Now remember, we're taking the limit of this whole thing (line above) divided by H, as h approaches zero.

that's where I don't know what to do...where do I go from there? Algebraically, I can't really simplify or cancel anything...

It's 3 am and I've been working on this question for over an hour, and I'm dead tired, so I do apologize if my post has broken any rules, or my attempt at a solution is not detailed enough...I simply do not have the energy to show more work than i already have. I would scan the work I've done and upload it, but my scanner doesn't work on this computer (64bit OS).

If you cannot answer the question or help because of that, then I do apologize for wasting anyones time. My basic question is, how do I prove that only using the definition of a derivative?

Again, thanks so much in advance, and I apologize for my lack of detail and perhaps a very unorganized post.
 
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  • #2
stripes said:
then I multiplied the numerator by the conjugate root. After doing so, I then expanded (x + h)4 = x4+4hx3+6h2x2+4h3x+h4

then i plugged it in and got 1 over the cubed root of x4+4hx3+6h2x2+4h3x+h4 MINUS 1 over the cubed root of x4.
one and upload it, but my scanner doesn't work on this computer (64bit OS).

My reaction would be "huh?" unfortunately. The way to approach this question is by direct substitution of f(x) into the limit expression. So,
[tex]
\lim_{h\rightarrow 0} \frac{f (x + h) - f (x)}{h} = \lim_{h\rightarrow 0} \frac{(x + h)^{-\frac{2}{3}} - x^{-\frac{2}{3}}}{h}
[/tex]

Next, expand using the binomial expansion:
[tex]
(x + h)^{-\frac{2}{3}} = x^{-\frac{2}{3}} (1 + \frac{h}{x})^{-\frac{2}{3}} = x^{-\frac{2}{3}} (1 + (-\frac{2}{3})(\frac{h}{x}) + \frac{(-\frac{2}{3})(-\frac{5}{3})}{2} (\frac{h}{x})^2 + ...)
[/tex]
Try to continue by yourself first before viewing the spoiler below.







This reduces the first limit expression to
[tex]
\lim_{h\rightarrow 0} x^{-\frac{2}{3}} ((-\frac{2}{3})(\frac{1}{x}) + \frac{(-\frac{2}{3})(-\frac{5}{3})}{2} (\frac{h}{x^2}) + ...) = (-\frac{2}{3}) x^{-\frac{5}{2}}
[/tex]
 

1. What is the definition of a derivative?

The derivative of a function f(x) at a point x is defined as the limit of the slope of the secant line passing through two points on the curve as the distance between the points approaches 0.

2. How do you find the derivative using the definition?

To find the derivative using the definition, we first need to find the slope of the secant line passing through two points on the curve. Then, we take the limit of this slope as the distance between the points approaches 0. This limit will give us the instantaneous rate of change at that point, which is the value of the derivative.

3. Why is it important to find the derivative using the definition?

Finding the derivative using the definition is important because it is the most fundamental and precise way to calculate the rate of change of a function at a specific point. It also helps us understand the behavior and properties of the function at that point.

4. Can the derivative always be found using the definition?

Yes, the derivative can always be found using the definition. However, in some cases, it may be difficult or time-consuming to compute the limit, and alternative methods such as using the rules of differentiation may be more efficient.

5. How does the definition of a derivative relate to the concept of limits?

The definition of a derivative is closely related to the concept of limits. In fact, the derivative is defined as the limit of the slope of the secant line, which is the same as the limit of the difference quotient. This relationship is important because it helps us understand the behavior of a function at a specific point and its rate of change.

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