#1
hello UG, i need some help on how to derivate this function: x^1/5 using the definition of the derivative... thx in advance
#5
Quote by Dirge Humani
(1/5)x^(-4/5)

Using the definition of a derivative would be a very long, drawn out process to type.

He got it. Maybe the question means use this Y= X^n so dy/dx= n.X^(n-1)
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#6
A derivative is the rate of change of a function as it's input value changes.
In this case, your example can be likened to this rule:

d/dx = (n)u^(n-1) (u')


So, your given function is:

f(x) = x^(1/5) . The derivative according to the given rule would be...

  • f(x) = x^(1/5)
  • (1/5) x^(1 - [1/5]) x (1)
  • (1/5) x^(-4/5)
  • Answer: f '(x) = 1/[5x^(4/5)]


Hope that makes sense.
#8
Quote by Dirge Humani
By the way guys, the definition of the deriviative is f'(x)=(f(x+h)-f(x))/h, where h is delta x.

That's true. There are also 36 Basic Differentiation Rules that come in handy when
"f'(x)=(f(x+h)-f(x))/h" becomes too cumbersome.
#9
Quote by Zaphikh
That's true. There are also 36 Basic Differentiation Rules that come in handy when
"f'(x)=(f(x+h)-f(x))/h" becomes too cumbersome.

Well in precalc, which I assume he is in, or else he probably wouldn't need to ask this, f'(x)=(f(x+h)-f(x))/h was the only definition of a derivative I was given originally. Then we learned the power rule, product rule, quotient rule, chain rule...etc.
#10
=lim (x^n - a^n)/(x - a)
x-->a
=lim (x - a)(x^n-1 + ax^n-2...+xa^n-2 + a^n-1)/(x - a)
x-->a
=lim (x^n-1 + ax^n-2...+xa^n-2 + a^n-1)
x-->a
= (a^n-1 + a^n-1...+a^n-1 + a^n-1)
=na^n-1

General rule for powers by first principles.
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#11
Quote by Dirge Humani
Well in precalc, which I assume he is in, or else he probably wouldn't need to ask this, f'(x)=(f(x+h)-f(x))/h was the only definition of a derivative I was given originally. Then we learned the power rule, product rule, quotient rule, chain rule...etc.

What about the sexy rule?

>.>

<.<

#12
Quote by Dirge Humani
By the way guys, the definition of the derivative is f'(x)=(f(x+h)-f(x))/h, where h is delta x.


It's the limit as h approaches 0 of everything on the right side of the above equation

You could also do it

f'(x)=lim((f(x)-f(a))/(x-a))

The limit is as x approaches a

I don't know which of these two ways your teacher wants you to do it
#13
using the definition of the derivative f'(x)=(f(x+h)-f(x))/h, where h is delta x.
#14
Quote by Deth_send
using the definition of the derivative f'(x)=(f(x+h)-f(x))/h, where h is delta x.

Well... what more do you need?
The template for your solution is sitting right before your eyes... use it!

If: f(x) = x, then f(x+h) = what? ...that's right ---> f(x+h) = x+h

So we have the rule that f '(x) = lim(h->0) [f(x+h) - f(x)] / (h)


What do you need to know? Well we know that f(x) = x^(1/5), so we can conclude that
f(x+h) = (x+h)^(1/5).

Plugging this in, you'll find that f '(x) = lim(h->0) [(x+h)^(1/5) - x^(1/5)] / (h)

Try working it out from there
#15
Definition of derivitive: a mathematical function that 99% of people will find absolutely useless in their lives. It is also ridiculously hard to get a hold of and requires an incredibly high brain-energy to results ratio. Some scientists say it is harmful to mental health to think about derivatives too much.

Using this definition, my advice is that you should slit your writs and write WTF!? in blood on your paper and hand it in.
#16
Quote by Dirge Humani
(1/5)x^(-4/5)

Using the definition of a derivative would be a very long, drawn out process to type.

This.
He also got it on the definition of a derivative.

For future reference...
I've taken engineering classes where we used a lot of calculus I-III.
You are NEVER going to use the exact definition, unless you are going to take some math history/origins class or take a hand at theoretical physics.

*EDIT*
Zap... he knows where it's at.