derivatives

A

ashley123456789

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Jan 4, 2007
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f(x)=x^1/3+1

a.)find first derivative
b.)find second derivative


*so far in trying to find the first derivative i got f1(x)=1/3x^-2/3. (i am not even sure if that is right and if it is im not sure what to do next.)

This is for a takehome test and i could really use the help so if anyone knows the answer it would mean alot. THANKSS.
 
You list the "question" at hand as being simplfy "f(x) = x<sup>1/3</sup> + 1", just an equation, with no instructions. But then you mention things like first and second derivatives, sign charts, etc, etc, saying then that you do not know what derivatives are.

First, without knowing what the instructions for this exercise might have been, there is little way to advise. Sorry.

Second, and much more important, if you are in a class that expects you to work with derivatives, but you've never studied them (how else would you not know what they are?), then you need to have a serious talk with your academic advisor regarding course placement. There is nothing we can do to rectify this situation, since the material you've missed, just from the calculus course, probably took two or three months to cover in the classroom.

My best wishes to you.

Eliz.
 
Here is a few derivative rules that could help in the future.

1. this is a basic rule that you use in you first derivative

\(\displaystyle \frac{d}{dx} x^n = nx^\left(n-1\right)\)

\(\displaystyle n=\) any number

2. Product Rule

\(\displaystyle \frac{d}{dx}[f(x)g(x)]=f'(x)g(x)+f(x)g'(x)\)

*note \(\displaystyle f(x)\) and \(\displaystyle g(x)\) are simply two functions in the equation such as \(\displaystyle y = x^2cos2x\) \(\displaystyle f(x)=x^2\) and \(\displaystyle g(x)=cos2x\)

3. Quotent Rule

\(\displaystyle \frac{d}{dx}[\frac{f(x)}{g(x)}]=\frac{f'(x)g(x)-f(x)g'(x)}{(g(x))^2}\)

4. Chain Rule

\(\displaystyle \frac{d}{dx}[f(f(g(x))]=f'(g(x))g'(x)\)
 
\(\displaystyle f(x) \ = \ x^{1/3}+1\)

\(\displaystyle f'(x) \ = \ \frac{1}{3}x^{-2/3}+0 \ = \ \frac{1}{3x^{2/3}}.\)

\(\displaystyle f"(x) \ = \ \ \bigg(\frac{1}{3}\bigg)\bigg(\frac{-2}{3}\bigg)x^{-5/3}} \ = \ -\frac{2}{9x^{5/3}}.\)
 
\(\displaystyle Take \ home \ test?\)

\(\displaystyle ashley \ posted \ this \ thread \ on \ 04/03/2007.\)

\(\displaystyle galactus, \ I \ wonder, \ is \ chrisr \ back \ with \ the \ nom \ de \ plume \ of \ JeffWest?\)
 
My apologies for the confusion. I have just started using this site and was using ashley's old thread to practice my formating. I didn't mean for it to come back to the front of the line. Now I know.

P.S. I am not this "chrisr" person either.
 
JeffWest said:
My apologies for the confusion. I have just started using this site and was using ashley's old thread to practice my formating. I didn't mean for it to come back to the front of the line. Now I know.

P.S. I am not this "chrisr" person either.
Click to expand...

"Preview" is good for seeing the outcome without posting :wink:
 
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