# Diff questions

• Jul 20th 2006, 01:21 PM
NeF
Diff questions
I want to double check these answers, this book doesnt supply answers.

1
$\displaystyle y= \frac{5x^4-3x}{x^2}\$
$\displaystyle y= 4x^2-2x^-1$
$\displaystyle \frac{dy}{dx}\ = 8x+2x^-2$
$\displaystyle 8x+\frac{2}{x^2}\$

2
$\displaystyle f(x)=(\frac{1}{x^2}+2)(x-3)$
$\displaystyle f(x)=(x^-2+2)(x-3)$
$\displaystyle f(x)= x^-1-3x^2+2x-6$
$\displaystyle f'(x)= x^-1-6x^-3+2$
$\displaystyle = \frac{-1}{x^2}\ + \frac{6}{x^3}+2$

3
$\displaystyle y= \frac{2\sqrt(x)}{x}$
$\displaystyle y= 2x^\frac{-1}{2}$
$\displaystyle \frac{dy}{dx} = -x^\frac{-3}{2}$
$\displaystyle = -\frac{1}{x^\frac{3}{2}}$

• Jul 20th 2006, 01:41 PM
topsquark
Quote:

Originally Posted by NeF
I want to double check these answers, this book doesnt supply answers.

1
$\displaystyle y= \frac{5x^4-3x}{x^2}\$
$\displaystyle y= 4x^2-2x^-1$
$\displaystyle \frac{dy}{dx}\ = 8x+2x^-2$
$\displaystyle 8x+\frac{2}{x^2}\$

$\displaystyle y = 5x^2 - 3x^{-1}$

Since you've evidently got the rules right, I'll let you finish it.

-Dan
• Jul 20th 2006, 01:42 PM
CaptainBlack
Quote:

Originally Posted by NeF
I want to double check these answers, this book doesnt supply answers.

1
$\displaystyle y= \frac{5x^4-3x}{x^2}\$
$\displaystyle y= 4x^2-2x^-1$
$\displaystyle \frac{dy}{dx}\ = 8x+2x^-2$
$\displaystyle 8x+\frac{2}{x^2}\$

It's not entirely clear what you are asking, but here I will assume that
you want the derivative of:

$\displaystyle y= \frac{5x^4-3x}{x^2}\$

First divide through on the right hand side by the $\displaystyle x^2$ at the bottom:

$\displaystyle y= 5x^2-3x^{-1}\$

Now differentiate using the rules for differentiating powers:

$\displaystyle \frac{dy}{dx}\ = 10x+3x^{-2}$

Now rearrange and simplify (back towards something like the form we started
from):

$\displaystyle \frac{dy}{dx}\ = 10x+\frac{3}{x^{2}}$

$\displaystyle \frac{dy}{dx}\ = \frac{10x^3+3}{x^{2}}$

RonL
• Jul 20th 2006, 01:50 PM
topsquark
Quote:

Originally Posted by NeF
I want to double check these answers, this book doesnt supply answers.

2
$\displaystyle f(x)=(\frac{1}{x^2}+2)(x-3)$
$\displaystyle f(x)=(x^-2+2)(x-3)$
$\displaystyle f(x)= x^-1-3x^2+2x-6$
$\displaystyle f'(x)= x^-1-6x^-3+2$
$\displaystyle = \frac{-1}{x^2}\ + \frac{6}{x^3}+2$

$\displaystyle f(x)= x^{-1}-3x^{-2}+2x-6$
$\displaystyle f'(x)= -x^{-2}+6x^{-3}+2$

Also, you might look into the product rule to do this one:
$\displaystyle f(x)=(\frac{1}{x^2}+2)(x-3)$
$\displaystyle f'(x) = \frac{-2}{x^3}(x-3) + \left ( \frac{1}{x^2}+2 \right ) \cdot 1$
$\displaystyle f'(x) = \frac{-2x + 3}{x^3} + \frac{1+2x^2}{x^2}$
etc, etc...

-Dan
• Jul 20th 2006, 01:51 PM
topsquark
Quote:

Originally Posted by NeF
I want to double check these answers, this book doesnt supply answers.

3
$\displaystyle y= \frac{2\sqrt(x)}{x}$
$\displaystyle y= 2x^\frac{-1}{2}$
$\displaystyle \frac{dy}{dx} = -x^\frac{-3}{2}$
$\displaystyle = -\frac{1}{x^\frac{3}{2}}$

This one looks okay. By the way, note the use of the {} in the Tex in the preceeding messages.

-Dan
• Jul 20th 2006, 01:52 PM
topsquark
Oh! Hi there Captain! :D

-Dan
• Jul 23rd 2006, 08:34 PM
NeF
Hectic, Thanks alot again for seeing all the stupid mistakes I am making. :D
• Jul 24th 2006, 11:04 AM
ThePerfectHacker
Quote:

Originally Posted by NeF
Hectic, Thanks alot again for seeing all the stupid mistakes I am making.

Does that mean you are going to kill yourself now!
• Jul 24th 2006, 11:07 AM
NeF
No sadly I am not :p , I want to waste more of your precious air...

That line refers to morons that do drugs and crap.
• Jul 24th 2006, 11:17 AM
ThePerfectHacker
Quote:

Originally Posted by NeF
That line refers to morons that do drugs and crap.

Then why do you have a picture of Raiden over there? You clan thing?
• Jul 24th 2006, 11:30 AM
NeF
Yep thats for my clan