Differentiation question

• Dec 10th 2008, 05:12 PM
timmy420
Differentiation question
Im reviewing for an exam and cant figure out this question..

what is y' of x^2/2 + 2/x^2 +pi (not sure pi symbol on here)

I know that pi is a constant so its prime is 0, and because this is an old test I know the answer is x^4-4/x^3, getting to that is another problem

the second term i changed to 2x^-2, which gives me the x^3 in the bottom of the answer, but i cant figure this out.

Another that is stumping me is y' of (1+2x)e^3x

Any help?
• Dec 10th 2008, 05:16 PM
fastcarslaugh
$y= x-4x^{(-3)}
$

I think
• Dec 10th 2008, 05:22 PM
timmy420
I have the answer the proff put on the old exam (answers only, no workings)

x^4 - 4
---------
x^3

I dont see where the x^4 came from

also how do you get your text to come up in the neat looking notation?
• Dec 10th 2008, 05:27 PM
fastcarslaugh
you rewrite the equation to $

y = \frac {x^{2}}{2} + 2x^{-2} + pi$

derivative of sum is sum of derivatives

so you derive each one.

As for the math notation stuff, just click that sigma next to youtube icon
I'm still new to this stuff
• Dec 10th 2008, 05:31 PM
timmy420
I was able to work it that far, and I see how -4 is part of the top, and x^3 is part of the bottom, I just cant figure how x^2/2 works out to be x^4. the derrivitive of x^2 should be 2x shouldnt it?

I just worked through the second question I posted (1+2x)e^3x, simple once i remembered to factor out the e^3x's
• Dec 10th 2008, 05:36 PM
fastcarslaugh
$y = 2x\frac{1}{2} + -4x^{-3}$

see how $2$ and $\frac {1}{2}$ cancel out.
• Dec 10th 2008, 05:46 PM
timmy420
Quote:

Originally Posted by fastcarslaugh
$y` = 2x\frac{1}{2} + -4x^{-3}$

see how $2$ and $\frac {1}{2}$ cancel out.

Thats what I keep working it out to as well. However the solution says it should be a x^4 as the first term, but best I can figure is x-4
----
x^3
• Dec 10th 2008, 05:54 PM
Ziaris
Quote:

Originally Posted by timmy420
Thats what I keep working it out to as well. However the solution says it should be a x^4 as the first term, but best I can figure is x-4
----
x^3

$\frac{x^4-4}{x^3}=\frac{x^4}{x^3}-\frac{4}{x^3}=x-4x^{-3}$. This is what you got, just in different form.