Posted by John on Monday, April 28, 2008 at 12:11am.
I have 3 calc problems that i have not been able to do please help me... i've been on here yesterday as well but no one was able to assist me with these particular ones...
1. If f(x)=5x^3–2/x^4
find f'(x)____
what i got was (4x^2+7x12)/x^8
2. The position function of a particle is given by s=t^3–4t^2–7t t>=0
where s is measured in meters and t in seconds. When does the particle reach a velocity of 10 m/s? ____
3. f(x)=2x^2+7x+3/ sqroot of x
Find f'(x)
i got ((x^(1/2))(4x+7)+1/2x^(3/2)(2x^2+7x+3))/x
i had asked earlier and Damon answered them but it turned out to be wrong

Calculus To Reiney  Reiny, Monday, April 28, 2008 at 8:52am
I assume you are referring to
http://www.jiskha.com/display.cgi?id=1209342655http://www.jiskha.com/
Damon answered your #1 question correctly according to the way you typed it, but now your denominator is different, so which way is it?
If you want it the way you typed it this time, f(x) = If f(x)=5x^3–2/x^4
then f'(x) = [x^4(15x^2)  (5x^32)(4x^3)]/x^8
take out a common factor of x^3 from the top
= x^3(15x^3  4(5x^32))/x^8
= (5x^2+8)/x^5
another way to do this is to rewrite the original
If f(x)=5x^3–2/x^4 as
f(x) = 5x^1  2x^4
then
f'(x) = 5x^2 + 8x^5
= 5/x^2 + 8/x^5
getting a common denominator and simplifying gives you the same answer of
(5x^3+8)/x^5
In the second question Damon made a typo by setting the derivative equal to 7 instead of 10.
If you had looked at his solution you should have noticed that, and should have been able to make the necessary changes.
for your third I suggest you write it as
f(x)=2x^2+7x+3/ sqroot of x
= f(x)=2x^2+7x+3/ x^(1/2) then do what I did in #1
so f(x) = 2x^(3/2) + 7x^(1/2) + 3x^(1/2)
f'(x) = 3x^(1/2) + (7/2)x^(1/2)  (3/2)x^(3/2)
I see (1/2)x^(3/2) as a common factor
so
f'(x) = (1/2)x^(3/2)[6x^2 + 7x  3]
= (6x^2 + 7x  3)/(2x^(3/2))
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