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January 31, 2015

January 31, 2015

Posted by **John** on Monday, April 28, 2008 at 12:11am.

1. If f(x)=5x^3–2/x^4

find f'(x)____

what i got was (4x^2+7x-12)/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 8:52amI 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^3-2)(4x^3)]/x^8

take out a common factor of x^3 from the top

= x^3(15x^3 - 4(5x^3-2))/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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