Posted by John on Monday, April 28, 2008 at 12:11am.
I assume you are referring to
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
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
f'(x) = -5x^-2 + 8x^-5
= -5/x^2 + 8/x^5
getting a common denominator and simplifying gives you the same answer of
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
f'(x) = (1/2)x^(-3/2)[6x^2 + 7x - 3]
= (6x^2 + 7x - 3)/(2x^(3/2))
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