Question

If f(x) = (6x^2 +5x+7)/square root of x, what is f'(x)?

Answer 1

now you have to take the derivative of f(x)
can you do it using
f'(x) = (u'v - v'u) / v^2
given that f(x) = u/v
?

Answer 2

I don't think so, I've never seen that before!

Answer 3

What I keep doing is changing the square root of x to x^-1/2, then bring it up, then I have (6x^2+5x+7)(x^-1/2) but then I must be doing something wrong because everything I try says its the wrong answer!

Answer 4

ok you can do it this way as well ..
lets to it together

Answer 5

If you do that, then you'll have to use the product rule... but you can use Quotient Rule.

Answer 6

I've tried distributing thex^-1/2, I've tried just finding the derivatives of each separate parentheses, but they don't work each time!

Answer 7

(6x^2+5x+7)(x^-1/2) = 6x^1.5 + 5x^0.5 + 7x^-0.5

Answer 8

I don't think we've learned the quotient rule, we've learned the sum rule, the general power rule, and the constant multiple rule!

Answer 9

\[\left({f \over g} \right)^\prime = {f'g - fg' \over g^2}\]

Answer 10

its ok.. you can do it as you said anlnt4 ..

Answer 11

when you distributing you get : (6x^2+5x+7)(x^-1/2) = 6x^1.5 + 5x^0.5 + 7x^-0.5 ?

Answer 12

now take the derivative of this

Answer 13

when I do that I get 9x^.5+5x^-.5+7x^-.5?

Answer 14

WAIT NO

Answer 15

waiting :P

Answer 16

I get 9x^.5+2.5x^-.5+3.5x^-.5 right?

Answer 17

your last term is incorrect ..

Answer 18

to the -1.5?

Answer 19

yes and it should be -3.5

Answer 20

be careful when you use the power rule with negative numbers ..

Answer 21

so it should be -3.5x^-1.5?

Answer 22

yes

Answer 23

9x^.5+2.5x^-.5-3.5x^-1.5

Answer 24

i think its good

Answer 25

IT IS! THANK YOU SOO MUCH. Actually I have one more problem if you wouldn't mind helping me on that one too? It seems you know your stuff!

Answer 26

im sorry but i have to go now .. :\

Answer 27

Alright that's fine! thanks anyways:)

Answer 28

yw