Question

USE DEFINITION OF DERIVATIVE to find f^(')x of the following..

Answer 1

\[f(x)= \frac{ 3x }{\sqrt{x+4}}\]

Answer 2

\[\frac{1}{h}\left(\frac{3(x+h)}{\sqrt{x+h+4}}-\frac{3x}{\sqrt{x+4}}\right)\] this is really going to suck

Answer 3

i have done it over about a thousand times and keep screwing it up…

Answer 4

you sure you have to use the definition? it is annoying enough using the quotient rule

Answer 5

yeah i have to use it. and whats so weird is the other questions he gave us to do with def'n were actually simple.

Answer 6

god ok lets see if we can do the algebra
lets ignore the \(\frac{1}{h}\) and put it back in when we can cancel it

Answer 7

\[\frac{3(x+h)}{\sqrt{x+h+4}}-\frac{3x}{\sqrt{x+4}}=\frac{3(x+h)\sqrt{x+4}-3x\sqrt{x+h+4}}{\sqrt{x+h+4}\sqrt{x+4}}\] is a good first step

Answer 8

ok i got that

Answer 9

lets not multiply out, and go straight to multiplying top and bottom by the conjugate \[3(x+h)\sqrt{x+4}+3x\sqrt{x+h+4}\]

Answer 10

leave the denominator in factored form

Answer 11

so i'm basically ignoring the denominator h in definition right now right?

Answer 12

yeah all the work is going to be up top

Answer 13

eventually we should get a factor of \(h\) in the numerator which will cancel the \(h\) in the denominator

Answer 14

if my algebra is right, when you multiply by the conjugate you should get
\[9(x+h)^2(x+4)-9x^2(x+h+4)\]

Answer 15

now more algebra to square
\[(9x^2+18xh+9h^2)(x+4)-9x^3-9x^2h-36x^2\]

Answer 16

yes, i got the same answer

Answer 17

ok this might work
now more multiplication
\[9x^3-18x^2h+9xh^2+36x^2+72xh+36h^x-9x^3-9x^2h-36x^2\]

Answer 18

damn typo!\[9x^3+18x^2h+9xh^2+36x^2+72xh+36h^x-9x^3-9x^2h-36x^2\]

Answer 19

now everything without an \(h\) should go away

Answer 20

and it does!

Answer 21

oh i see another typo
\[9x^3+18x^2h+9xh^2+36x^2+72xh+36h^2-9x^3-9x^2h-36x^2\]

Answer 22

i think i get
\[9x^2h+9xh^2+72xh+36h^2\] let me know if that looks good

Answer 23

cancel an \(h\) to get rid of the one in the denominator and get
\[9x^2+9xh+72x+36h\]

Answer 24

ok working through it… not as fast as you.. haha
give me a minute

Answer 25

take you time
we still have the denominator to compute

Answer 26

\[(\sqrt{x+h+4}\sqrt{x+4}\left(3(x+h)\sqrt{x+4}+3x\sqrt{x+h+4}\right)\] is the denomiator
then replace \(h\) by \(0\) and see what you get

Answer 27

actually that part is easy
it is
\[\sqrt{x+4}\sqrt{x+4}\left(3x\sqrt{x+4}+3x\sqrt{x+4}\right)\]
or
\[(x+4)(6x\sqrt{x+4})\]

Answer 28

so as soon as i get rid of h in numerator and denominator i sub in limit number right?

Answer 29

yes
i.e. replace all other \(h\)'s by \(0\)

Answer 30

all that is left in the numerator is \(9x^2+72x\)

Answer 31

gotta run
i think this is right
good luck

Answer 32

i got it pretty much thanks so much!!!

Answer 33

\[\frac{9x^2+72x}{6(x+4)\sqrt{x+4}}\] you can cancel a 3
maybe there is algebra mistake, but it is more or less right

Answer 34

yw