Question

Calc 1 question...need to find derivative of 1/sqrt x
I know the answer is -(1/2x^3/2) but no clue how to show the work - I suck when radicals and fractions together. I need to use the (f(x+h)-f(x))/h method. Any help would be greatly appreciated =]

Answer 1

you have to do this by hand right? using
\[\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}\]?

Answer 2

yeah :)

Answer 3

I've done like 20 problems already but none involved a radical/fraction combo like this

Answer 4

start out with
\[f(x+h)-f(x)=\frac{1}{\sqrt{x+h}}-\frac{1}{\sqrt{x}}\]

Answer 5

then subtract, just like you would with normal fractions to get
\[\frac{\sqrt{x}-\sqrt{x+h}}{\sqrt{x}\sqrt{x+h}}\]

Answer 6

now rationalize the numerator, multiply top and bottom by \(\sqrt{x}+\sqrt{x+h}\) leave the denominator in factored form

Answer 7

getting a bit confused reading the text as opposed to seeing it on paper, would you mind if i upload image of what I have down so far on paper? I think I've done what you have said but stuck on last part rationalizing the numerator

Answer 8

by which i mean write
\[\frac{(x-x-h)}{\sqrt{x}\sqrt{x+h}(\sqrt{x}+\sqrt{x+h})}\]

Answer 9

i hope that last step was clear
then \(x-x=0\) so numerator is just \(-h\)
divide all this by \(h\) and the numerator becomes \(-1\)

Answer 10

then replace \(h\) by 0 and you are done

Answer 11

since that picture i multipled top and bottom by sqrt (x+h) +sqrt x

Answer 12

that is the step before you rationalize the numerator

Answer 13

ok right, then multiply as you said

Answer 14

but i would write it as multiplying by \(\sqrt{x}+\sqrt{x+h}\) of course it doesn't matter, but it is easier to see what you will get

Answer 15

got yah

Answer 16

okay so the top should be canceled out now with rationalizing...just confused on what the bottom should look like now

Answer 17

wish there was a way for both people to draw on here so I could see in real time...deciphering the steps is throwing me off a little

Answer 18

are you having trouble viewing the code? it should be look just like math

Answer 19

yeah its showing up as \frac{(x-x-h)}{\sqrt{x}\sqrt{x+h}(\sqrt{x}+\sqrt{x+h})}\]

Answer 20

\[\frac{(x-x-h)}{\sqrt{x}\sqrt{x+h}(\sqrt{x}+\sqrt{x+h})}\]is what you should have, without the \(h\) in the denominator
you get
\[\frac{-1}{\sqrt{x}\sqrt{x+h}(\sqrt{x}+\sqrt{x+h})}\]

Answer 21

refresh your browser

Answer 22

now wonder it was hard to read

Answer 23

wow there we go lo

Answer 24

now I can see, let me try now :D

Answer 25

now it should be clear
review the steps, i think they are all there

Answer 26

just unclear how you got (x-x-h) on the numerator? I thought when you rationalize the numerator it becomes 1?

Answer 27

\[(\sqrt{x}-\sqrt{x+h})(\sqrt{x}+\sqrt{x+h})=x-(x+h)=-h\]

Answer 28

then when you divide by \(h\) you get a \(-1\) in the numerator

Answer 29

ah okay I see now, wow thank you so much for the help...just started calculus 1 last week after just having finished pre-cal this summer so this derivative stuff is killing me :(...always sucked with radicals aswell

Answer 30

is there a reason though that wolfram shows the answer as\[-(\frac{ 1 }{ 2x ^{3/2} })\]

Answer 31

that is the answer

Answer 32

you end up with
\[\frac{-1}{\sqrt{x}\sqrt{x+h}(\sqrt{x}+\sqrt{x+h})}\] and then let \(h=0\) and get
\[\frac{-1}{\sqrt{x}\sqrt{x}(\sqrt{x}+\sqrt{x})}\]
\[\frac{-1}{2x\sqrt{x}}=-\frac{1}{2x^{\frac{3}{2}}}\]

Answer 33

oh! okay

Answer 34

thank you very much :D