Question

Never been sure how to handle double square roots. d/dx.

Answer 1

\[\frac{ dy }{ dx }(\sqrt{x+\sqrt{x}}\]

Answer 2

I assume chain rule as usual, and outer function is sqrt(u)

Answer 3

yup

Answer 4

\[\frac{1}{2\sqrt{x+\sqrt{x}}}\times \frac{d}{dx}[x+\sqrt{x}]\]

Answer 5

Correct.

Note that the derivative of \(\sqrt{x}\) is \(\dfrac{d}{dx} x^{\frac{1}{2}} = \dfrac{1}{2} x^{-\frac{1}{2}} = \dfrac{1}{2}\cdot \left(\dfrac{1}{\sqrt{x}}\right) = \dfrac{1}{2\sqrt{x}}\)

Answer 6

So then.

\[\sqrt{x+\sqrt{x}}(1+\frac{ 1 }{ 2 }x^-1/2)\]

Answer 7

forget that damned one half!!

Answer 8

forget it forget it forget it
what is \(8\times 7\) ?

Answer 9

O_o

Answer 10

Okay uhh

Answer 11

this simple point i am trying to make is this: if you can remember \(8\times 7=56\) then you can remember that \(\frac{d}{dx}[\sqrt{x}]=\frac{1}{2\sqrt{x}}\)

Answer 12

Yeah i got that part now. But for the end of the chain rule the g'(x). How do i take the derivative of x+sqrt(x)

Answer 13

so while your class mates are writing nonsense like
\[\sqrt{x}=x^{\frac{1}{2}}\]
\[\frac{d}{dx}x^{\frac{1}{2}}=\frac{1}{2}x^{\frac{1}{2}-1}=...\] you are done in a nano second

Answer 14

Cant you just....\[\large \frac{d}{dx}[x+x^{1/2}]^{1/2} \cdot \frac{d}{dx}(x^{1/2})\]?..

Answer 15

:'\

Answer 16

the derivative of \(x\) is 1 and the derivative of \(\sqrt{x}\) is ...

Answer 17

really just memorize it, like you know that \(a^2-b^2=(a+b)(a-b)\) it never changes, no need to reinvent the wheel every time

Answer 18

FASTERRR @SgtSausage

Answer 19

I was just telling him to note down that \(\sqrt{x} = \dfrac{1}{2\sqrt{x}}\) so he doesn't have to write it again-and-again. I wrote the intermediate steps too.

Answer 20

the derivative of the square root of x is one over two square root of x

Answer 21

Yeah i'm starting to catch on.

Answer 22

so if \[\large \sqrt{x} = \frac{1}{2\sqrt{x}}\] if we take \[\large u= x+\sqrt{x}\] then wouldnt we get the same thing? \[\large \frac{d}{dx} (u)^{1/2} = \frac{1}{2\sqrt{u}}\]

Answer 23

Correct so just add that on to the end multiplied by

Answer 24

\[\frac{ 1 }{ 2\sqrt{x+\sqrt{x}} }(\frac{ 1 }{ 2\sqrt{x} })\]

Answer 25

Correct? I assume that can simplify as well.

Answer 26

Well wait they both aren't u... One is u the other is the derv of u??

Answer 27

Thats the chain rule....

Answer 28

\[\large \frac{1}{2\sqrt{u}}\cdot \frac{1}{2\sqrt{x}}\]

Answer 29

I typoed.

Answer 30

I know its fine. no need to fix. Just am i right what i posted above ^^

Answer 31

Yep.

Answer 32

And can it simplify more?

Answer 33

Yep.

Answer 34

Thats where i'm stuck

Answer 35

oh no!

Answer 36

careful here

Answer 37

then "inside" function is not \(\sqrt{x}\) it is \(x+\sqrt{x}\)

Answer 38

Okay that makes no sense, why does it jump inside the other square root????

Answer 39

Oh i forgot to take the derivative once more i believe.

Answer 40

let me write it out
\[\frac{d}{dx}[\sqrt{x+\sqrt{x}}]=\frac{1}{2\sqrt{x+\sqrt{x}}}\times \frac{d}{dx}[x+\sqrt{x}]\]
\[=\frac{1}{2\sqrt{x+\sqrt{x}}}\times \left(1+\frac{1}{2\sqrt{x}}\right)\]

Answer 41

Yeah...i messed up.\[\large \frac{1}{2\sqrt{u}}\cdot \frac{d}{dx}(u)' \cdot \frac{1}{2\sqrt{x}}\]

Answer 42

\[\frac{d}{dx}[\sqrt{f(x)}=\frac{1}{2\sqrt{f(x)}}\times f'(x)\]

Answer 43

Yeah i follw you satelight. Just asking how to simplify the end there.

Answer 44

@Jhannybean that was not the mistake
the mistake was forgetting that the inside piece is \(x+\sqrt{x}\) so the derivative is \(1+\frac{1}{2\sqrt{x}}\) rather than just \(\frac{1}{2\sqrt{x}}\)
only the 1 was missing

Answer 45

everything else you had was correct
just forgot the 1

Answer 46

Yeah i have it correct on my paper now. No worries guys i understand, just trying to simplify. Gets pretty messy :/

Answer 47

That's what the u prime stood for. the derivative of [x+sqrt{x}]

Answer 48

i wouldn't mess with "simplifying" it
leave it alone, it is simple enough as it is

Answer 49

Oh i didn't think i could leave it like that. Thanks you're right. : ) Appreciate all the help everyone.

Answer 50

: ) Thanks again

Answer 51

Np

Answer 52

Um actually i'm pretty sure thats not right. You added another term. The final 1/2(sqrt x) shouldn't be there.

Answer 53

Oh you know what, i thought there was anotehr sqrt{x} there lol nvm!!

Answer 54

haha its all good, i got the right answer now and understand it.!

Answer 55

YAY!!!