Question

Find the derivative of the function g(x)=(sqrt3-x) directly from the definition. (For this question, are we using the limits?)

Answer 1

is it sqrt(3-x)?

Answer 2

\[g(x)=\sqrt{3-x}\]

Answer 3

yeap as h -> 0

Answer 4

okay..lets get to work

Answer 5

So far I got this:
\[\lim_{h \rightarrow 0}\frac{ \sqrt{3-(x+h)} -\sqrt{3-x} }{ h }\]

Answer 6

I've gotten that much as well

Answer 7

hmmm I don't see it coming out cleanly

Answer 8

hmmmm...

Answer 9

I am hanging on other question still so I will see how it goes.

Answer 10

hmmm.....

Answer 11

\(\large\color{slate}{\displaystyle\lim_{h \rightarrow ~0}\frac{\sqrt{3-(x+h)}-\sqrt{3-x}}{h}}\)

my first thought is to rationalize the numerator

Answer 12

for any (a-b) conjugate is (a+b).....

Answer 13

you don't want me to do this problem for you, do you ?

Answer 14

I did try rationalizing the numerator, but I'm not getting it correctly. If you don't mind, can you show me how?

Answer 15

\(\large\color{black}{ \displaystyle \large\color{slate}{\displaystyle\lim_{h \rightarrow ~0}\frac{\sqrt{3-(x+h)}-\sqrt{3-x}}{h}} }\)

\(\large\color{black}{ \displaystyle \large\color{slate}{\displaystyle\lim_{h \rightarrow ~0}\frac{ \color{red}{(}\sqrt{3-(x+h)}-\sqrt{3-x}\color{red}{)~\times (\sqrt{3-(x+h)}+\sqrt{3-x})} }{h\color{red}{~\times (\sqrt{3-(x+h)}+\sqrt{3-x})} }} }\)

Answer 16

see how I am multiplying times a conjugate on top and bottom ?

Answer 17

yes..

Answer 18

can you simplify the top for me please?

Answer 19

what if it was [ sqrt(A)-sqrt(B) ] * [ sqrt(A) + sqrt(B) ] , what would you then tell me ?

Answer 20

The square roots get cancel out

Answer 21

and what would you then be getting ?

Answer 22

(A-B)(A+B)

Answer 23

please try again

Answer 24

(g-h)(g+h)=g^2-h^2
is this correct ?

Answer 25

No....

Answer 26

that is a rule, called a difference of squares.

Answer 27

you can prove why it is true, simply be expanding and canceling out the terms... you will end up with g^2-h^2.

Answer 28

Oh yeah... Sorry I forgot

Answer 29

so then,
\(\large\color{black}{ \displaystyle (\sqrt{A}+\sqrt{B})(\sqrt{A}-\sqrt{B})=(\sqrt{A~})^2-(\sqrt{B~})^2}\)

Answer 30

am I correct ?

Answer 31

Yes. final anwer is A-B

Answer 32

yes, so
\(\large\color{black}{ \displaystyle (\sqrt{a}-\sqrt{b})(\sqrt{a}+\sqrt{b})=a-b }\)

Answer 33

Now, look at this (where all this discussion about difference of squares started).

\(\large\color{black}{ \displaystyle \large\color{slate}{\displaystyle\lim_{h \rightarrow ~0}\frac{ \color{red}{(}\sqrt{3-(x+h)}-\sqrt{3-x}\color{red}{)~\times (\sqrt{3-(x+h)}+\sqrt{3-x})} }{h\color{red}{~\times (\sqrt{3-(x+h)}+\sqrt{3-x})} }} }\)

Answer 34

do you see this \(\large\color{black}{ \displaystyle (\sqrt{a}-\sqrt{b})(\sqrt{a}+\sqrt{b}) }\) form in the numerator ?

Answer 35

Yes. It would be 3-(x+h)-3-x?

Answer 36

yes, there you go....
and that simplifies to what ?

Answer 37

oh, wait you are wronf

Answer 38

3-(x+h)-(3-x)

like this.

Answer 39

see why is it like this with the parenthesis ?

Answer 40

you are subtracting the entire b, and your entire b is 3-x, (not just x or just 3)
so you are subtracting the (3-x).

and that is why we get

a - b

(3-(x+h)) - (3-x)

Answer 41

understand why we need to write in parenthesis 3-(x+h)-(3-x) ?

Answer 42

Yes... We need to distribute negative sign

Answer 43

@SolomonZelman .....wowww...I GOT IT..:)...cool man..

it gets simplified to -h...

h and h get cancelled...

apply de limits....

the answer comes...:)

Answer 44

yes, sidarth, then you plug in zero for h (which is valid from that point - i.e. without making denom. undefined)

Answer 45

3-(x+h)-(3-x) simplifies to ?

Answer 46

@SolomonZelman ...yeah...:)

Answer 47

Oh! I got the final answer! \[\frac{ -1 }{ 2\sqrt{3-x} }\]

Answer 48

\(\large\color{black}{ \displaystyle \large\color{slate}{\displaystyle\lim_{h \rightarrow ~0}\frac{ \color{red}{(}\sqrt{3-(x+h)}-\sqrt{3-x}\color{red}{)~\times (\sqrt{3-(x+h)}+\sqrt{3-x})} }{h\color{red}{~\times (\sqrt{3-(x+h)}+\sqrt{3-x})} }} }\)

\(\large\color{black}{ \displaystyle \large\color{slate}{\displaystyle\lim_{h \rightarrow ~0}\frac{ -h }{h\color{red}{~\times (\sqrt{3-(x+h)}+\sqrt{3-x})} }} }\)

\(\large\color{black}{ \displaystyle \large\color{slate}{\displaystyle\lim_{h \rightarrow ~0}\frac{ -1 }{\sqrt{3-(x+h)}+\sqrt{3-x} }} }\)

\(\large\color{black}{ \displaystyle =\large\color{slate}{\displaystyle\frac{ -1 }{\sqrt{3-(x+0)}+\sqrt{3-x} }} }\)

\(\large\color{black}{ \displaystyle =\large\color{slate}{\displaystyle\frac{ -1 }{\sqrt{3-x}+\sqrt{3-x} }} }\)

\(\large\color{black}{ \displaystyle =\large\color{slate}{\displaystyle\frac{ -1 }{2\sqrt{3-x} }} }\)

Answer 49

coirrect

Answer 50

(( Same way you would normally differentiate and do the chain rule for "3-x". ))

Answer 51

oh okay! Thanks for explaining it to me! ^_^

Answer 52

\(\large\color{black}{ \displaystyle \frac{d}{dx}\left[ \sqrt{3-x}~\right] ~=~(3-x)^{1/2 \color{red}{~~-1}}~\color{blue}{ \times \frac{d}{dx}\left[ 3-x\right]} }\)

\(\large\color{black}{ \displaystyle \frac{d}{dx}\left[ \sqrt{3-x}~\right] ~=~(3-x)^{-1/2 }~\color{blue}{ \times (-1)} }\)

\(\large\color{green}{ \displaystyle \frac{d}{dx}\left[ \sqrt{3-x}~\right] ~=~\frac{-1}{\sqrt{3-x}}}\)

Answer 53

in red I decremented the power by 1, and blue is the chain rule for the part inside the square root.

Answer 54

yw

Answer 55

do you know derivatives though, have you learned them ?

Answer 56

anyway... have fun (I guess)