Question

Im pretty sure this might be simple but i tend to struggle with equations that include square roots so if anyone can help me thatd be lovely please...step by step if possible and explain if possible...

Use Implicit Differentiation to find the derivative

\[\sqrt{x}+\sqrt{y}=1\]

please help...

Answer 1

So for square root, recall that you can rewrite it with a rational exponent,\[\large\rm \sqrt{x}=x^{1/2}\]Applying power rule shows us how to differentiate this,\[\large\rm \frac{d}{dx}x^{1/2}\quad=\frac{1}{2}x^{-1/2}\quad=\frac{1}{2x^{1/2}}\quad=\frac{1}{2\sqrt x}\]

Answer 2

But, I strongly recommend that you take the time to memorize this derivative.
Square root shows up sooo often, just put it to memory.

Derivative of square root is `one over two square roots`
\[\large\rm \frac{d}{dx}\sqrt x=\frac{1}{2\sqrt x}\]

Answer 3

\[\large\rm \sqrt{x}+\sqrt{y}=1\]So we're going to differentiate each side with respect to x.

Answer 4

\[\large\rm \frac{d}{dx}\left(\sqrt{x}+\sqrt{y}\right)=\frac{d}{dx}1\]What do you get on the right side? :)

Answer 5

i actually knew the rational exponent but never figured that the derivative went to \[\frac{ 1 }{ 2\sqrt{x} }\] thank you..and do you think that maybe you could possibly explain by differentiating y instead? its only because thats how i need to do it and i dont wish to become confused but if not its okay @zepdrix

Answer 6

The `chain rule` is the most difficult of the basic differentiation rules to get a grasp of. Make sure you spend a lot of time practicing it.

Realize, that will any problem, we can always apply the chain rule up to the variable of differentiation.

Example:\[\large\rm \frac{d}{dx}(x)^2\]The power rule tells me how to differentiate, but I still need to apply chain rule to the inner function after that,\[\large\rm \frac{d}{dx}(x)^2\quad=2(x)^1\frac{d}{dx}x\]We can also think of this result in this way,\[\large\rm \frac{d}{dx}(x)^2\quad=2(x)^1\frac{dx}{dx}\]But the problem is,
dx/dx has no real significance.
It's measuring how much x changes, as x changes.\[\large\rm \frac{d}{dx}(x)^2\quad=2(x)^1(1)\]So we get this result, ya?\[\large\rm \frac{d}{dx}x^2=2x\]Hopefully this weird explanation isn't too long-winded :)
But I feel this might help understand why we end up with these weird dy/dx looking things when we do implicit differentiation.

So notice that when we differentiate something involving x,
with respect to x,
that very last chain is completely unnecessary.

Answer 7

That is not the case when differentiating something involving y though.

Answer 8

\[\large\rm \frac{d}{dx}\sqrt{x}\quad=\frac{1}{2\sqrt x}\frac{d}{dx}x\]\[\large\rm \frac{d}{dx}\sqrt{x}\quad=\frac{1}{2\sqrt x}\cdot 1\]

Answer 9

Same process, but now this chain produces something important that doesn't simply go away.\[\large\rm \frac{d}{dx}\sqrt{y}=\frac{1}{2\sqrt{y}}\frac{d}{dx}y\]

Answer 10

\[\large\rm \frac{d}{dx}\sqrt{y}=\frac{1}{2\sqrt{y}}\frac{dy}{dx}\]

Answer 11

Questions? :)
It can be a lot to take in heh

Answer 12

No no you're good i am following what you are saying c: its after differentiating to the equation you displayed above that i have a problem isolating \[\frac{ dy }{ dx }\] to solve for it

Answer 13

Oh i see :3

Answer 14

Did you remember to differentiate the right side?
Derivative of our constant 1 gives us 0, ya?

Answer 15

Yes c:

Answer 16

\[\large\rm \sqrt{x}+\sqrt{y}=1\]\[\large\rm \frac{1}{2\sqrt x}+\frac{1}{2\sqrt y}\frac{dy}{dx}=0\]So this is just basic Algebra from this point.
We have a few different routes we can take.

Answer 17

I don't like fractions,
I would probably multiply both sides by the stuff in the denominators,\[\large\rm 2\sqrt x \sqrt y\left(\frac{1}{2\sqrt x}+\frac{1}{2\sqrt y}\frac{dy}{dx}\right)=0\]\[\large\rm \sqrt y+ \sqrt x \frac{dy}{dx}=0\]Then how be subtract sqrt y to the other side,\[\large\rm \sqrt x\frac{dy}{dx}=-\sqrt y\]And then divide the sqrt x over, ya? :)

Answer 18

Then how bout* we* subtract sqrt y

wow those typos today ;c

Answer 19

So then the answer becomes \[\frac{ -\sqrt{y} }{ \sqrt{x}}\]?

Answer 20

Yes, good.
And if you like, you can combine the roots,\[\large\rm \frac{dy}{dx}=-\sqrt{\frac{y}{x}}\]

Answer 21

Holy cheese its! thats amazing, you have no idea how thankful i am for your guidance because i honestly struggle so much in calculus because of this and tomorrow we have a test over implicit differentiation and im pretty sure square root will show up and i will be eternally grateful because i can understand the concept of it ^.^

Answer 22

np,
yay we did it \c:/

Answer 23

tomorrow?? :O
Oh boy, keep studying!

Answer 24

Are you comfortable with your chain rule?

Like,
do you understand how to find the derivative of something like this?\[\large\rm \frac{d}{dx}\sqrt{x^2+2x}\]

Answer 25

i know ive been studying the worksheets we received and it has the answers on it but i like to understand how a person gets the answer so i learn as well and well now i understand one of the problems c: Thanks to you

Answer 26

And sadly...no. i failed that test horrendously so im still trying to work on it its only cause the class is fast paced and well she doesnt necessarily teach during tutoring either so i tend to ask here and there but still struggle for the most part with chain rule

Answer 27

So we decided that the derivative of sqrt(x) is this,\[\large\rm \frac{d}{dx}\sqrt{x}\quad=\frac{1}{2\sqrt x}\]When the "stuff" under the root is more than just x, we `must` apply chain rule.\[\large\rm \frac{d}{dx}\sqrt{stuff}\quad=\frac{1}{2\sqrt{stuff}}\cdot(stuff)'\]multiplying by the derivative of the inner function.

Answer 28

Okay im following slightly...proceed.

Answer 29

So we apply square root derivative, and chain,\[\large\rm \frac{d}{dx}\sqrt{x^2+2x}=\frac{1}{2\sqrt{x^2+2x}}\cdot (x^2+2x)'\]The square root of the stuff turns into one over two square roots of the stuff.
Chain rule tells us to make a copy of the inner function,
take its derivative,
and multiply.

That prime that I wrote shows that we still need to take that derivative.
The prime is sort of like ... "setting up" the chain.
This should feel familiar if you've gone over Product and Quotient Rules at this point. "setting up" the derivative can be very helpful before actually taking them.

To resolve the chain, we just power rule a couple times, ya?\[\large\rm \frac{d}{dx}\sqrt{x^2+2x}=\frac{1}{2\sqrt{x^2+2x}}\cdot (2x+2)\]

Answer 30

Okay that makes sense c:

Answer 31

Wait..so if were using power rule to find the derivative of the copy of the inner function do we also do the same to the original inner function(underneath the square root)?

Answer 32

Nooo :O
When differentiating "the outside",
the inside of THAT doesn't change.

Good question though.

Answer 33

Okay cool, just verifying c:

Answer 34

So...once you take the derivative of the "outside" what follows in order to actually solve the equation?

Answer 35

Its the solving part that gets me because i dont necessarily like square root and fractions applied together...they're my weakness with implicit -.-

Answer 36

Well this chain rule is producing these derivative things, y' or dy/dx if you like.
With implicit differentiation, your goal is to solve for this y'.

It's going to seem extra confusing when you have MULTIPLY y' floating around.
Example:

Answer 37

\[\large\rm \sqrt{x^2+2x}+y^2=y+1\]Differentiating with respect to x,\[\large\rm \frac{2x+2}{2\sqrt{x^2+2x}}+2yy'=y'+0\]The first term is just the example from before.
Now you're expected to solve for this y'.
You need to apply some fancy Algebra skills.
Let's subtract 2yy' from each side,\[\large\rm \frac{2x+2}{2\sqrt{x^2+2x}}=y'-2yy'\]And then factor a y' out of each term on the right,\[\large\rm \frac{2x+2}{2\sqrt{x^2+2x}}=y'(1-2y)\]And then divide both sides by (1-2y),\[\large\rm \frac{2x+2}{2(1-2y)\sqrt{x^2+2x}}=y'\]That would be an example of "solving".

Answer 38

Wow! o.O Thank you, how do you understand all of this? isnt it confusing because i know for me it truly is

Answer 39

Haha :)
I'm math major, I really truly miss this fun easy math.
It becomes a lot less fun as you go on further XDD

Answer 40

If you're not a math major, then it will probably always be a little bit weird.
But after you've done tons and tons and tons of problems,
these steps seem really simple after a while ^^

Answer 41

Youre a Math major? Man no wonder this is so simple to you XD I still got a long way for the harder stuff to come but yes! Alot of practice it shall take

Answer 42

I have another example just in case you'd like to see :3

Answer 43

Example 2:\[\large\rm \sqrt{x^2+y^2}=y-1\]Differentiating, setting up our chain rule,\[\large\rm \frac{1}{2\sqrt{x^2+y^2}}(x^2+y^2)'=y'-0\]resolving chain rule gives us,\[\large\rm \frac{1}{2\sqrt{x^2+y^2}}(2x+2yy')=y'-0\]Divide the 2's out maybe,\[\large\rm \frac{1}{\sqrt{x^2+y^2}}(x+yy')=y'-0\]Distribute the fraction to each term in the brackets,\[\large\rm \frac{x}{\sqrt{x^2+y^2}}+\frac{y}{\sqrt{x^2+y^2}}y'=y'\]Again let's subtract the (stuff)y' term to the other side,\[\large\rm \frac{x}{\sqrt{x^2+y^2}}=y'-\frac{y}{\sqrt{x^2+y^2}}y'\]Again, factor out the y',\[\large\rm \frac{x}{\sqrt{x^2+y^2}}=y'\left(1-\frac{y}{\sqrt{x^2+y^2}}\right)\]And then divide by that big mess to isolate your y',\[\large\rm \frac{x}{\left(1-\frac{y}{\sqrt{x^2+y^2}}\right)\sqrt{x^2+y^2}}=y'\]And obviously that answer can be written in a much much nicer way, but let's not worry about that.

Answer 44

Examples are good c: heh

Answer 45

Examples are rich i must say c: An for you to actually give so many makes you extremely pro

Answer 46

But i must say that you seem to enjoy this c: Thank god you helped me

Answer 47

"No crying in calculus"
lol silly teacher :3

Answer 48

Yes...she is very fond of that quote because everytime she teaches us something that seems easy even though we know its not we always tell her "thats it im going to cry.." and well yeah she laughs because she's been there but because shes also a math major well...she finds it silly that we complain XD

Answer 49

hah

Answer 50

Well i would like to thank you again for the help and examples you have given me but i shall now try and get some rest for the *uses sarcasm* Amazing test that i am so looking forward to tomorrow or later technically...but eh Anyways, thank you and hopefully if i happen to have another question about something i can go to you for help again c: But yeah

Answer 51

ok np \c: have good night

Answer 52

I shall, Good Night and thanks a million times again hah