Question

differentiate:
y=arctan(sqrt:(1-x/1+x))

Answer 1

oo that's a doozy :U
getting confused on the arctan derivative part of it or what?

Answer 2

\[\Large\bf\sf y\quad=\quad \arctan\left(\sqrt{\frac{1-x}{1+x}}\right)\]

Answer 3

haha yeah its scary

Answer 4

it is not scary

Answer 5

yes thats correct @zepdrix

Answer 6

\[\Large\bf\sf y=\arctan(\color{royalblue}{x})\]\[\Large\bf\sf y'=\frac{1}{1+(\color{royalblue}{x})^2}\]So we'll have to keep this derivative in mind.

Answer 7

ok

Answer 8

\[\Large\bf\sf y=\arctan\left(\color{royalblue}{\sqrt{\frac{1-x}{1+x}}}\right)\]
\[\Large\bf\sf y'=\frac{1}{1+\left(\color{royalblue}{\sqrt{\frac{1-x}{1+x}}}\right)^2}\cdot\color{orangered}{\left(\sqrt{\frac{1-x}{1+x}}\right)'}\]

Answer 9

We've got a couple things going on here,
the derivative of arctan is probably what you expect, right?
The entire argument is stuffed into that square.

But we also need to apply the chain rule.
So we need to take the derivative of that orange thing.

Answer 10

yes and would you then say that the square roots are changed to the 1/2 and -1/2?

Answer 11

then quotient rule?

Answer 12

yah derivative of square root, then chain into quotient rule.
very good.

Answer 13

\[\Large\bf\sf y'=\frac{1}{1+\left(\color{royalblue}{\sqrt{\frac{1-x}{1+x}}}\right)^2}\cdot\color{orangered}{\left(\frac{1}{2\sqrt{\dfrac{1-x}{1+x}}}\right)}\cdot\color{green}{\left(\dfrac{1-x}{1+x}\right)'}\]

Answer 14

brb i need moar chocolate milk! :O

Answer 15

mmm

Answer 16

haha ok idk what ya did there in orange

Answer 17

You can apply the power rule `if you want`.
But that's a derivative that is really really worth memorizing.

derivative of sqrt is `one over 2sqrts`.\[\Large\bf\sf \frac{d}{dx}\sqrt x\quad=\quad \frac{1}{2\sqrt x}\]

Answer 18

oh never seen that one before, only the power rule! will write this one down!, so then we finally take the derivative of the last one in green which is -1?

Answer 19

for your quotient rule? Hmm no you should** get something quite different from -1

Answer 20

\[\Large\bf\sf \left(\frac{1-x}{1+x}\right)'\quad=\quad \frac{(1-x)'(1+x)-(1-x)(1+x)'}{(1+x)^2}\]

Answer 21

-2x?

Answer 22

\[\Large\bf\sf =\quad \frac{(-1)(1+x)-(1-x)(1)}{(1+x)^2}\]
\[\Large\bf\sf =\quad \frac{-1-x-1+x}{(1+x)^2}\]

Answer 23

oops forgot the bottom: -2x/(1+x^2)

Answer 24

i mean the bottom is (1-x)^2

Answer 25

ok good.
And it looks like the x's cancel on top, yes?
Leaving us with -2

Answer 26

@Stew.a.r.t. I know this isn't exactly the same problem you're working on. But look at the steps at least. Might help you through yours.

Answer 27

-x+x on top right?
Make sure you distributed your negative signs correctly.

Answer 28

umm i know im wrong but wouldit be -2x/(1+2x-x)?? sorry its been awhile since ive done algebra lol

Answer 29

Don't expand out the square in the bottom. Leave it alone.

I still don't understand how you're getting -2x on top.
You should be getting -2.

Answer 30

After taking derivative, your numerator should be,

(-1)(1+x) - (1-x)

did you end up with that?

Answer 31

yes over (1+x)^2

Answer 32

Distribute the first negative,

-1-x - (1-x)

distribute the `second negative`

-1-x -1+x

did you forget about the second negative?

Answer 33

oh ok so the top would be -2 cause the x's cancel

Answer 34

\[\Large\bf\sf y'=\frac{1}{1+\left(\color{royalblue}{\sqrt{\frac{1-x}{1+x}}}\right)^2}\cdot\color{orangered}{\left(\frac{1}{2\sqrt{\dfrac{1-x}{1+x}}}\right)}\cdot\color{green}{\dfrac{-2}{(1+x)^2}}\]Ok good! :)

Answer 35

It can probably be simplified a little bit.. but uhhhh whatever.
That's a fine answer.

Answer 36

WOW thats crazy sorry it took me awhile lol...so you wouldn't have to simplify at all?

Answer 37

I dunno, I'm not your teacher lol.
Really depends on what your teacher wants.

Like notice that we have a square root being squared in our tangent derivative.
And we're diving by 2 and multiplying by 2 in a few different spots.

It can be cleaned up a little bit.
Nothing to fuss over though :o

Answer 38

ok i dont think he would mind if i left it like this but just to clarify so the twos inthe two different spots could you cancel them into a 1 and -1?

Answer 39

thanks a bunch!!! :))

Answer 40

\c:/