Question

Differentiate and simplify.

Answer 1

a)\[y=\frac{ \sqrt{x}-x }{ x^2 }\]
b)\[y=\sqrt{\frac{ x }{ 1+x^2 }}\]

Answer 2

Have you used the chain rule?

Answer 3

Yes, but I still didn't get the answer right yet...

Answer 4

Okay, I'll help you now. This one is a bit complicated.

Answer 5

So apply the chain rule and let \[\frac{ x }{ 1+x^2 }\] to be letter U.

Now we'll get \[\frac{ d }{ du } (\sqrt{u}) \frac{ d }{ dx } \frac{ x }{ 1+x^2 }\]
So now \[\frac{ d }{ du } (\sqrt{u}) \] apply the power rule \[0.5u^0.5-1 = \frac{ 1 }{ 2\sqrt{u} }\]

now use the quotient rule \[[\frac{ d }{ dx }(x)(1+x^2)-\frac{ d }{ dx }(1+x^2)x] / (1+x^2)^2\]

Simplify and you'll end up with \[\frac{ (1-x^2)-2x^2 }{ (1+x^2)^2 } = \frac{ 1-x^2 }{ (x^2+1)^2}\]

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

Substitude back u = x/(1+x^2)^2

I simplified and got \[\frac{ 1-x^2 }{ 2\sqrt{x} (x^2+1)^(5/2)}\] as the final answer.


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Answer 6

Hi @esam2

Answer 7

hi :)

Answer 8

so which one do you want to look at

Answer 9

both

Answer 10

Yes, both please... I tried so many times and I couldn't get them... :(

Answer 11

Ok let's look at a first
we are going to just focus right now on rewriting a
(we will differentiate a later)
\[y=\frac{\sqrt{x}-x}{x^2}=\frac{\sqrt{x}}{x^2}-\frac{x}{x^2}\]
are you cool with separating the fraction like this?

Answer 12

yes

Answer 13

\[\sqrt{x}=x^\frac{1}{2} \\ \text{ so we can also rewrite a little more } y=\frac{x^\frac{1}{2}}{x^2}-\frac{x}{x^2}\]
now it is necessary for you to remember all those things you learned in your math classes before

you know like the law of exponents

Answer 14

yes

Answer 15

if you have a quotient and the same base on top and bottom
you can do this:
\[\frac{x^a}{x^b}=x^{a-b}\]

Answer 16

\[y=x^{\frac{1}{2}-2}-x^{1-2}\]

Answer 17

you know how i got that 1 right
x is x to the first

Answer 18

anyways we need to simplify both 1/2-2 and 1-2

Answer 19

\[\frac{1}{2}-2=\frac{1}{2}-\frac{4}{2}=\frac{-3}{2} \\ 1-2=-1 \]
so the final way we will rewrite y is like this:
\[y=x^\frac{-3}{2}-x^{-1}\]
now we just use power rule for both terms when we wish to differentiate

Answer 20

you wanna give that ago
that is differentiating our problem we rewrote just now

Answer 21

Oh! You made it a little easier to solve it

Answer 22

yeah

Answer 23

we could have straight up quotient rule it

Answer 24

but i thought rewriting it first would make things easier

Answer 25

So i got -3/2x^-5/2+x^-2?

Answer 26

brilliant
\[y'=\frac{-3}{2}x^{\frac{-5}{2}}+x^{-2} \\ \text{ or without negative exponents } \\ y'=\frac{-3}{2 x^\frac{5}{2}}+\frac{1}{x^2}\]

Answer 27

I got it! :D
it is \[\frac{ 2\sqrt{x}-3 }{ 2\sqrt{x^5} }\]

Answer 28

for some reason I got logged out :(

Answer 29

and congrats

Answer 30

I had did the quotient rule for you but before I could post anything I got logged out

Answer 31

but it was unnecessary work anyways

Answer 32

anyways you want to look at the second problem too correct?

Answer 33

\[y=\sqrt{\frac{ x }{ 1+x^2 }} \]
we could use some other tricks to differentiate here
but I don't like you heard of implicit or log differentiation
right?

Answer 34

it doesn't matter
we will just straight up chain rule it

Answer 35

before we go on to the second problem
do you understand the following:
\[y=\sqrt{g(x) } \\ y=((g(x))^\frac{1}{2} \\ y'=\frac{1}{2}(g(x))^{\frac{1}{2}-1} \cdot g'(x) \\ y'=\frac{1}{2}(g(x))^\frac{-1}{2} \cdot g'(x) \\ y'=\frac{1}{2} \frac{1}{(g(x))^\frac{1}{2}} g'(x) \\ y'=\frac{1}{2}\frac{g'(x)}{(g(x))^\frac{1}{2}}\]

Answer 36

Hi again! I got logged out too. I think there are techical issues with the website. Okay, give me few minutes to read over these notes...

Answer 37

my OpenStudy page keeps refreshing

Answer 38

I think I got the following above.

Answer 39

so you know your g(x) in this case
=x/(1+x^2)

Answer 40

replace the g(x)'s in my formula above with x/(1+x^2)
so you do this:
\[y'=\frac{1}{2} \frac{\frac{d}{dx}(\frac{x}{1+x^2})}{(\frac{x}{1+x^2})^2}\]

Answer 41

since we have a whole bunch of fractions I will rewrite that a little so it looks prettier to you:
\[y'=\frac{1}{2} [\frac{d}{dx} (\frac{x}{1+x^2})] \cdot \frac{1}{(\frac{x}{1+x^2})^2} \\ y'=\frac{1}{2} [\frac{d}{dx}(\frac{x}{1+x^2})] \cdot (\frac{1+x^2}{x})^2\]

Answer 42

@freckles I got it! Thanks for your help! ^_^