Question

Find the Differential of: "y = ln[sqrt.(1+x^2)]"

Any and all help is greatly appreciated!

Answer 1

Hmm so where you stuck? :d

Answer 2

\[\large\rm y=\ln\left[\sqrt{1+x^2}\right]\]

Answer 3

Hmm I don't see a t anywhere :d

Answer 4

Lols, sorry. My book has t's. I was doing x's on here to make it "easier". My bad. yeah, that's supposed to be an "x" ;)

Answer 5

broski.
I strongly recommend you commit this derivative to memory.
It shows up sooo often, its really worth a spot in your brain.\[\large\rm \frac{d}{dx}\sqrt{x}=\frac{1}{2\sqrt x}\]
The derivative of square root is `one over two square roots`.
Don't bother with power rule on that guy.

Answer 6

\[\large\rm y=\ln\left[\sqrt{1+x^2}\right]\]First step, derivative of log, and "set up" chain,\[\large\rm dy=\frac{1}{\sqrt{1+x^2}}\cdot d\sqrt{1+x^2}\]

Answer 7

Gotcha! I did that:)

Answer 8

Apply square root derivative, and "set up" the next chain,\[\large\rm dy=\frac{1}{\sqrt{1+x^2}}\cdot \frac{1}{2\sqrt{1+x^2}}\cdot d (1+x^2)\]

Answer 9

Wait... the last term there. The d(1+x^2).. what's that? Is that dx?

Answer 10

Hmm no.
We're not to dx just yet :)

Answer 11

you can simplify the expression.. then it would be quite easy
ln x^a = a ln x
that will remove the sq rt
and differentiation will become easy

Answer 12

Yes, that's also a good idea :)
Saves you a couple steps.

Answer 13

As second step, you took this derivative:\[\large\rm d\sqrt{stuff}\quad=\quad \frac{1}{2\sqrt{stuff}}\cdot d(stuff)\]Gotta differentiate that stuff on the inside by chain rule, ya?

Answer 14

I'm good on the first two pieces of that... but why do we need to take the derivative of (sqrt.(1+x^2) twice?

Answer 15

We took the derivative of log first.
Then square root second.
Then (1+x^2) third.

Answer 16

so...
\[d (\ln(stuff)) = 1/(stuff) * Chain \Rule of net stuff * d(stuff)\] ?

Answer 17

Gotcha... Okay..

Answer 18

And then I guess derivative of the inner most function gives us,\[\large\rm dy=\frac{1}{\sqrt{1+x^2}}\cdot \frac{1}{2\sqrt{1+x^2}}\cdot (2x)dx\]Ya? :o

Answer 19

Yeppers, I'm there:)

Answer 20

And then uhhh, simplify if you can.

Answer 21

I think I see now... I didn't know about doing the chain rule last half part err... "twice". Damn, my prof is so.... irritating. He never does examples like we have in the homework!! >:/

Answer 22

Oh you have one of those? :C grr

Answer 23

That makes sense though, I get it now:)

Answer 24

Where do you go to school? :)

Answer 25

Divu's tip was quite helpful if you want to try that one at some point.\[\large\rm \ln\left[(1+x^2)^{1/2}\right]\quad=\quad \frac{1}{2}\ln\left[1+x^2\right]\]And that will save you the work of one of the chains.

Answer 26

University of Central Florida.
I hate it :o school is so harrrrd.

Answer 27

True, that looks slick.
Oh, nice! My friend goes to Emery-Riddle School of Aviation down in Florida, and I've been wishing I could be in that weather right now. I go to the University of Minnesota, Twin Cities, and it's been like, 50-60 degrees... I'm missing my shorts and tan already :(

Answer 28

Emery? :o
Ooo nice school.

Yah the weather can get pretty rough down here sometimes...
I had to throw an extra shirt on due to a light breeze.
Brrrr! I think it was in the low 70s :(

Answer 29

Haha, low 70's;) If only... I wish it would be that nice even through November! I've been thinking about snow, and where I'm going to put my bike where it won't get drifted in between classes!
Yeah, Emery Riddle sounds super sweet. My friend was flying a plane on literally, the first day of class!! It's insane, and also pricey..

Answer 30

Woahhh :O fun!

Answer 31

Alright, so if I have:
\[y = \frac{1-v^2}{1+v^2}\]
And I'm looking to get the "differential" again, is quotient rule the best bet?

Answer 32

Umm.. quotient rule is fine I guess.
I would prefer long division though.\[\rm \frac{1-v^2}{1+v^2}\quad=\quad \frac{-(v^2-1)}{v^2+1}\quad=\quad \frac{-(v^2+1-2)}{v^2+1}\quad=\quad -\frac{v^2+1}{v^2+1}+\frac{2}{v^2+1}\]
And then from this point,\[\large\rm =-1+2(v^2+1)^{-1}\]You can simply apply power rule.
If those Algebra steps are way confusing though,
you can just stick with quotient ^^

Answer 33

That little trick works way better on problems like this:\[\large\rm \frac{x^2}{x^2+1}\]Maybe it's a little confusing for your particular problem,
since we had to mess with the negatives so much.

Answer 34

Yeah, that was a lil' over my head.. I'm going to finish doing Quotient Rule. What do you get as the final answer?

Answer 35

\[\large\rm dy=\frac{-4v^3}{(1+v^2)^2}dv\]

Answer 36

Sec, checking my work :D

Answer 37

Don't the 2v^3 's cancel out, and there are two (-2v)'s? :/

Answer 38

Mmm let's just find out I guess :d

Answer 39

:D

Answer 40

I'm like, 90% sure lols.

Answer 41

\[\large\rm y=\frac{1-v^2}{1+v^2}\]
\[\large\rm dy=\frac{(-2v)(1+v^2)-(2v)(1-v^2)}{(1+v^2)^2}dv\]This step ok?

Answer 42

Good there:) I got that too:)

Answer 43

Ah yes :) I pulled out too many negatives I think.
Good call.

Answer 44

-4v on top then?

Answer 45

Yeppers man:)

Answer 46

smoooooth! ୧ʕ•̀ᴥ•́ʔ୨

Answer 47

If...
\[y = e^{x/10}\]
Then...
\[dy = \frac{x}{10}lne \] ?

Answer 48

Hmm where did this magical log come from?
I only see one on the right side.

Answer 49

Remember derivative for e^x?
Gives you the same thing back, ya?\[\large\rm \frac{d}{dx}e^{x}=e^{x}\]And we'll generalize it a little more for chain rule,\[\large\rm \frac{d}{dx}e^{stuff}=e^{stuff}~\frac{d}{dx}stuff\]

Answer 50

I remember that, about d/dx of e^x! Ah.. so I don't need to take the natural log of both sides or anything??

Answer 51

Logarithmic differentiation?
Only if `both the base and exponent are x stuff`.

\(\large\rm y=f(x)^{g(x)}\qquad \text{<--- Apply Logarithmic Differentiation}\)
Example:
\(\large\rm y=x^{cos(x)}\)

Answer 52

I mean, yes you should go through the log differentiation process at least one time to justify the exponential derivative,
but after that first time, it's a nice shortcut to remember.

Here is the more general rule though:\[\large\rm \frac{d}{dx}a^x=a^{x}\ln(a)\]So NORMALLY, you have to multiply by the log of the base.
But ln(e) is just 1, so we generally skip that on the e^x derivative.

Answer 53

Gotcha... so, it should be like this?
\[dy = e^{\frac{x}{10}}*\frac{(1*10) - (x*0)}{100}\]

Answer 54

Yes, you COULD do quotient rule.
Lemme state this as I did for the exponential rule:
Apply quotient rule only if `the numerator and denominator are both x stuff`.

Answer 55

Instead do this:\[\large\rm \frac{x}{10}=\frac{1}{10}x\]And apply your power rule or whatever.

Answer 56

\[\large\rm \frac{d}{dx}ax=a\]ya? :)

Answer 57

True! :)

Answer 58

\[\large\rm y=e^{x/10}\]\[\large\rm dy=e^{x/10}\cdot d~\left(\frac{x}{10}\right)\]So what do you get for that chain? And final answer?

Answer 59

Confused? :o

Answer 60

Final answer: .1(e^(x/10))*dx :)

Answer 61

hmmm

Answer 62

My book says it's right... ? :/

Answer 63

Oh you're being sneaky, .1 = 1/10,
Ok fine fine

Answer 64

yeah, sorry:)

Answer 65

Interesting thread, can you find the differential for the following?You may have to use a substitution!
\[y=\tan^{-1}(\frac{a}{\sqrt{x^2-a^2}})\]
\[y=\sin^{-1}(2x \sqrt{1-x^2})\]

Answer 66

I'm sorry, but I am jus trying to get something on paper for my assignment in Calc that's due today... or tomorrow, depends on wether "Today" starts once I sleep, or now, since it's 2am lols. I would love to challenge myself, but I have been stuck on a pellet-ton of problems, and am just trying to get something down for all of them, so I can get some credit.. Sorry:/

Answer 67

Sh1t = pellet... ? why?

Answer 68

ya the filter word is hilarious lol

Answer 69

1 = i

Answer 70

Oh best of luck, feel free to try them later if you want to, I think these problems would be in your textbook anyway if you read on further into your book

Answer 71

Thank you!! I certainly will take a peek, once the fire's put out here lols. :)

Answer 72

zepdrix you are gr8 thanks so much for helping

Answer 73

@amonoconnor shіt

Answer 74

What's up now @imqwerty? Sorry, I went to bed last night.

Answer 75

haha i jst tacked the filter :D