Question

FInd dy/dx

Answer 1

\[y= \ln \sqrt{x ^{2} + 4}\]

Answer 2

error ?

Answer 3

I'm just trying to figure this out ... my Calculus teacher put this on the study guide so our study group is trying to figure it out. :(

Answer 4

I think the best way to start would be to simplify the expression first.\[\large y=\ln\left[(x^2+4)^{1/2}\right] \qquad \rightarrow \qquad y=\frac{1}{2}\ln\left[x^2+4\right]\]
Does that step make sense? :o

Answer 5

We got to that step, :)

Answer 6

Do you know the derivative of \(\large y=\ln\left(\text{stuff}\right)\) ? :)
Err maybe it'd be better ask if you know the derivative of \(\large \ln x\).

Answer 7

1/x?

Answer 8

Wait, I think we have it.

Answer 9

Yep, use chain rule.

Answer 10

Oh ok c:

Answer 11

So from there, we can write, \[\frac{ 1 }{ 2 }(\frac{ 1 }{ x + 1 }) = \frac{ 1 }{ 2(x+1) }\]

Answer 12

Hmm

Answer 13

Oh oops

Answer 14

Wrong problem

Answer 15

Oh lol c:

Answer 16

\[\frac{ 1 }{ 2 }(\frac{ 1 }{ x^{2} + 4})= \frac{ 1 }{ 2(x ^{2} + 4)}\]

Answer 17

start with \(\frac{1}{2}\ln(x^2+4)\) and take the derivative of that

Answer 18

Mondays ... -_____-

Answer 19

Uh oh looks like you forgot the chain rule D: Hmm

Answer 20

\[\large y'=\frac{1}{2}\frac{1}{x^2+4}\color{royalblue}{(x^2+4)'}\]

Answer 21

After you took the derivative of the natural log, you need to multiply by the derivative of the inner function :O See the blue part?

Answer 22

Okay

Answer 23

using the chain rule
the derivative of \(\ln(f(x))\) is \(\frac{f'(x)}{f(x)}\)

Answer 24

but get rid of that annoying square root at step one

Answer 25

SO the derivative becomes x

Answer 26

oh what @zepdrix already said

Answer 27

And then multiplied by the rest of the problem, \[\frac{ x }{ x ^{2} + 4}\]

Answer 28

The derivative of the blue part? :O Hmm I think it's 2x.
Oh you simplified it! :) Yes good job!

Answer 29

That was rather simple after all. What can I say? Mondays, I guess. -___-. Well thank you! So much!

Answer 30

Yah those darn mondays c: lol