Question

Differentiate: ln√(9-t^2 )
Differentiate: ln√(9-t^2 )
@Mathematics

Answer 1

So by the chain rule would i firs differentiate 9-t^2?

Answer 2

Then differentiate \[(D _{x}[9-t ^{2}])^{1/2}\]

Answer 3

Then differentiate: ln(Dx[(Dx[9−t2])1/2])

Answer 4

I would first rewrite the problem.

\[\ln((9-t^{2})^{1/2})\]

Then using rule of logs, you can take down the exponent to make.

\[1/2 \ln(9-t^{2})\]

Then we differentiate. When taking derivative of ln, it's just 1/x, rule of logs. So...

\[(1/2) * (1/(9-t^{2}) * (2t)\]
*remember to use the chain rule when taking the derivative of ln(9-t^2).

So clean it up and you have...

\[2t/(18-2t^{2})\]

----

I may be wrong. This is what I got though....

Answer 5

Ah ok

Answer 6

That rule of logs loves to escape my brain

Answer 7

Factor out 2 on the bottom and we have t/(9-t^)

Answer 8

Also, another thing I always forget.

Remember the derivative of e^x is e^x. So important to remember.

Answer 9

differenetiate means to go down as opposed to integrate right?

Answer 10

\[[ln√(9-t^2 )]'=\frac{[\sqrt(9-t^2)]'}{\sqrt(9-t^2)}\]is a nasic start to it

Answer 11

*basic start to it

Answer 12

\[[\sqrt(9-t^2)]'=\frac{[9-t^2]'}{2\sqrt(9-t^2)}\] is anohter part to it

Answer 13

By differentiate I mean take the derivative of, thats what it means no?

Answer 14

\[[9-t^2]'=-2t\]is the last bit

Answer 15

i get so many terms in my head that i get them mixed up alot; they need less d words and more j words perhaps lol

Answer 16

Hmm, so t/(9-t^) isnt the solution?

Answer 17

Cause wolfram agrees with t/(9-t^): http://www.wolframalpha.com/input/?i=differentiate%3A+ln%E2%88%9A%289-t%5E2+%29

Answer 18

dunno, I havent put it all together yet and simplified

Answer 19

So where did I mess up? That's what I want to know. As for what amistre is doing, i have no idea? Can you explain what you are doing? How did you do your first step?

Answer 20

\[\frac{[\sqrt(9-t^2)]'}{\sqrt(9-t^2)}\]
\[\frac{\frac{[9-t^2]'}{2\sqrt(9-t^2)}}{\sqrt(9-t^2)}\]
\[\frac{[9-t^2]'}{{2\sqrt(9-t^2)}*\sqrt(9-t^2)}\]
\[\frac{[9-t^2]'}{2(9-t^2)}\]
\[\frac{-2t}{2(9-t^2)}\]
\[\frac{-t}{9-t^2}\]

Answer 21

the first step is just to diff the ln|u| into its basic part: du/u

Answer 22

then its just a matter of following the chain rule for the successive diffs

Answer 23

hoodrych's methodology seems a bit simpler

Answer 24

(1/2) * (1/(9-t^{2}) * (2t) <-- should be -2t

Answer 25

[(√9−t^2)]′ / (√9−t^2
If you're just rewriting du/u then where is the ln on the numerator?

Answer 26

Good catch amistre.

Answer 27

there doesnt need to be an ln in the numerator; im just following the ln' rule

Answer 28

The ln` rule is

d/dx ln(x) = 1/x

That's not what you did?

I'm not saying you're wrong, just trying to understand.

Answer 29

\[[\ ln|f(u)|\ ]'=\frac{f'(u)}{f(u)}\]

Answer 30

they make alot of different formats for the same basic rule, if you always remember to just chain out your argument, youll do fine

Answer 31

So the f`(u)/f(u) is the same thing as the rule I said? (the 1/x)???

Answer 32

\[\frac{d}{dx}ln(x)=\frac{dx/dx}{x}=\frac{1}{x}\]

Answer 33

your simply used to ignoring the dx/dx since its equal to 1

Answer 34

yes, its the same thing you did :)

Answer 35

aaaaaaaaaaand, click *

Thanks bud.

Answer 36

youre welcome ;)