Question

find dy.dx

y=ln(lnx)

Answer 1

*dy/dx

Answer 2

The derivative of ln (u) is du/u.

Let u = ln x

So the derivative of ln (u) = du/u = 1/(x ln(x))

Answer 3

Yup

Answer 4

where did you get the extra x from? :o

Answer 5

That's from the derivative of ln(x) with respect to x (Think: Chain Rule).

Answer 6

The derivative of ln(anything) is 1/anything times the derivative of that anything

The derivative of ln(ln x) is 1/ln x times the derivative of ln x (wjich is 1/x)

so the dserivative is 1/ln x multiplied by 1/x

= 1/(x ln(x))

Answer 7

Makes sense?

Answer 8

is it like a rul that the derivative of ln x is 1/ x? .-.

Answer 9

If you want the derivative of ln(any function)....then it is
1/that function x derivative of that function

The dervative of ln(sin x) = 1/sinx x cos x = cos x/sin x

Answer 10

aaaa, okay(:

Answer 11

thanks!!!!

Answer 12

cool

Answer 13

So, what is the derivative of ln(x^2 + 5x)??

Answer 14

1/x^2 +5x ....? >.<

Answer 15

not really....you started off right....

The derivative will be 1/(x^2 + 5x) multiplied by (2x+5)

Answer 16

The derivative of ln(u) is 1.u multiplied by the derivative of u

Answer 17

So, the derivative of ln(x^2 + 5) = (2x+5)/(x^2 + 5x) Agree?

Answer 18

ohhhhhh! yes yes I remember that now!

Answer 19

another one! :)

Answer 20

Try one more.........what is the derivative of y = ln(x sin(x))?

Answer 21

(Oh, that one's a pain!)

Answer 22

cos x/ xsinx ?

Answer 23

jk ..

Answer 24

there's a uv withing the problem right?

Answer 25

multiplication?

Answer 26

Not really........

The derivative of ln(x sin(x)) is 1/(x sin(x)) multiplied by the derivative of x sin(x).

The derivative of x sin(x) is found using the product rule...it is x cox(x) _ sin x

Answer 27

Yeah, it's Chain Rule AND Product Rule in 1.

Answer 28

The derivative of x sin(x) is x cos(x) + sin (x).

Answer 29

So the final answer will be [x cos(x) + sin (x)]/(x sin(x))

Answer 30

Agree?

Answer 31

Note: You can also do some nifty simplification...

Answer 32

Make it look nice:

\[= \frac{ 1 }{ x } +\cot (x)\]

Answer 33

Yes. But we have to make sure that soccergirl13 understand the procedure it finding the derivative of ln(x sin(x)). Then we can simplify our answer at the end.

Answer 34

what happened to the cos x in the denominator ? ? :o

Answer 35

where's there a cos x in the denominator?

Answer 36

I know, I went a step further since you already posted the derivative....

Answer 37

you said that the derivative of x sin(x) is x cos(x) + sin (x).

Answer 38

I never said that!

Answer 39

I said that to find the derivative of x sin(x) you need to use the product rule.

and you get x cos(x) + sin (x)

Answer 40

OHHHHHHHHHH

Answer 41

And that part ends up in the numerator

Answer 42

SEE WHAT YOU DID THERE

Answer 43

LOL

Answer 44

OKAY

Answer 45

OK..Soccergirl13...one last problem for you...show me you understand!

Answer 46

you wouldn't find the derivative until multiplying to the 1/u

Answer 47

okay(:

Answer 48

What is the derivative of y = ln(x/2)?

Answer 49

(This is where log rules come in handy...)

Answer 50

Well?

Answer 51

ahhhhh. :(

Answer 52

1/2??

Answer 53

is the deriv of x/2 1/2?

Answer 54

1/2 is the derivative of x/2

Answer 55

ok

Answer 56

So the final answer, would be 1/(x/2) divided by 1/2

Answer 57

*multiplied*

Answer 58

sorry...multiplied by 1/2

Answer 59

Final answer is 1/x after you simplify that expression.

Answer 60

no prob -- I got your back. :)

Answer 61

Thanks, Dance

Answer 62

The 2's canceledz?

Answer 63

yes

Answer 64

Yeah--it's crazy--but Soccergirl, do you remember the rules for logs?

Answer 65

what rules? D: lol

Answer 66

Since:
\[\ln(\frac{ a }{ b }) = \ln(a) - \ln(b)\]

You could've rewritten that as:

\[\ln(\frac{ x }{ 2 }) = \ln(x) - \ln(2)\]
and the "ln(2)" disappears since it's just a constant...

Answer 67

Then it becomes just the simple derivative of ln(x)

Answer 68

ohhh yeahhhhhh!

Answer 69

i see that

Answer 70

OK...One last one for Soccergirl.............and that will be it. If you get it wrong, I will be disappointed. So get it right.

Answer 71

ahhhD: ok

Answer 72

Last question for SoccerGirl........

What is the derivative of y = ln(sqrt(x))? Simplify your answer.

Answer 73

question is for SoccerGirl.

Answer 74

Simplifying logs that way comes up A LOT! For instance if you see something like:

\[\frac{ d }{ dx }[\ln(x^{4})]\]

You can simplify that as:

\[\frac{ d }{ dx }[4\ln(x)] = 4*\frac{ d }{ dx }[\ln(x)] = \frac{ 4 }{ x }\]

and forget all that chain rule stuff....

Answer 75

Oh, whoops (I didn't see your last question...)

Answer 76

SoccerGirl..you there? working on that last question?

Answer 77

1/2 x^ -1/2 / sqrt of x ....?

Answer 78

Yes. Now simplify that.

Answer 79

........... D: sqrt of x/ 2x^1/2 ...?

Answer 80

Try again.

Answer 81

:'''( I don't remember this alg 2 stuff!!!

Answer 82

(1/2) (x^(-1/2)) = (1/2) (1/sqrt x)
becuase x^(-1/2) = 1/x^(1/2) = 1/sqrt(x)

Answer 83

So the numerator, 1/2 x ^(-1/2) = 1/(2 sqrt(x))

so, your final answer should be 1/(2x).

Answer 84

Review that. But at least you got the derivative right! Way to go!

Answer 85

But remember your log rules!! You could've simplified that to:

(1/2)log(x)
first and then taken the derivative...

Answer 86

ahh, okay i din't know that the negative exponent could be replaced by a 1/ ...

Answer 87

/: sorry for not getting it right lol

Answer 88

a^(-n) = 1/a^n

5^(-2) = 1/(5^2) = 1.25

x^(-1/2) = 1/(x^(1/2)) = 1/sqrt(x)

Answer 89

you are so helpful :) thank you!

Answer 90

SoccerGirl.............you've come a long way.

Now, you can go to the section in your text on derivatives of ln (u)...work on problems. Parctice makes perfect. If you need any help, come back for more!

Answer 91

You're welcome.

Answer 92

thank you so much!! you're so kind. i do have just another question(:

Answer 93

OK. Shoot.

Answer 94

for y=logv 2 (1/x)

Answer 95

What is that v..and what is that 2?

Answer 96

it's like beneath the log lol

Answer 97

log 2 basically

Answer 98

The 2 is probably base 2. What is that v? I'm not understanding that v.