Question

Logarithmic differentiation? What is that and how do I use it?

Answer 1

I guess you care for an example?

Answer 2

Yes, I have a calculus homework question on it.

Answer 3

\[y=2^x\] That would be a pretty common one, where you intend to use logarithmic properties (mainly natural log) to be able to get a function that is easier to differentiate.

\[ \ln y = x\ln 2 \]

Answer 4

\[\sqrt{1-x}\sqrt{1-2x}\sqrt{1-3x}\]

Answer 5

well you can apply logarithmic differentiation here, although I must confess, that's the last thing I would do *laughs*.

Answer 6

The question requires you to use it :L

Answer 7

given say ; \(\int f(x)\ dx\)

logging is such that we do

y = f(x) ; ln both sides

ln(y) = ln(f(x)) ; then do the deed

Answer 8

Well if the question requires it, deliver it (-; Here you have to remember a few of your log identities. that's all.

Answer 9

i never remember anything, so i tend to have to make it up anew each time :)

Answer 10

Deriving the equations, yay.

Answer 11

Oh, this is some fun stuff. Logarithmic differentiation is useful for finding the derivatives of functions that are in exponential form but are not easy to derive on their own, such as the problems here ( http://www.math.ucdavis.edu/~kouba/CalcOneDIRECTORY/logdiffdirectory/LogDiff.html).
Take the first problem as an example:
\[y=x ^{x}\]
That looks hella hard to differentiate, right? Wrong. We can utilize the properties of logarithms to make this easier to deal with. Take the natural log of both sides to start:
\[\ln y = \ln (x^{x})\]
which is equivalent to
\[\ln y = x \ln x\]
Use the chain rule to further derive the equation.
\[(1/y) dy/dx = 1 + \ln x\]
Multiply by y to get the answer:
\[dy/dx = y(1 + \ln x)\]
and substitute in known values (remember, y = x^x)
\[dy/dx = x^{x}(1 + \ln x)\]

Look through some of the other problems on that page, on Wikipedia, or on Wolfram MathWorld for some more insights.

Answer 12

http://tutorial.math.lamar.edu/Classes/CalcI/LogDiff.aspx

for some light reading

Answer 13

Huh. Makes more sense.

Answer 14

Alright. Thanks!

Answer 15

The website amistre64 linked is amazing.I pretty much taught myself through Calculus II on that website. Utilize it frequently.

Answer 16

Wait a sec.

Answer 17

How am I supposed to use it for
y=\[\sqrt{1-x}\sqrt{1-2x}\sqrt{1-3x}\]

Answer 18

Rewrite the function in exponential form:
\[y = (1-x)^{1/2}(1-2x)^{1/2}(1-3x)^{1/2}\]
and play around with it from there.

Answer 19

Wait a sec...

Answer 20

If we use
y=\[\sqrt{1-x}\sqrt{1-2x}\sqrt{1-3x}\]

Answer 21

ln y=
ln\[\sqrt{1-x}\sqrt{1-2x}\sqrt{1-3x}\]

Answer 22

(1/y)*(dy/dx)=
1/\[\sqrt{1-x}\sqrt{1-2x}\sqrt{1-3x}\]

Answer 23

dy/dx=
y/\[\sqrt{1-x}\sqrt{1-2x}\sqrt{1-3x}\]

Answer 24

That's not true. You have to apply the chain rule and I don't even want to get into that mess that you're about to get in.

Answer 25

Oh my goodness...

Answer 26

Isn't it ln (\[\sqrt{1-x}\sqrt{1-2x}\sqrt{1-3x}\])

Answer 27

Oh... good point

Answer 28

Never forget the properties of logarithms and how the exponent and root are related. Ever. You'll make extensive use of these properties throughout your mathematics career.

Answer 29

Mind if I latex it out real quick?

\[ \Large \ln y = \ln(1-x)^\frac{1}{2} + \ln(1-2x)^\frac{1}{2}+\ln(1-3x)^\frac{1}{2} \]

Answer 30

^That's what you want going on

Answer 31

Oh yea :L

Answer 32

and then take it to \[\ln y = (1/2)\ln (1-x) + (1/2)\ln(1-2x) + (1/2)\ln(1-3x)\] and derive that/.

Answer 33

log xy=log x+log y

Answer 34

:L

Answer 35

Yeah. Isn't that much easier than brute differentiation?

Answer 36

Pretty much.

Answer 37

Hmm

Answer 38

Can we derive the equation you wrote?

Answer 39

That turns out to be like 1/2(1-x)+1/2(1-2x)+1/2(1-3x).

Answer 40

then *y