Quick Math Q.

[JJChicken]

Differentiate log (x/k)

Thanks ATOT :beer:

 

[rudeguy]

42

 

[KLin]

I differentiate it to be the log of x over k.

 

[Leros]

I did this on my calc. I got log(e)/x.

Its weird that it doesn't depend on k.

 

[Cyco]

Originally posted by: rudeguy
42

There is no other answer.

 

[GML3G0]

Originally posted by: Leros
I did this on my calc. I got log(e)/x.

Its weird that it doesn't depend on k.

Unless k is a variable, this is correct. I got 1/(x ln10), which is an equivalent expression.

Otherwise, it would be (k-x)/(xkln10). // Scratch that. Bad math. It would have to be a partial derivative, unless one of the variables could be expressed in terms of the other.

 

[JJChicken]

Originally posted by: GML3G0

Originally posted by: Leros
I did this on my calc. I got log(e)/x.

Its weird that it doesn't depend on k.

Unless k is a variable, this is correct. I got 1/(x ln10), which is an equivalent expression.

Otherwise, it would be (k-x)/(xkln10).

Well K is a constant, and that's what I got. I think my lecturer might have made a mistake in his notes, look at this:

d1=[(log (x/k))]/C

d(d1)/dx = 1/(Cx)

only variable is x

my answer is d(d1)/dx = 1/(Ckx)

 

[JJChicken]

Originally posted by: Barack Obama

Originally posted by: GML3G0

Originally posted by: Leros
I did this on my calc. I got log(e)/x.

Its weird that it doesn't depend on k.

Unless k is a variable, this is correct. I got 1/(x ln10), which is an equivalent expression.

Otherwise, it would be (k-x)/(xkln10).

Wait i miss-read u guys, how come it doesn't depend on k?

 

[JJChicken]

by the way when I say log, i mean log(e) not log(10)

 

[DanDaManJC]

first change the base

log (x/k) = ln((x/k)/10)

then you can just differentiate using the ln(x) rule for differentiation, ofc assuming differentiating with respect to x and x is the variable

edit: ohhh ln, not log10, iono then couldnt you just use the ln(x) differentiating rule dealio?

 

[GML3G0]

Originally posted by: Barack Obama
by the way when I say log, i mean log(e) not log(10)

What do you mean log(e). Is the expression a base e logarithm? If so, the answer is just 1/x. The 1/k cancels out when you use the chain rule.

 

[Leros]

Originally posted by: GML3G0

Originally posted by: Barack Obama
by the way when I say log, i mean log(e) not log(10)

What do you mean log(e). Is the expression a base e logarithm? If so, the answer is just 1/x. The 1/k cancels out when you use the chain rule.

log(e) means log e base 10

 

[JJChicken]

sorry i meant Log natural. Thanks GML3G0, can't believe I was so stupid not to see that.

 

[ShotgunSteven]

Originally posted by: Barack Obama
sorry i meant Log natural. Thanks GML3G0, can't believe I was so stupid not to see that.

I don't think anyone here will have trouble believing it.

 

[JJChicken]

Originally posted by: ShotgunSteven

Originally posted by: Barack Obama
sorry i meant Log natural. Thanks GML3G0, can't believe I was so stupid not to see that.

I don't think anyone here will have trouble believing it.

That's true, I was once contemplating buying a Mac 😛

 

[DrPizza]

Why not use properties of logs BEFORE taking the derivative?

ln(x/k) = ln(x) - ln(k)
and, since k is a constant, the derivative of ln(k) = 0

Thus, it's retardedly simple (when you realize you can break it up that way).

If by log, you meant log base 10, then you can still break it up into two logs.
and the derivative of log base 10 (x); via change of base to ln(x)/ln(10), is 1/(x(ln10))

Originally posted by: Leros
I did this on my calc. I got log(e)/x.

Its weird that it doesn't depend on k.

It shouldn't be, when you consider the properties of logs. If you were graphing y=ln(x/5) it would be the same as the graph of ln(x) shifted down ln(5) units. And, shifting a graph in the vertical direction does not affect the shape of the graph, nor the instantaneous rate of change at any x-value. But (*sigh*) such knowledge is "obsolete" in an age where everyone overly-relies on graphing calculators. Hence, there is little to no thought put into what graphs of functions look like. Perhaps another way of explaining it is "y=x² +5x -k" - the derivative does not depend on k. It merely shifts the parabola in the vertical direction, not affecting the instantaneous rate of change at any x value (slope of the tangent line.)


Also, in B.O.'s post of the lecturer's solution, since it was 1/C * (log(x/k)), and his answer is 1/(Cx), then it appears that the "log" notation was referring to natural log, rather than log base 10. That's almost starting to get annoying as "log" is more and more commonly being used to mean "natural log", but "log" on a calculator is "log base 10." You have to realize what context you're in.

And, lastly to address the answer of "log(e)/x", it's the same as 1/(xln10); the 1/(ln10) part, via change of base formula become 1/(log(10)/log(e)) = log(e)/log(10) = log(e)/1

 

[Fenixgoon]

Originally posted by: DrPizza
And, shifting a graph in the vertical direction does not affect the shape of the graph, nor the instantaneous rate of change at any x-value. But (*sigh*) such knowledge is "obsolete" in an age where everyone overly-relies on graphing calculators. Hence, there is little to no thought put into what graphs of functions look like.

i can't tell you how often i use LN in math.. but i still don't know the shape of it 😛

 

[Tiamat]

Originally posted by: DrPizza
Why not use properties of logs BEFORE taking the derivative?

ln(x/k) = ln(x) - ln(k)
and, since k is a constant, the derivative of ln(k) = 0

Thus, it's retardedly simple (when you realize you can break it up that way).

Yup, this is the simplist way and most easy to see. Always take advantage of log rules.