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Differentiating logarithms and expoentials

by PlasmaSphere
Tags: differentiating, expoentials, logarithms
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PlasmaSphere
#1
Jan3-08, 09:31 AM
P: 78
1. Differenciate the following functions

i) y = x2ln(4x)

ii) y = ln (x + 1)/x

iii) y = ln (x2 - 1)1/2


2. Laws for differentiating logs and exponentials



3. I did some of the more easy one's, these ones just got me stumped.

i) i think you would use the product rule. so u = x2 and v = ln(4x)

du/dx = 2x and dv/du = 1/x

which gives; x2.1/x + ln(4x).2x = x + 2x ln(4x) ????

ii) I would start using the quotient rule, but i'm not sure where the ln comes in to it.

iii) Chain rule? i cant see what you have to break it into.

Are there any easy ways to learn how to differenciate logarithms and exponentials?, i dont want to learn each individual case off by heart, that would take a while. I cant seem to get my head around them.
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CompuChip
#2
Jan3-08, 09:37 AM
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i) Why the question marks? Looks fine to me.
ii) I'd use the quotient rule too. What does it say about the derivative of ( f(x)/g(x) ) ? What are f(x) and g(x) in this case?

iii) Start by y = u^1/2, u = ln(x^2 - 1). You're not done using the chain rule though.

As a general rule of thumb, always differentiate from the outside inwards. In iii), you see an expression to the power 1/2, so you start by deriving u^1/2. To do that, you need the chain rule again, start outward: define v such that it comes down to differentiating ln v, etc.

What you should learn by heart is:
- differentiating e^x gives e^x
- differentiating ln x gives 1/x
- product rule
- chain rule
If you apply them in the right order, they will always work.
PlasmaSphere
#3
Jan3-08, 09:59 AM
P: 78
Quote Quote by CompuChip View Post
As a general rule of thumb, always differentiate from the outside inwards. In iii), you see an expression to the power 1/2, so you start by deriving u^1/2. To do that, you need the chain rule again, start outward: define v such that it comes down to differentiating ln v, etc.

What you should learn by heart is:
- differentiating e^x gives e^x
- differentiating ln x gives 1/x
- product rule
- chain rule
If you apply them in the right order, they will always work.
thanx, that should help a lot. Just a couple more questions before i attempt them;

for number ii), what i dont get is what the ln applies to. does it only apply to the top part of the fraction? or the bottom too? I think the way they have written it in my textbook has confused me.

I haven't worked out how to do the proper script yet, but in my book the ln is multilplied by the whole fraction;

y = ln ((x + 1)/ x) so, is that equivalent to y = ln(x + 1)/x ?

you cant differenciate ln by itself, as that doesn't mean anything (i think), thats why i'm getting confused, i dont know which term applies to the ln from the fraction.

rocomath
#4
Jan3-08, 10:10 AM
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P: 1,754
Differentiating logarithms and expoentials

ii. use quotient rule ... and no those 2 are not equivalent forms

[tex]y=\ln{\frac{x+1}{x}}[/tex]

vs

[tex]y=\frac{\ln{(x+1)}}{x}[/tex]

The first one, is in the form of [tex]\ln{\frac{a}{b}}[/tex] which can be broken down into [tex]\ln{a}-\ln{b}[/tex] while the second is simply "ln" divided by x.
CompuChip
#5
Jan3-08, 10:12 AM
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That's a matter of putting parentheses to get an unambiguous expression. I would read
ln (x + 1) / x as
[tex]\frac{\ln(x + 1)}{x}[/tex]
and not as
[tex]\ln\left( \frac{x + 1}{x} \right) [/tex]
which I would write as
ln( (x+1) / x)

If you can't make out which is meant in your textbook, then you have a bad textbook
But you can do them both, if you want to practice. The only difference is in the order in which you apply the chain rule and the quotient (or product) rule.
PlasmaSphere
#6
Jan3-08, 10:26 AM
P: 78
[tex]\ln\left( \frac{x + 1}{x} \right) [/tex]
thats the one i meant. give me a min for the answers, i'm working out how to use the proper scripting this time to avoid confusion
PlasmaSphere
#7
Jan3-08, 10:41 AM
P: 78
still got a problem with ii)

[tex]y=\ln{\frac{x+1}{x}}[/tex]

to use the quotient rule i need to define what u and v are. I still dont know what i would put them as. just by looking at it i would say that [tex]u=x+1[/tex] and [tex]v=x[/tex] but then i am leaving out the [tex]ln[/tex] completely when i apply the rule to the variables, which i presume you shouldn't do
rocomath
#8
Jan3-08, 10:45 AM
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P: 1,754
You do not need u and v. No quotient rule is needed at all since it is simply ln(a/b). So let's make things easier by breaking it down using the logarithmic formula/rule.

[tex]y=\ln{\frac{x+1}{x}}[/tex] Breaks down to ...

[tex]y=\ln{(x+1)}-\ln{x}[/tex] Now differentiate as you would normally.

If you had ...

[tex]y=\frac{\ln{(x+1)}}{x}[/tex] then you would need the quotient rule.

u=ln, v=x
CompuChip
#9
Jan3-08, 11:14 AM
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Or, alternatively, you could solve
[tex]y = \ln \frac{x+1}{x}[/tex]
by first applying the chain rule to y = ln u with u = (x+1)/x. This gives (1/u) * u', where you can work out u' using the quotient rule.

But rocophysics way is a lot easier, if you have the ln(a/b) = ln(a) - ln(b) rule available
PlasmaSphere
#10
Jan3-08, 11:24 AM
P: 78
Quote Quote by rocophysics View Post
You do not need u and v. No quotient rule is needed at all since it is simply ln(a/b). So let's make things easier by breaking it down using the logarithmic formula/rule.

[tex]y=\ln{\frac{x+1}{x}}[/tex] Breaks down to ...

[tex]y=\ln{(x+1)}-\ln{x}[/tex] Now differentiate as you would normally.

If you had ...

[tex]y=\frac{\ln{(x+1)}}{x}[/tex] then you would need the quotient rule.

u=ln, v=x
Cheers, that makes it clearer.

[tex]y=\ln{(x+1)}-\ln{x}[/tex]

[tex]dy/dx={\frac{1}{x+1}}.{\frac{-1}{x}} = {\frac{-1}{x(x+1)}}[/tex]

i think thats right. The book doesn't have the answers for this section.
rocomath
#11
Jan3-08, 11:32 AM
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P: 1,754
it was y = lna "minus" lnb

so you need to work it separately
PlasmaSphere
#12
Jan3-08, 11:51 AM
P: 78
ok, so if [tex]y=\ln{(x+1)}-\ln{x}[/tex]

because differentiating ln x gives 1/x;

[tex]dy/dx={\frac{1}{x+1}}-{\frac{1}{x}}[/tex] or have i got that wrong?

i hate these questions, i find it hard to conceptualise exactly what i am doing, eveything starts losing its meaning the more i try to work it out.
rocomath
#13
Jan3-08, 11:58 AM
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P: 1,754
It's correct. Just remember that the derivative of ln is [tex]y'=\frac{1}{f(x)}f'(x)[/tex] that's it! Don't let it get you down.
CompuChip
#14
Jan3-08, 02:49 PM
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The idea is that eventually it gets more meaning.
But don't worry, everyone I tried to teach this so far had trouble with it, so I think it is one of the hardest parts of mathematics. On the other hand, it's a really ubiquitous one so a new world will open for you once you master it.
Gib Z
#15
Jan3-08, 06:49 PM
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If you want a quicker way for doing questions like iii) remember that instead of having to use the chain rule a second time, you can just bring that exponent down. ie [itex]a\log_b c = \log_b c^a[/itex]
CompuChip
#16
Jan4-08, 03:57 AM
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Gib Z, that is:
[tex]a \log_b(c) = \log_b(c^a)[/tex]
Again, it is not clear if the question means
[tex]\left( \ln(x^2 + 1) \right)^{1/2} = \sqrt{ \ln (x^2 + 1) }[/tex]
or
[tex]\ln\left( (x^2 + 1)^{1/2} \right) = \ln\left( \sqrt{x^2 + 1} \right)[/tex].


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