differentiation question... (1 Viewer)

[peeasoup]

basically, the differential of (1.1) to the power n = In(1.1).(1.1)^n
i dont get it ?

 

[withoutaface]

1.1 is a constant

:. d/dx(1.1)=0

 

[paper cup]

1.1 is a constant

:. d/dx(1.1)=0

to the power of n though. hmm oh yeh but it cancels out anyway. are you sure you have the right question?

 

[mojako]

I dont get the question

but btw, differential is not the same as derivative.
say y = 2x
derivative (dy/dx) = 2
differential (dy) = 2 dx

 

[peeasoup]

ok, the questions answer - working out, blah blah. Then is goes (NB d(1.1)^n / dn = In(1.1).(1.1)^n)

it wants this to be differentiated...........-810 000(1.1)^n
there answer to it is -810 000 In (1.1).(1.1)^n

I thought that (1.1)^n differentiated would be n(1.1)^(n-1)...

I dont know. Maybe im reading it wrong, who knows. This is question 10. b. iv. of the 1998 2 unit test........
so if you got one of those success one hsc books like me, check it out..

 

[Xayma]

(1.1)ⁿ=e^{ln (1.1)ⁿ}
=e^{nln (1.1)}
d/dn (e^{nln (1.1)})=ln (1.1)*e^{nln (1.1)}
=(1.1)ⁿ*ln (1.1)

 

[Freedom_Dragon]

I dont get the question

but btw, differential is not the same as derivative.
say y = 2x
derivative (dy/dx) = 2
differential (dy) = 2 dx

Derivative/Differentiate/Differential is all the same me think.

Eg:

Find the derivative of
y=cosx
y'=-sinx

Differentiate
y=(1+2x)^32
y'=64(1+2x)^31

Write down the Differential coefficient of
y=xcosx
y'=-xsinx+cosx

Its kinda like saying *chocolo baby* in 3 different language. ^^

 

[peeasoup]

ohh thanks Xayma. But why exactly was it done that way ? ?
Can you do it this way - differentiated would be n(1.1)^(n-1) ? ?? ? ?

 

[Estel]

You can't because the variable is the index.

 

[mojako]

Looks like my previous post isn't 100% correct.

differentiate: a verb to mean find the derivative of
derivative: a noun
differential: a noun, in itself meaning an infinitesimal change in a variable, written as d(something)
Note: I might be wrong.. differential might mean dy only, where y=f(x) (with dx not being a differential)
differential equation: any equation containing differential
So dy=2dx is a differential eqn, and dy/dx=2 is also a differential equation.
But one thing I'm sure is that if the derivative of y=2x is "2", the differential of y=2x is definitely not "2", it's likely to be "2dx" (assuming you can say "differential of y=2x".

Regarding differential equation:
dy/dx=2 tells the instantaneous *rate* at which y is changing per 1 unit change in x... like instantaneous velocity being the rate of change of displacement per 1 unit change in time
dy=2dx tells the *amount* of change in y when x changes by dx (remembering dx is a quantity.. probably 0.000000000000000000000000000000001 units).

differential coefficient... now this one is interesting coz ive never seen it in any papers (not that ive seen the word "differential" on any HSC papers)... where did you find it?
but it means the derivative.
for example, if you have y=2x, the differential equation can be written as dy=2dt, and the coefficient on the RHS is 2.

Someone can correct me if I'm wrong ^_^

 

[peeasoup]

I just assumed differential was another way of say to differentiate. But i reckon that to differeniate, find the derivative, and find differential is 'kinda' similar.
Like y=2x differentiated is 2
the derivative is 2
but if they were gonna use differential, that answer would be dy/dx=2.
But im just guessing, i dont remember seeing the word differential in anything.

 

[Xayma]

ohh thanks Xayma. But why exactly was it done that way ? ?
Can you do it this way - differentiated would be n(1.1)^(n-1) ? ?? ? ?

As estel said, you can't because the variable is the index.

You can however differentiate y=x^k with respect to x in the same fashion (although I dont see why someone would).

y=x^k
y=e^{ln (x)^k}
=e^{k*ln (x)}
dy/dx=k/x*e^{k*ln (x)}
=k/x*x^k
=kx^k-1

 

[paper cup]

(1.1)ⁿ=e^{ln (1.1)ⁿ}
=e^{nln (1.1)}
d/dn (e^{nln (1.1)})=ln (1.1)*e^{nln (1.1)}
=(1.1)ⁿ*ln (1.1)

heh apparently...
d/dx a^x = a^x In a
I just found that question, 98 paper q 10. lol. thank you xayma, and comb your hair. sorry, spamming, my bad.

 

[withoutaface]

I tend to use implicit diff:

y=a^x
lny/lna=x
1/ylna=dx/dy
ylna=dy/dx
(a^x)lna=dy/dx

< /spam>

EDIT: This is not part of the 2 or 3u course, so please disregard this post because you don't need this level of understanding

 

[Xayma]

Really useful if that was taught at the 2unit or even 3unit level (not that I don't use it when I can).

 

[withoutaface]

Really useful if that was taught at the 2unit or even 3unit level (not that I don't use it when I can).

I thought implicit diff was 3u...

 

[Xayma]

The closest we get is:

^dy/_dx=^dy/_du*^du/_dx

Not in any form like the above.

 

[mojako]

The closest we get is:

^dy/_dx=^dy/_du*^du/_dx

Not in any form like the above.

that is called:
"the technique of adding an extra variable"
or the chain rule, if you prefer that word...

implicit diff.. I think the idea of implicit diff is apparent in the binomial identity stuff [ remember Rench's thread about (1+x)^n(1+x)^n=(1+x)^2n ]
basically it is about differentiating LHS and RHS separately.

so you can say that 3U students learn implicit differentiation implicitly, while 4U students learn it explicitly.

 

[Xayma]

Yeah I know its the chain rule but that is the closest to it.

Meh I learnt it out of the maths ext 2 forum so Im happy.

 

[JamiL]

I dont get the question

but btw, differential is not the same as derivative.
say y = 2x
derivative (dy/dx) = 2
differential (dy) = 2 dx

dy/dx=2 is the same as dy=2dx... think of it as a fraction(even thou its not, jus a different way of righting it like y' or m)