differentiation question... (1 Viewer)

[Estel]

So what's all this fuss about needing to know implicit in 2/3u?

"I would imagine so for 4 unit, but since the proof involves implicit differentiation, I doubt you'd need to know it for 2 or 3 unit." - Slide Rule

??? Is this referring to something else?

 

[withoutaface]

I incorrectly called my method implicit.

 

[Estel]

ahh ok

(word min)

 

[Slidey]

Oh also, thats the correct proof.
you're not using the version of the thing to be proven.
y=e^(xlna) is y=a^x

You said it yourself. y=e^(xlna) is y=a^x, so giving the derivative of y=e^(xlna) without proving it is the derivative, in order to prove the derivative of y=a^x, is not allowed.

y=a^x.
This simply means that y is a function such that you have a constant, a, with a rule "to the power of x" applied to it, where x is some variable or function of a variable (and hence, still a variable).

Now,

y=e^(xlna)... what do we have?
We have y, which is a function such that a constant, e, has the rule "to the power of x times a constant, lna" applied to it. THIS IS STILL IN THE FORM "a constant to the power of a variable", or y=a^x.

Now, of course, if you had previously proven what the derivative of e to the power of a variable was, you could certainly use the proof Estel gave.

d/dx(ln[e^{xlna}])=d/dx(xlna)=lna
d/dx(ln[e^{xlna}])=d/dx(lnu), where u=e^{xlna}
d/dx(lnu)=(1/u)*u'=1/(e^{ax})*d/dx(e^{ax})=lna (above)

So, d/dx(e^{ax})=lna.e^{ax}
Q.E.D

I tried to prove this using implicit differentiation and ended up with this (so I did the above instead):

y=e^{ax}
lny=ax
d/dx(lny)=d/dx(ax)
y'/y=a
y=y'/a
So, d/dx(e^{ax})=(e^{ax})/a

Where'd I go wrong?

 

[JamiL]

In your proof, you take the reciprocal of both sides of the equation, treating dy/dn as a fraction. Can you do this?

when my teacher showed us it, i asked the same question cos he told us they wont fractions and we couldn treat them like fractions, but dy/dx=1/(dx/dy)
he told us we could
like when u have dy/dn*dn/dx=dy/dx, some teacher will say the dn's will cancel out, that is wrong they realy dont but as u can see this is the chain rule and it works. the proof that i gave u is a valid 1

 

[JamiL]

In your proof, you take the reciprocal of both sides of the equation, treating dy/dn as a fraction. Can you do this?

when my teacher showed us it, i asked the same question cos he told us they wont fractions and we couldn treat them like fractions, but dy/dx=1/(dx/dy)
he told us we could
like when u have dy/dn*dn/dx=dy/dx, some teacher will say the dn's will cancel out, that is wrong they realy dont but as u can see this is the chain rule and it works. the proof that i gave u is a valid 1

 

[mojako]

Before I start, I need to say that 2U students should really ignore this junk and ignore this thread except the first few posts.

y=e^{ax}
lny=ax
d/dx(lny)=d/dx(ax)
y'/y=a
y=y'/a
So, d/dx(e^{ax})=(e^{ax})/a

the last line is wrong.
you should have have y' = a*y, that's it and the second last line is not necessary.
If you wanna differentiate the second last line for whatever reason,
d/dx (y' /a) is not y/a.. it will be y''/a

d/dx(ln[e^{xlna}])=d/dx(xlna)=lna
d/dx(ln[e^{xlna}])=d/dx(lnu), where u=e^{xlna}
d/dx(lnu)=(1/u)*u'=1/(e^{ax})*d/dx(e^{ax})=lna (above)

So, d/dx(e^{ax})=lna.e^{ax}
Q.E.D

the third line from the bottom is wrong.
the conclusion is wrong too.
d/dx(e^{ax})=a*e^{ax}

Remember that Estel was proving the derivative of y=a^x, not y = e^(ax).
a^x = [e^(lna)]^x = e^(xlna)

Now, of course, if you had previously proven what the derivative of e to the power of a variable was, you could certainly use the proof Estel gave.

That is a quotable result. It will never be asked to be proven.
But you might be asked to prove the derivative of y=a^x.

when my teacher showed us it, i asked the same question cos he told us they wont fractions and we couldn treat them like fractions, but dy/dx=1/(dx/dy)

dy/dx is a fraction taken as a limit.
delta(x) is represents a change in x
the limit of delta(y)/delta(x) as delta(x) becomes infinitely small is written as dy/dx.
so dy/dx is not a fraction in the sense that dy is not delta(y) and dx is not delta(x) and dy and dx are notations which aren't invented to represent quantities.
But it can be perfectly treated as fraction in any problems in any HSC maths courses (I mean to say that I can't guarantee it can always be treated as fraction in any situation since I'm still in year 12 and have limited knowledge).

below is a proof of the chain rule in case it adds to your understanding... and to scare you as well
dy/dx
= lim delta(y)/delta(x) as delta(x)->0
= lim [delta(y)/delta(u) * (delta(u)/delta(x)] as delta(x)->0 (multiplying top and bottom by delta(u) which is a quantity)
= lim delta(y)/delta(u) as delta(u)->0 * lim delta(u)/delta(x) as delta(x)->0 (since when delta(x)->0, delta(u)-> as well)
= dy/du * du/dx (by definition of the notation da/db)

 

[Jase]

Cure for insomnia. Math debate.

 

[Xayma]

when my teacher showed us it, i asked the same question cos he told us they wont fractions and we couldn treat them like fractions, but dy/dx=1/(dx/dy)
he told us we could
like when u have dy/dn*dn/dx=dy/dx, some teacher will say the dn's will cancel out, that is wrong they realy dont but as u can see this is the chain rule and it works. the proof that i gave u is a valid 1

As mojako said you can treat it as a fraction for the entire course as dy=lim (δy-->0) δy which is a quantity that is near 0.

In fact many uses you would be required to.

Eg if you are given dx/dt=kx

you have to manipulate it to
dx/x=k dt
∫ dx/x=∫ k dt
ln x=kt+C
x=e^kt+C
x=e^kt*e^C
=Ae^kt

Where A is a constant.

 

[Slidey]

Yea, I was going to correct my first proof but the boards were down. Thanks for explaining my mistake in the second one, though, Mojako. I new it was something simple, otherwise the laws of maths would come crashing down.

Proof that d/dx(a^x)=lna.a^x:

d/dx(ln(a^x))=d/dx(xlna)=lna ... (1)
d/dx(ln(a^x))=d/dx(lnu), where u=a^x,
d/dx(lnu)=u'/u=d/dx(a^x)/(a^x) ... (2)
Since both (1) and (2) equal d/dx(ln(a^x)), we conclude (1)=(2):
d/dx(a^x)/(a^x)=lna, multiplying both sides by a^x:
d/dx(a^x)=lna.a^x

Q.E.D.

By implicit differentiation:

y=a^x
lny=xlna
d/dx(lny)=d/dx(xlna)
y'/y=lna
y'=y.lna
And since y=a^x,
y'=lna.a^x

Q.E.D.

Both methods can also be used to prove the derivative of e^x is e^x.

Your proof of the chain rule is interesting, but I'll have to think about it some more to understand it.

d/dx is sort of weird. For example, d/dx * y=dy/dx, so one might incorrectly conclude d/dx * y^2=dy/dx * y. Blah.

 

[JamiL]

implicit differentiation is it HSC 3u, or prelim 3u or wot... cos i have no idea wot it is. ive done all the 2u course and all prelim 3u and im confused... n isn this a 2u thread, fuk the 3u shit... ill do it next year
Ps im accell 4 2u mathematics (not 3u) thats y i have done the whole course

 

[Estel]

Implicit is 4U.

 

[mojako]

As mojako said you can treat it as a fraction for the entire course as dy=lim (δ-->0) δy which is a quantity that is near 0.

lim (δ-->0) ??
looks interesting, hehe...
no I don't think you can have lim delta->0
because delta is not a quantity.
need to be delta(x) or delta(something).

d/dx is sort of weird. For example, d/dx * y=dy/dx, so one might incorrectly conclude d/dx * y^2=dy/dx * y.

(d/dx * y) * y = dy/dx * y
d/dx * (y^2) = 2y * dy/dx
But never use the * because it's not a product as I understand it. It should be
d(y^2)/dx
the thing inside (..) goes with and is inseparable from the d.

But.. second derivative.. d^2y/(dx)^2.. hmm... it looks like a product
need someone more expert to comment on this.

 

[Slidey]

Oh, I know d/dx*y^2 isn't d/dx * y, but it's all very strange to me.

I haven't covered any calculus at all in class yet; I just teach it to myself, so hopefully it'll all become clear when my teacher explains it.

 

[Slidey]

Eg if you are given dx/dt=kx

you have to manipulate it to
dx/x=k dt
∫ dx/x=∫ k dt
ln x=kt+C
x=e^kt+C
x=e^kt*e^C
=Ae^kt

Where A is a constant.

Interesting. This is exponential growth and ecay stuff, I think, right?

 

[mojako]

Oh, I know d/dx*y^2 isn't d/dx * y, but it's all very strange to me.

I haven't covered any calculus at all in class yet; I just teach it to myself, so hopefully it'll all become clear when my teacher explains it.

oh, that explains, haha
so wad have u been doing in yr11 though?
do many schools postpone calculus until yr 12?

 

[Xayma]

lim (δ-->0) ??
looks interesting, hehe...
no I don't think you can have lim delta->0
because delta is not a quantity.
need to be delta(x) or delta(something).

Yeah I forgot the y after δ
Fixed.

 

[Slidey]

oh, that explains, haha
so wad have u been doing in yr11 though?
do many schools postpone calculus until yr 12?

In year 11 2u we did:
Trigonometry
Algebra
Locus and parabola
Quadratics (identities, discriminant, et cetera)
Coordinate geom
Geom
3u:
Circle geom
Trig, including 2u radian work
Polynomials
Divison of an interval, quadratic inequalities, et cetera
Harder 2 unit

I think that's it. Personally, though, I've been going through Fitzpatrick 2u doing the questions and learnign stuff.

 

[Xayma]

Interesting. This is exponential growth and ecay stuff, I think, right?

Yep it is exponential growth and decay.

Where the rate of change is directly related to the amount of a substance. Thats the 2unit form, the 3 unit form is dx/dt=k(x-b) where k and b are constants. In this case rate of change is directly proportional to the difference between the amount of substance and b.

 

[JamiL]

thats 3u growth n decay thou, 2u only is dx/dt, dv/dt and where dx/dt=Velocity and dv/dt=accell but in 3u hsc course dont u do dx/da or dx/dv or sum shit like that