differentiation question... (1 Viewer)

[JamiL]

i think it might be y=1.1^n
find dy/dn
therefore dy/dn= 1.1^n * ln1.1
proof..
y=1.1^n
n=ln y/ln1.1 (deff of log)
ie dn/dy = 1/yln1.1
therefore dy/dx = y ln1.1
y=1.1^n thou
therefor dy/dx = 1.1^n * ln1.1
if thats not it i got no idea wot the hell u r talkn about

 

[Slidey]

"Find an equation of the tangent to the graph of (x^2+y^2)^3 = 8x^2y^2 at the point (-1, 1)"

(x^2+y^2)^3 = 8x^2y^2
D{(x^2+y^2)^3}=D{8x^2y^2}
3*D{x^2+y^2}*(x^2+y^2)^2=8(D{x^2}y^2+D{y^2}x^2)
3*(2x +2yy')*(x^2+y^2)^2=8(2xy^2+2yx^2 * y')
6x*(x^2+y^2)^2 + 6yy'*(x^2+y^2)^2 = 16xy^2 + 16yx^2 * y'
6x*(x^2+y^2)^2 - 16xy^2 = (16yx^2 - 6y(x^2+y^2)^2)y'
y' = (6x[x^2+y^2]^2 - 16xy^2)/(16yx^2 - 6y[x^2+y^2]^2)

m=y'
At point (-1,1), y' is:
(6(-1)[(-1)^2+(1)^2]^2 - 16(-1)(1)^2)/(16(1)(-1)^2 - 6(1)[(-1)^2+(1)^2]^2)=
(-24 + 16)/(16 - 24)=1
tangent is:
f(x)=x+b
f(-1)=-1+b=1
b=2
so the equation fo the tangent is:
f(x)=x+2

 

[Li0n]

I thought implicit diff was 3u...

i wish :/

 

[mojako]

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)

it is a kind of fraction

well some ppl use the word differential form to mean equations like dy=2dx or dy=2xdx, so they can tell which one is which I guess.

 

[Xayma]

i think it might be y=1.1^n
find dy/dn
therefore dy/dn= 1.1^n * ln1.1
proof..
y=1.1^n
n=ln y/ln1.1 (deff of log)
ie dn/dy = 1/yln1.1
therefore dy/dx = y ln1.1
y=1.1^n thou
therefor dy/dx = 1.1^n * ln1.1
if thats not it i got no idea wot the hell u r talkn about

That is wrong.

You are differentiating with respect to n not x.

You went from dn/dy to dy/dx. It should have been dy/dn.

 

[Slidey]

i think it might be y=1.1^n
find dy/dn
therefore dy/dn= 1.1^n * ln1.1
proof..
y=1.1^n
n=ln y/ln1.1 (deff of log)
ie dn/dy = 1/yln1.1
therefore dy/dx = y ln1.1
y=1.1^n thou
therefor dy/dx = 1.1^n * ln1.1
if thats not it i got no idea wot the hell u r talkn about

Let 11/10=a
Let y=a^n
lny=nlna
n=lny/lna
D{n}=D{lny}/lna
1=(y'/y)/lna
1=(y'/[ylna])
y'=ylna
Since y=a^n,
y'=a^n * lna
Remember a is a constant, namely 1.1.

Though my method is different to Justin's, so dunno what's up there.

 

[Li0n]

you guys are prooving the formula, you dont need to
all you need to know is that
d/dx(a^x) = lna.a^x

 

[mojako]

you guys are prooving the formula, you dont need to
all you need to know is that
d/dx(a^x) = lna.a^x

some ppl dont wanna memorise that formula

 

[Li0n]

ah i see, meh

 

[Slidey]

Proving > memorising.

 

[JamiL]

That is wrong.

You are differentiating with respect to n not x.

You went from dn/dy to dy/dx. It should have been dy/dn.

yer sorri, as u can see i started with dy/dn n then habbit hit in n i wrote dy/dx... same shit thou, it is right

 

[JamiL]

you guys are prooving the formula, you dont need to
all you need to know is that
d/dx(a^x) = lna.a^x

actualy u need 2 no the proof...

 

[Slidey]

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.

Althoguh I reckon you SHOULD know it.

 

[Li0n]

Why waste 5 or 6 lines of working out though.

Btw the ONLY question ive seen where you need to find the derivative of a function a^x is that quesiton 10 fish thing (i think it was 98 or 00), which is what i think this whole thread started from --> 1.1ⁿ

 

[Estel]

Slide Rule:

let y =a^x
then y =e^(xlna)
dy/dx = lna.a^x

 

[JamiL]

i dunno wot the fuk implistic diff is, but look at my proof, it contained using log n diff log... thats all 2u work, it is so easy 2 prove, like in 2u we need to no the proof for y=e^x, y'=e^x, it not that hard

 

[Slidey]

Slide Rule:

let y =a^x
then y =e^(xlna)
dy/dx = lna.a^x

Proof that 1+1=2
2=1+1
.'. 1+1=2

'Cause if I'm not mistaken, you used just a specific version of the rule you were trying to prove to prove it. Circular logic. y=e^(xlna) is still in the form y=a^n, you still need to prove y=a^n.

Jamil: Implicit is quite easy. When you have something like x^2+y^2=1, both variables are a function of eachother. So what you do is differentiate with respect to one of them and then use the chain rule on the other, keeping in mind that the other is a function of the other.

Example: d/dx(x^2+y^2)=d/dx(1)
=2x+2y'y=0
y'=-x/y

Another way of looking at it: for the function [f(x)]^2, if we differentiate it with respect to x, it becomes 2.f'(x).f(x).

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

 

[mojako]

Yet another way to look at implicit diff:
you just differentiate both sides independently with respect to x. (or to y if that's what you need)

dy/dx is indeed a fraction... some ppl might disagree but it can be treated as fraction.

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

 

[Estel]

I'm lost as to what you're all trying to achieve now.

withoutaface hasn't used implicit as far as I can tell.

 

[withoutaface]

I didn't use implicit, rather I differentated with respect to y, then reciprocated the dx/dy to make dy/dx and substituted in a^x for y.