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Differentiation Help! (3 Viewers)

nightweaver066

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I'm thinking along the lines of manipulating dy/dx expression so that it's in terms of x and y, then differentiate (would need to use implicit) which *may* yield the required equation, depending if you manipulated it correctly.
 
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I'm thinking along the lines of manipulating dy/dx expression so that it's in terms of x and y, then differentiate (would need to use implicit) which *may* yield the required equation, depending if you manipulated it correctly.
Yeah, I've done that to prove differential equations before

no idea how to apply it to this question though
 

RealiseNothing

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I'm thinking along the lines of manipulating dy/dx expression so that it's in terms of x and y, then differentiate (would need to use implicit) which *may* yield the required equation, depending if you manipulated it correctly.
What I've done is sub in x=1, which makes all the ln(x) go to 0, and all the x^2, x^3, etc go to 1, which yields an equation in terms of a, b, and c only.

But I can't figure out where to go from here.
 
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What I've done is sub in x=1, which makes all the ln(x) go to 0, and all the x^2, x^3, etc go to 1, which yields an equation in terms of a, b, and c only.

But I can't figure out where to go from here.
Ah that's alright

no need to lose sleep over it (literally) :p
 

deswa1

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What I've done is sub in x=1, which makes all the ln(x) go to 0, and all the x^2, x^3, etc go to 1, which yields an equation in terms of a, b, and c only.

But I can't figure out where to go from here.
What I just tried was divide both sides by x^2 and then use implicit differentiation. You quickly end up with:
<a href="http://www.codecogs.com/eqnedit.php?latex=x^2(\frac{d^2y}{dx^2})-2y=12x^3@plus;9x^2" target="_blank"><img src="http://latex.codecogs.com/gif.latex?x^2(\frac{d^2y}{dx^2})-2y=12x^3+9x^2" title="x^2(\frac{d^2y}{dx^2})-2y=12x^3+9x^2" /></a>

Which is so close to the required form. Maybe something along this line would work but I don't know tbh.
 

RealiseNothing

what is that?It is Cowpea
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What I just tried was divide both sides by x^2 and then use implicit differentiation. You quickly end up with:
<a href="http://www.codecogs.com/eqnedit.php?latex=x^2(\frac{d^2y}{dx^2})-2y=12x^3@plus;9x^2" target="_blank"><img src="http://latex.codecogs.com/gif.latex?x^2(\frac{d^2y}{dx^2})-2y=12x^3+9x^2" title="x^2(\frac{d^2y}{dx^2})-2y=12x^3+9x^2" /></a>

Which is so close to the required form. Maybe something along this line would work but I don't know tbh.
Sub in the values of y etc:



Divide through by x^2:









Seems legit lol.
 
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The 2 questions before this one involved implicit differentiation.

I guess it could possibly work here somehow
 

deswa1

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Found what I did wrong- read the question as 3x^3 not x^3. Will retry

EDIT: Still doesn't work. You end up with the same answer as above (though with the RHS as 9x^2 + 4x^3 so there actually is equality)
 
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nightweaver066

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isn't implicit in 4u
It's not just something for 4U, it's a technique we occasionally use to makes things easier.

Implicit differentiation is just an application of chain rule, and i think it would be applicable to rates of change questions and possible some sketching questions in 3U.
 

deswa1

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It's not just something for 4U, it's a technique we occasionally use to makes things easier.

Implicit differentiation is just an application of chain rule, and i think it would be applicable to rates of change questions and possible some sketching questions in 3U.
Yeah it is 4U but its not very hard and I reckon 3U students should learn it because its heaps useful at times (especially when you have something to do with a circle as it avoids having to take the root and differentiate a root which is annoying).
 

RealiseNothing

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ok what I did was differentiate everything 3 times (lol).

This left me with no ln(x) and the only 'x' left were in the denominator.

So by subbing in stuff like x=1, I ended up with a few different equations in terms of a, b, and c only. Then just use simultaneous equations.

Not sure if this was actually shorter though lol.
 

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