Maths C3 - Differentiation... Help?? Watch

X_IDE_sidf
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(Original post by Philip-flop)
Oh right I see. So that gives me...
 \frac{dy}{dx} = \frac{v du}{v^2 dx} - \frac{u dv}{v^2 dx}

But then how do I go from this to?...
 \frac{dy}{dx} = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2}
Marks may well be deducted if you treat \frac{du}{dx} as fractions like that.

 \frac{dy}{dx} = \frac{v}{v^2}\frac{du}{dx} - \frac{u}{v^2}\frac{dv}{dx}

They are Leibniz notation not fractions.
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Philip-flop
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#42
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(Original post by X_IDE_sidf)
Marks may well be deducted if you treat \frac{du}{dx} as fractions like that.

 \frac{dy}{dx} = \frac{v}{v^2}\frac{du}{dx} - \frac{u}{v^2}\frac{dv}{dx}
What? How else are you meant to prove it?
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X_IDE_sidf
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(Original post by Philip-flop)
What? How else are you meant to prove it?
I'm probably being a tad pedantic but you shouldn't treat \frac{dy}{dx} as a fraction and combine it like normal fractions, this is because it isn't a fraction.

Just keep the multiplication obviously separate.
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Philip-flop
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So I'm a little confused with how to these...

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I would know how to do these, if for example I take part (a), I would let  y = ln(x+1) and then work from there. But how do I do this when I have to let  f(x) = ln(x+1) ??
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RDKGames
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(Original post by Philip-flop)
So I'm a little confused with how to these...

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I would know how to do these, if for example I take part (a), I would let  y = ln(x+1) and then work from there. But how do I do this when I have to let  f(x) = ln(x+1) ??
y=\ln(x+1) is introducing a variable y which is dependent on x. When you have f(x)=\ln(x+1) there is absolutely no difference except from the fact that you do not have a y-variable anymore, you simply have a function through which you plug in a number, x, and get a number out.

If it makes it easier for you, you may change f(x) to y and work from there, you will get the same results.
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pereira325
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(Original post by Philip-flop)
So I'm a little confused with how to these...

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I would know how to do these, if for example I take part (a), I would let  y = ln(x+1) and then work from there. But how do I do this when I have to let  f(x) = ln(x+1) ??
yeah.  y= f(x) .
If  y= ln(x+1), let u = x+1, hence y= ln(u), now work out du/dx then dy/du.  ln(x) = 1/x, so ln(u)= 1/u. Then multiply and work out dy/dx
Also, imo c3 ch.8 is a nice way to end the book! Compared to the complexity of parts in ch.7, lol.
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Philip-flop
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(Original post by RDKGames)
y=\ln(x+1) is introducing a variable y which is dependent on x. When you have f(x)=\ln(x+1) there is absolutely no difference except from the fact that you do not have a y-variable anymore, you simply have a function through which you plug in a number, x, and get a number out.

If it makes it easier for you, you may change f(x) to y and work from there, you will get the same results.
Oh yeah!! Panic over Thanks RDKGames !!

So for part (a) I would do exactly what I've been doing...

 f(x) = ln(x+1)

let...  t=x+1

This implies that...  \frac{dt}{dx} = 1 and that...  \frac{d[f(x)]}{dt} = \frac{1}{t}

Using the chain rule...
f'(x) = \frac{d[f(x)]}{dt} \times \frac{dt}{dx}

 f'(x) = \frac{1}{t} \times 1

 f'(x) = \frac{1}{t}

 f'(x) = \frac{1}{x+1}

(Original post by pereira325)
yeah.  y= f(x) .
If  y= ln(x+1), let u = x+1, hence y= ln(u), now work out du/dx then dy/du.  ln(x) = 1/x, so ln(u)= 1/u. Then multiply and work out dy/dx
Also, imo c3 ch.8 is a nice way to end the book! Compared to the complexity of parts in ch.7, lol.
Thank you!!

Yeah Chapter 7 was very difficult for me
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Philip-flop
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Hi everyone, you'll be pleased to know that I've finally got to the end of the Edexcel C3 Modular Maths Textbook!!

The bad news is... I'll be starting C4 sometime next week

Thanks to everyone who has helped me, I seriously appreciate it!!
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Naruke
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(Original post by Philip-flop)
Hi everyone, you'll be pleased to know that I've finally got to the end of the Edexcel C3 Modular Maths Textbook!!

The bad news is... I'll be starting C4 sometime next week

Thanks to everyone who has helped me, I seriously appreciate it!!
Congrats!

Make sure you do some past papers or you'll get rusty!
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RDKGames
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(Original post by Philip-flop)
Hi everyone, you'll be pleased to know that I've finally got to the end of the Edexcel C3 Modular Maths Textbook!!

The bad news is... I'll be starting C4 sometime next week

Thanks to everyone who has helped me, I seriously appreciate it!!
Well done.

I expect a "Maths C4 - Partial Fractions... Help??" thread soon.
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Philip-flop
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(Original post by Naruke)
Congrats!

Make sure you do some past papers or you'll get rusty!
Yeah Im going to try and do at least one past paper every two weeks, and go over some questions from the modular maths textbooks too
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Philip-flop
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(Original post by RDKGames)
Well done.

I expect a "Maths C4 - Partial Fractions... Help??" thread soon.
Hahaha!! This made me chuckle! Am I that predictable?

Thanks for always helping me out though man! I would still be stuck on half the C3 content if it wasn't for you!!
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Philip-flop
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So I've just looked over all my notes made from C3... Can anyone tell me what the following forms are, and when they might be needed/applied? I remember Zacken mentioning these to me and I found them being very useful. But I can't remember for what topic/s.

 (a+b)^2 = a^2 + 2ab + b^2

 (a-b)^2 = a^2 - 2ab + b^2
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RDKGames
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(Original post by Philip-flop)
So I've just looked over all my notes made from C3... Can anyone tell me what the following forms are, and when they might be needed/applied? I remember Zacken mentioning these to me and I found them being very useful. But I can't remember for what topic/s.

 (a+b)^2 = a^2 + 2ab + b^2

 (a-b)^2 = a^2 - 2ab + b^2
Those don't link to any specific topics - they're just basic, general expansions.

(x+y)(x-y)=x^2-y^2 can be a more useful technique to use at times, but again, it doesn't tie into any particular topic - it's just general subject knowledge. You can use this when you see a difference of two squares.
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Philip-flop
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(Original post by RDKGames)
Those don't link to any specific topics - they're just basic, general expansions.

(x+y)(x-y)=x^2-y^2 can be a more useful technique to use at times, but again, it doesn't tie into any particular topic - it's just general subject knowledge. You can use this when you see a difference of two squares.
Yeah I also made note of the difference of two squares too as it is used quite frequently throughout C3
I just thought...  (a+b)^2 = a^2 + 2ab + b^2 and  (a-b)^2 = a^2 - 2ab + b^2 might have had a name to it. One way I remember it is that it's similar to the form of Completing the Square

Thanks a lot RDKGames !!
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Philip-flop
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Ok so I know this isn't a differentiation question but I didn't know what one of my threads to post in. Part (c) is so tricky I really don't even know how to start

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Zacken
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(Original post by Philip-flop)
Ok so I know this isn't a differentiation question but I didn't know what one of my threads to post in. Part (c) is so tricky I really don't even know how to start
So you have x(t) = De^{-\frac{1}{8}t}. Then part(a) gets you to say that at t=0 you have x(0) = 10 = D. So this determines x(t) = 10e^{-t/8}.

Now in (c) you want the time T satisfying x(T) = 3. So all they want is for you to solve the equation e^{-T/8} = 3 which can be done by taking logarithms of both sides.

Which bit in particular is tripping you up?
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Philip-flop
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(Original post by Zacken)
So you have x(t) = De^{-\frac{1}{8}t}. Then part(a) gets you to say that at t=0 you have x(0) = 10 = D. So this determines x(t) = 10e^{-t/8}.

Now in (c) you want the time T satisfying x(T) = 3. So all they want is for you to solve the equation e^{-T/8} = 3 which can be done by taking logarithms of both sides.

Which bit in particular is tripping you up?
Ok, so for part (c)...

I know that after the second dose of the drug is given  t=5 where t is the time in hours.

But I need to find  T , the extra time (in hours) when  x , the amount of drug in the bloodstream is equal to 3mg meaning that  x = 3

Therefore...
 10 e^{- \frac{1}{8} (5 + T)} + 10 e^{- \frac{1}{8} (T) } = 3

But then I just get stuck from there
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Zacken
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(Original post by Philip-flop)
Ok, so for part (c)...

I know that after the second dose of the drug is given  t=5 where t is the time in hours.

But I need to find  T , the extra time (in hours) when  x , the amount of drug in the bloodstream is equal to 3mg meaning that  x = 3

Therefore...
 10 e^{- \frac{1}{8} (5 + T)} + 10 e^{- \frac{1}{8} (T) } = 3

But then I just get stuck from there
Okay, I was tired last night. So from your equation: 10(e^{-5/8 - T/8} + e^{-T/8}) = 10e^{-T/8}(e^{-5/8} + 1) = 3 which is easily solvable
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Philip-flop
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(Original post by Zacken)
Okay, I was tired last night. So from your equation: 10(e^{-5/8 - T/8} + e^{-T/8}) = 10e^{-T/8}(e^{-5/8} + 1) = 3 which is easily solvable
Oh yeah!! A simple indices rule then taking out a common factor could have given me...
 10 e^{- \frac{1}{8} (5 + T)} + 10 e^{- \frac{1}{8} (T) } = 3

 10e^{-T/8}(e^{-5/8} + 1) = 3

then I could divide both sides by 10 and then take logs to both sides, then times both sidesby -8 to give...

 T = -8 ln ( \frac{3}{10( e^{- \frac{5}{8}} +1) } )

which gives...

 T = 13.1 to 3sf


Thanks Zacken
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