Maths C3 - Differentiation... Help??

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.

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

What? How else are you meant to prove it?

So I'm a little confused with how to these...



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 $f(x) = ln(x+1)$ ??

So I'm a little confused with how to these...



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.

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.

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!!

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!

Well done.

I expect a "Maths C4 - Partial Fractions... Help??" thread soon.

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.

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?

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