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Hard A-level Maths Differentiation Quenstions

Place to post THE HARDEST Y13 Differentiation A-level maths quenstions

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Reply 2
Avatar for mqb2766
mqb2766
21
Original post by CaptainDuckie
E8F85C92-3FB1-4E9A-8CD3-E9097B0A57E9.jpg.jpeg

Ahhh... there is a left opening bracket after ln and before x.
Tbh, that was the hardest thing, otherwise its just a bit tedious.
(edited 3 years ago)
Original post by mqb2766
Ahhh... there is a left opening bracket after ln and before x.
Tbh, that was the hardest thing, otherwise its just a bit tedious.



Yeah, I’d say just tedious. Other than that, should be straight forward for a normal student who’s learnt the principles👍
Not even tedious - took 3 lines.
Original post by ghostwalker
Not even tedious - took 3 lines.



With big pages...
Original post by CaptainDuckie
With big pages...


:no:
Original post by ghostwalker
:no:



How did you manage? I’m guessing you missed some working out and did some parts in your head?
Reply 8
Avatar for Notnek
Notnek
21
Something like this would cause problems for a lot of A Level students:

ddx((sin⁡x)cos⁡x)\displaystyle \frac{d}{dx}\left(\left({\sin x}\right)^{\cos x}\right )
Original post by ghostwalker
:no:

Thing is, it's a "show", so if you do *too* much of it mentally you haven't actually shown anything. (When I was prepping for Cambridge at at my peak for this kind of thing, I remember a somewhat similar (easier) "show" question from an S-level paper where literally the first line I wrote was the given answer and so I had to restart and actively put in a couple of intermediate steps).
Original post by CaptainDuckie
How did you manage? I’m guessing you missed some working out and did some parts in your head?

You don't have to do much in your head, the difficulty is mainly keeping track (mentally) where you are in the question. But the actual calculations you can mainly just write down.
Original post by DFranklin
You don't have to do much in your head, the difficulty is mainly keeping track (mentally) where you are in the question. But the actual calculations you can mainly just write down.



I agree, it took me 8 lines, rather not risk silly errors
Original post by CaptainDuckie
How did you manage? I’m guessing you missed some working out and did some parts in your head?


The deriviative took one line.

Substituting x=0 a second line.

Tidying up to the desired result the final line.

Edit: I think that shows full working.
(edited 3 years ago)
Original post by ghostwalker
The deriviative took one line.

Substituting x=0 a second line.

Tidying up to the desired result the final line.

Edit: I think that shows full working.



How did you get the derivative as one line, let me get some tips man👍
Original post by ghostwalker
The deriviative took one line.

Substituting x=0 a second line.

Tidying up to the desired result the final line.

Edit: I think that shows full working.

To be clear, I wasn't suggesting you wouldn't have shown full working in 3 lines, just that I think that's about as short as you can go.

I also took 3 lines, divided exactly the same way. If I'd just wanted an answer I could have got it into 2 lines, but I wouldn't have expected full marks in a "show that".
Original post by DFranklin
To be clear, I wasn't suggesting you wouldn't have shown full working in 3 lines, just that I think that's about as short as you can go.

I also took 3 lines, divided exactly the same way. If I'd just wanted an answer I could have got it into 2 lines, but I wouldn't have expected full marks in a "show that".



Yeah man, I’m pretty sure Questions like these on an A level paper want you to prove how you got your answer methodologically.

How would you get the derivative as one line though?
Original post by CaptainDuckie
How did you get the derivative as one line, let me get some tips man👍

I'm going to try to explain the thought process, as well as what you actually need to write. The lines that are what I'd actually write will all start with dydx=\dfrac{dy}{dx} = , and you'll see that at each step I'm only adding to what was previously written, so it can be part of the same (single) line of working.

So, we start with y=2{e2x+3ln⁡[x+(ex+1)2]}2y = 2 \left\{e^{2x} + 3 \ln \left[x + (e^x+1)^2\right]\right\}^2

We'll differentiate as y=2f(x)2y=2f(x)^2 (so y′=4f′(x)f(x)y' = 4f'(x)f(x)). So we start by writing down:

dydx=4(\dfrac{dy}{dx} = 4(

From here I want to put in the derivative of f(x)=e2x+3ln⁡[x+(ex+1)2]f(x) = e^{2x} + 3 \ln \left[x + (e^x+1)^2\right].

I can write down the derivative of e^2x:

dydx=4(2e2x\dfrac{dy}{dx} = 4(2e^{2x}

Now I need the derivative of 3ln⁡g(x)=3g′(x)g(x)3 \ln g(x) = 3 \frac{g'(x)}{g(x)} (where g(x) = x + (e^x + 1)^2). Obviously I can diff x immediately and put g(x) in the denominator:

dydx=4(2e2x+31+x+(ex+1)2\dfrac{dy}{dx} = 4( 2e^{2x} + 3 \dfrac{ 1 + }{x + (e^x+1)^2}

I still need to add the derivative of (e^x+1)^2 (which is 2e^x (e^x+1) by the chain rule) into the numerator:

dydx=4(2e2x+31+2ex(ex+1)x+(ex+1)2\dfrac{dy}{dx} = 4( 2e^{2x} + 3 \dfrac{ 1 + 2 e^x (e^x+1)}{x + (e^x+1)^2}

At this point I can close the bracket and multiply by f:

dydx=4(2e2x+31+2ex(ex+1)x+(ex+1)2)(e2x+3ln⁡[x+(ex+1)2])\dfrac{dy}{dx} = 4\left( 2e^{2x} + 3 \dfrac{ 1 + 2 e^x (e^x+1)}{x + (e^x+1)^2}\right)\left(e^{2x} + 3 \ln \left[x + (e^x+1)^2\right]\right)

(Note: it's a lot more tricky doing this and typesetting at the same time that doing it straight - I found + fixed one error, but if there are others, it's more an artifact of typesetting it than that it's inherently an error prone method).
(edited 3 years ago)
Original post by DFranklin
From y=2{e2x+3ln⁡[x+(ex+1)2]}2y = 2 \left\{e^{2x} + 3 \ln \left[x + (e^x+1)^2\right]\right\}^2

We'll differentiate as y=2f(x)2y=2f(x)^2 (so y′=4f′(x)f(x)y' = 4f'(x)f(x)). So we start by writing down:

dydx=4(\dfrac{dy}{dx} = 4(

From here I want to put in the derivative of f(x)=e2x+3ln⁡[x+(ex+1)2]f(x) = e^{2x} + 3 \ln \left[x + (e^x+1)^2\right].

Note: posting this half finished so you can see I'm going to answer, but a full answer is going to take more time.


Right, so it’s more of a direct method, didn’t really think you could do it all in two lines though lol
Evaluate ddx(xx)\dfrac{d}{dx} (x^x)
(edited 3 years ago)
Original post by RDKGames
Evaluate ddx(xx)\dfrac{d}{dx} (x^x)

A classic.

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