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Edexcel A2 C4 Mathematics June 2015 - Official Thread

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Original post by Ripper Phoenix
HOW DOES LN(2X) Become 1/x when differentiated?

let u = 2x
du/dx = 2

y = ln u
dy/du = 1/u


dy/dx = dy/du * du/dx
so 1/u * 2
2/u
sub u back in
2/2x
the twos cancel out so becomes 1/x
can someone explain question 6iii on June 2014, I dont understand how you integrate the y's???
Original post by Ripper Phoenix
HOW DOES LN(2X) Become 1/x when differentiated?


Using the chain rule:

let U=2X

Therefore DU/DX =2

D(LN(U)/DX = 1/U=1/2X

Therefore, D(LN(2X)/DX = 2/2X = 1/X

another way of looking at it is the way that when you differentiate ln it' always differentiates to f'(x)/f(x), which in this case is 2/2X = 1/X
Original post by Ripper Phoenix
HOW DOES LN(2X) Become 1/x when differentiated?


let u=2xu=2x

y=lnuy=lnu

dudx=2\dfrac{du}{dx}=2

dydu=1u\dfrac{dy}{du}=\dfrac{1}{u}

dudx∗dydu = 2u\dfrac{du}{dx} * \dfrac{dy}{du}\ =\ \dfrac{2}{u}

Sub uu back in

22x = 1x \dfrac{2}{2x}\ =\ \dfrac{1}{x}
for implicit differentiation, when do you make the denominator or numerator equal to zero? How do you know which one to do?
Original post by Moniii16081997
Could someone help me with this question please. Thank you.
ImageUploadedByStudent Room1434403036.403476.jpg


Posted from TSR Mobile


For the first bit you the volume of a cone is 1/3pir2h
given that the base is a right angled triangle this must mean the r=h to form 1/3pih3 thats my reasoning behind it.

For b, you know that dv/dt = 70

and by differentiating v you know that dv/dh = pih2
you can then use connected rates of change dh/dt = dh/dv * dv/dt
hence dh/dt (1/pih2)(70)

and for c, just split the variables and solve the differential equation
Reply 1806
Avatar for jmuz
jmuz
2
HOW DO YOU INTEGRATE

(5^4y ) - WITH RESPECT TO Y
Original post by maccisha
yeah, and if it is the same e.g. x^2 on top and x^2 on the bottom it can be done as A + B/denominators


Cheers! So if it's got say x^4 on top and x^2 on bottom, it can be done as Ax^2 + Bx + C + D/denominators
Original post by Bustamove
nah.. it builds up.. I only gave 2 examples.. I haven't told you about my failure on the 6 marker for question 5 (the one for finding the coordinates, only managed to get to the differentials) and my trig identities... I couldn't solve it because I used Tan2A = 2Tan/1-Tan^2A.... don't know why I did that.. I can't believe I didn't used Tan2A = Sin2A / Cos2A..
and when finding the solutions, I used the CAST diagram, but then I don't know why my brain didn't compute, but I didn't draw it facing negative.. I thought it was positive even though the angle I got was negative... (if that makes sense)

anyway, what's done is done.. my best case scenario is actual 56-57 based on Arsey's markscheme... I could have saved my self and secured my self a definite A even with an extra 1 or 2 marks if I had not shot myself in the foot.... every mark counts...


well i think a good nights sleep would still definitely get you that A/A* overall, best of luck!
Original post by Phenylethanone
for implicit differentiation, when do you make the denominator or numerator equal to zero? How do you know which one to do?


What do you mean?
Original post by H0PEL3SS
Between 13 and 14, so I'd go with Arsey's.
56-58 for the A
63-65 for the A*,
70-72 for the 100 (I hope :s-smilie:)


I would agree with this.
56
63
70

People on here seem to forget sometimes that the people on TSR aren't the only ones who took the exam.
Original post by jmuz
HOW DO YOU DO

(5^4y ) - WITH RESPECT TO Y



Differentiate to get (5^4y)ln5 then multiply by differential of 4y using chain rule
Original post by Phenylethanone
for implicit differentiation, when do you make the denominator or numerator equal to zero? How do you know which one to do?


If you mean when you have rearranged to get dy/dx
If you're interested in a tangent parallel to the x-axis, or the gradient being zero, the numerator is zero
If you're interested in one parallel to the y-axis, the gradient is undefined; therefore you can set the expression on the denominator equal to zero
Screenshot_2015-06-16-01-33-57.pngGUYS! I think this is helpful for all of youll asking for the formulae we might need
Original post by ember8
well i think a good nights sleep would still definitely get you that A/A* overall, best of luck!


Yea, thanks :smile: I'll try not to mess it up this time :colondollar:
best of luck to you too! :smile:
Original post by jmuz
HOW DO YOU INTEGRATE

(5^4y ) - WITH RESPECT TO Y


By substitution; let x=54yx = 5^{4y}
Original post by imnoteinstein
Screenshot_2015-06-16-01-33-57.pngGUYS! I think this is helpful for all of youll asking for the formulae we might need


Don't they give these to us in the question? they usually give us one of these formulas if it's related to the question? like the june 2014 one?
(edited 8 years ago)
Original post by RobynDear
can someone explain question 6iii on June 2014, I dont understand how you integrate the y's???


Split the variables, and rearrange, this bits probably the longest part of the question, but you should then get.

2Sin2yCosy dy = ex dx

Now let U = SinY

du/dy = Cosy

du = Cosy dy (which is coincidentally the last bit of the equation)

So the equation now becomes 2u2 du

Integrate with respect to U to get 2/3U3 sub U back in and we get 2/3Sin3Y = ex + c
Original post by Paraphilos
By substitution; let x=54yx = 5^{4y}


then what?
Original post by Mutleybm1996
then what?


Then, log⁡(x)=4ylog⁡(5)\log(x) = 4y\log(5). Now upon implicit differentiation, you should get a suitable substitution for dydy.

Some have stated general results for the behaviour of the function f(x)=axf(x) = a^x; I don't know what you have been taught so I am assuming that all you guys know is integration by substitution.

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