maths problem

colm-ccfc84

from the 97 higher level paper 1.
q 6 b (ii)
let f(x) = sin4x + cos4x
find the derivative of f(X) and express it in the form ksinpx

Note where i typed sin4x and cos4x they are supposed to be to the power of 4 not by 4, I don't know how to type the to the power of.

Mushy

I havent thought much of this through, but could ya not do it as two chain rules where it becomes f '(x)=4sin3x(cosx) + 4cos3x(-sinx). then multiply this out and work your way towards an answer.

NOTE: Im also in the same situation where the 3 before the X should be a "power of".

I also take it that the P that you have to get it in the form of is a power, is it? Im too lazy to go upstairrs and check

JC 2K3

For future reference, x^2 = x to the power of 2(shift+6)

As for your q(Written in the way I like dealing with Qs like this):

y = (sinx)^4 + (cosx)^4

dy/dx = 4cosx(sinx)^3 - 4sinx(cosx)^3
=4cosxsinx((sinx)^2 - (cosx)^2)
=2sin2x(1/2 - 1/2cos2x -1/2 -1/2cos2x)
=-2sin2xcos2x
=-sin4x

k=-1 ; p=4

Done quickly, is it right?


Oh and btw, you'll never progress in maths if you just ask for the solution. If you're asking for help here in future try to explain why exactly you don't understand the question and perhaps type up your attempt at the question. The best way to learn is to get something wrong.

JakeTheMsitake

From the same paper, same question part (c):

If sin y = 1/2(1-x^2) for -sqrt3 < x < sqrt3

calculate the value of a and value of b when:

(dy/dx)^2 = a/(3-x^2) - b/(1+x^2), a,b elements of Nzero (N with the 0 subscript)

And yes, the above answer is correct.

JC 2K3

Bah, I was stumped last night with this and then I realised that by (dy/dx)^2 they don't mean a second derivative, they mean simply dy/dx squared, makes life a lot easier... this is what I got:

siny = 1/2(1 - x^2)
y = sin^-1((1 - x^2)/2)

dy/dx = 1/sqrt(4 - (1 - x^2)^2)*-2x
=-2x/sqrt(3 + 2x^2 - x^4)

(dy/dx)^2 = 4x^2/(3 + 2x^2-x^4)

(dy/dx)^2 = a/(3-x^2) - b/(1+x^2)
=(a + ax^2 - 3b + bx^2)/(3 + 2x^2 - x^4)

cancelling out the bottom of both expressions for (dy/dx)^2 gives us:
4x^2 = a - 3b + x^2(a + b)

which means:
a - 3b = 0
a = 3b

a + b = 4

4b = 4

b = 1
a = 3

Crumbs

JC 2K3 wrote:

Now, the answer in the EdCo book is a = 3, b = 1, and you can get this answer by making a very simple mistake. When differentiating y = sin^-1((1 - x^2)/2) it's easy to get: dy/dx = 1/sqrt(4 - (1 - x^2)^2)*-2x, which is incorrect, but probably what the person writing the EdCo book did.

Can anyone verify this?

I'm with EdCo on this. I think that when differentiating the arcsin function, you can work it out in either of two ways, each giving the same answer.

1 / sqrt(4 - (1-x^2)^2) * -2x

or

1 / sqrt(1 - ((1-x^2)/2)^2) * -x

In the first method, we treat our initial function as sin^-1 u/a, meaning we use the a^2 bit in our derivative (ie. 4) and then only use the chain rule on the numerator but in the second method, we treat our initial function as just sin^-1 u and so use the chain rule on the whole thing.

JC 2K3

Hmm... yeah, you make sense... my bad, I'm rusty with differentiation.

*EDITS POST