Tricky Trig Differentiation

thomas49th

1. The problem statement, all variables and given/known data

x = 5sin(2y +6) find dy in terms of x

2. Relevant equations

3. The attempt at a solution

dx/dy = 8 cos (2y+6)

dy/dx = 1 / (8cos(2y+6)

but according to the mark scheme the final answer is

+ or - 1 / (2 sqrt[16/x^2])

i dont see how on earth they got there :O

Can anyone? Thanks :)

dynamicsolo

The result they give is more directly found by inverting the original function so that it becomes an inverse-sine function: that would account for the 1/sqrt form of their answer.

As for your method, it may be better to use implicit differentiation to solve for dy/dx, rather than finding dx/dy and using the reciprocal. BTW, I don't see how you get an '8' there -- wouldn't the Chain Rule give you 10?

thomas49th

i though i should use the chain rule aswell but, but the mark scheme doesn't seem to use it.

here is the paper. it's question 4 ii

http://eiewebvip.edexcel.org.uk/Repo...e_20050620.pdf

here is the mark scheme

http://eiewebvip.edexcel.org.uk/Repo...s_20050618.pdf

wierd huh?

dynamicsolo

Having worked this through now, I believe you have a typo here:

Quote by thomas49th

x = 4 sin(2y +6) find dy in terms of x

This approach:

dx/dy = 8 cos (2y+6)

dy/dx = 1 / [8 cos(2y+6)]

does work and gives the same result as implicit differentiation, but only because your function f(y) is continuous in y everywhere. Such a method using reciprocal derivatives should be used with some caution...

How they get their answer is by a substitution. You know that sin(2y+6) = (x/4). How would you express cos(2y+6) in terms of x? (And I believe you also have a typo in your copy of their answer that you presented...)

thomas49th

yes yes your right

x = 4sin(2y +6) find dy in terms of x

+ or - 1 / (2 sqrt[16 - x^2])

substitution... ohh that sounds like a better method. how do i do that here?

y = (sin^-1(x/4) - 6)/2

is that right? What do i subistute in?

dynamicsolo

I don't know what stage you are at in rules of differentiation. Have you done derivatives of inverse trig functions? The relevant rule is

d/dx [sin^-1 (u)] = [ 1 / sqrt( 1 - u^2 ) ] · (du/dx) .

For our function, u = x/4 .

Alternatively, you can use your result

dy/dx = 1 / [8 cos(2y+6)] ,

together with

sin(2y + 6) = (x/4) and

sin u = sqrt( 1 - u^2 ) ,

to get to the given answer.

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thomas49th	i though i should use the chain rule aswell but, but the mark scheme doesn't seem to use it. here is the paper. it's question 4 ii http://eiewebvip.edexcel.org.uk/Repo...e_20050620.pdf here is the mark scheme http://eiewebvip.edexcel.org.uk/Repo...s_20050618.pdf wierd huh?

dynamicsolo Recognitions: Homework Help	I don't know what stage you are at in rules of differentiation. Have you done derivatives of inverse trig functions? The relevant rule is d/dx [sin^-1 (u)] = [ 1 / sqrt( 1 - u^2 ) ] · (du/dx) . For our function, u = x/4 . Alternatively, you can use your result dy/dx = 1 / [8 cos(2y+6)] , together with sin(2y + 6) = (x/4) and sin u = sqrt( 1 - u^2 ) , to get to the given answer.