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# C3 differetation question watch

1. Given that x=4sin(2y+6) find dy/dx in terms of X

What i did was found dx/dy to get 8cos(2y+6).. then found dy/dx as 1/8cos(2y+6)... however im not sure where to go from here. The mark scheme gets arcsins from somewhere>?

help plz
2. Yep, so far; so good! But it wants the answer in terms of x; your answer is in terms of y. Now, using the original equation (x=4sin(2y+6)) get 2y+6 on its own. That'll get you your arcsins!
3. (Original post by jakezg)
Given that x=4sin(2y+6) find dy/dx in terms of X

What i did was found dx/dy to get 8cos(2y+6).. then found dy/dx as 1/8cos(2y+6)... however im not sure where to go from here. The mark scheme gets arcsins from somewhere>?

help plz
Hello,

I would try taking the inverse of both sides.

arcsin (x/4) = 2y + 6

then differentiate. The formula for arcsin is in the formulae book for edexcel (FP2).
4. (Original post by jakezg)
Given that x=4sin(2y+6) find dy/dx in terms of X

What i did was found dx/dy to get 8cos(2y+6).. then found dy/dx as 1/8cos(2y+6)... however im not sure where to go from here. The mark scheme gets arcsins from somewhere>?

help plz

x=4sin(2y+6)

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

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

Using: Sin^2(2y+6)+cos^2(2y+6) = 1

cos^2(2y+6) = 1 - Sin^2(2y+6)

We know that x=4Sin(2y+6), therefore x/4 = Sin(2y+6) and so
(x/4)^2 = Sin^2(2y+6)

Hence, dy/dx = 1/8Root1-(x/4)^2
5. What's arcsines C3??
6. its the inverse sine function. you know, sin^(-1)x

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