X = 4 sin (2y + 6), find dy/dx in terms of x

Please help me, I am stuck in a rut with this … I have a brain drain and the urge to burn this question.

2. Originally Posted by pablohacker

X = 4 sin (2y + 6), find dy/dx in terms of x

Please help me, I am stuck in a rut with this … I have a brain drain and the urge to burn this question.
dy/dx[x] = dy/dx[4*sin(2y + 6)]
1 = 2cos(2y + 6)*dy/dx
dy/dx = 1/(2cos(2y + 6))
dy/dx = 1/2*sec(2y + 6)

3. Originally Posted by pablohacker

X = 4 sin (2y + 6), find dy/dx in terms of x

Please help me, I am stuck in a rut with this … I have a brain drain and the urge to burn this question.
There are a number of ways of doing this, one of which is to rewrite this as:

y = (1/2) arcsin(x/4) -3

then

dy/dx = (1/8) 1/sqrt(1-(x/4)^2)

RonL

4. Originally Posted by pablohacker

X = 4 sin (2y + 6), find dy/dx in terms of x

Please help me, I am stuck in a rut with this … I have a brain drain and the urge to burn this question.
I just noticed that. You need to solve this problem so that there are no y's in the differential equation?

If so, we can solve the original problem for y, then differentiate in terms of x:

x = 4sin(2y + 6)
y = 1/2*arcsin(x/4) - 3

dy/dx[y] = dy/dx[1/2*arcsin(x/4) - 3]
dy/dx = 1/(8sqrt(1 - (x/4)^2))

5. Originally Posted by ecMathGeek
dy/dx[x] = dy/dx[4*sin(2y + 6)]
1 = 2cos(2y + 6)*dy/dx
dy/dx = 1/(2cos(2y + 6))
dy/dx = 1/2*sec(2y + 6)

d/dx [x] = d/dx [4 sin(2y+6)]

so:

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

rearranging:

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

but you will still have to rewrite the right hand side as a function of x.

RonL

6. Originally Posted by CaptainBlack

d/dx [x] = d/dx [4 sin(2y+6)]

so:

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

rearranging:

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

but you will still have to rewrite the right hand side as a function of x.

RonL
oops

Thank you CaptainBlack. I know better but for some reason I felt like writting dy/dx.

7. can some one please explain what happens to the arcsin(x/4) because its change has got me confused.

8. Originally Posted by pablohacker
can some one please explain what happens to the arcsin(x/4) because its change has got me confused.
the derivative of arcsin(u) = u'/sqrt(1 - u^2)

so since the derivative of x/4 is 1/4, arcsin(x/4) becomes

d/dx (arcsin(x/4)) = (1/4)*1/sqrt(1 - (x/4)^2)

9. why is the derivative of arcsin u u'/sqrt(1-u^2), how is this got.

10. Originally Posted by pablohacker
why is the derivative of arcsin u u'/sqrt(1-u^2), how is this got.
i guess you can prove it using the limit definition of a derivative, but i don't remember if that's the way i was taught it. it's actually a well-known fact that everyone just takes for granted. it's like saying the derivative of sine is cosine, no one asks how that is, because its so common. derivatives of arcsine, arccos, arctan, etc should be in the back of your calculus book or wherever the formula page is, check it out. if you're still skeptical, try the limit definition to prove it (there may be an easier way, don't remember).

11. Originally Posted by Jhevon
i guess you can prove it using the limit definition of a derivative, but i don't remember if that's the way i was taught it. it's actually a well-known fact that everyone just takes for granted. it's like saying the derivative of sine is cosine, no one asks how that is, because its so common. derivatives of arcsine, arccos, arctan, etc should be in the back of your calculus book or wherever the formula page is, check it out. if you're still skeptical, try the limit definition to prove it (there may be an easier way, don't remember).
The way you derive it is quite simple (using a nice theorem)*

sin (asin x) = x on (-1,1)
Take derivative and use chain rule.

*)If f is invertible differenciable function then f^{-1} is a differenciable function (on its domain ... and ... whatver).