Question

Derivative of y = sqrt(sin x +sqrt(sin x) + sqr(sin x).......)

Answer 1

\[d/dx ( \sqrt{\sin x +\sqrt{\sin x +\sqrt{\sin x..........}}}\] Medals will be awarded.

Answer 2

I think it is equal to \[d/dx(\sin x ^{1/2(n)}) \] where n is equal to the number of \[\sqrt{\sin x}\] But I think I forgot about the adding part I think its multiplication in n times

Answer 3

Nope. Nice try though. Think in another direction.

Answer 4

Keep in mind this is an infinite series. There's no finite "sines."

Answer 5

Hint: Use implicit differentiation.

Answer 6

And please call in others. If the person you call in can solve it, you guys can both get a medal.

Answer 7

okay. But what is Implicit Differentiation? its my first time seeing that kind of word . xD

Answer 8

Basically it's a type of differentiation dealing with implicit functions (functions containing both x and y). Example\[xy = c\]\[ (xy)' = xy' + y = 0 \] y' = dy/dx = -x/y.

Answer 9

@TuringTest @Zarkon @Hero @Callisto @robtobey @radar @ujjwal @Neemo Care to try to tackle this problem.

Answer 10

\[y' = \frac {\cos x}{2y -1}\]Is that correct?

Answer 11

\[y = \sqrt {\sin x + \sqrt {\sin x + \sqrt {\sin x +...}}}\]\[y^2 = \sin x + y\]\[2y y' = \cos x + y'\]\[y'(2y - 1) = \cos x\]\[y' = \frac {\cos x}{2y -1}\]

Answer 12

This reminds me of a problem back in algebra class ;)

What does \[\sqrt {2 + \sqrt { 2 + \sqrt {2 + ...}}}\] converge to?

Answer 13

Absolutely correct sir.

Answer 14

*If you're asking me what it converges to, it's 2 I believe.