Question

\[y=\sqrt{sinx+\sqrt{sinx+\sqrt{sinx+\infty}}}\]

Answer 1

What is the value of \(\frac {dy}{dx} \)
\[a)\frac{Cosx}{(2y-1)}\]\[b) \frac{sinx}{(2y-1)}\]\[c) \frac{2y}{(cosx-1)}\]\[d) \frac{2y}{(sinx-1)}\]

Answer 2

Do you really want the answer?

Answer 3

Yes , I want to know How its done

Answer 4

Okay, but maybe you want to reconsider again. Can you see some substitution that might work here?

Answer 5

Um ..substituting sinx ?

Answer 6

No, I think I must; I shall give you the procedure but it would have been nice if you'd figured it out on your own.
\[\Large \mathsf{y = \sqrt{\sin{x} + y}}\]

Answer 7

\[\frac{dy}{dx}= \frac{dy}{dx}+\frac{d}{dx}\sqrt{sinx}\]\[\frac{cosx}{2\sqrt{sinx}} + y'(x) \]

Answer 8

this is a murderous application of the "chain rule". its desidned to get you familiar with the process i believe.

Answer 9

given a function within a function within a function, you just derive it like peeling an onion.

y = f(g(h(...(x))))

y' = f'(gh...x) * g'(h...x)*h'(....x)*...*

Answer 10

for example:

y = sin(cos(ln(x)))

y' = cos(cos(ln(x))) * (-sin(ln(x))) * 1/x

Answer 11

Okay , i'll just try it

Answer 12

to dbl check yourself, you might wanna plug it into the wolframalpha.com

Answer 13

I tried Wolframalpha but it gave me some other answer which is not in the choices