## anonymous 4 years ago $y=\sqrt{sinx+\sqrt{sinx+\sqrt{sinx+\infty}}}$

1. anonymous

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)}$

2. anonymous

Do you really want the answer?

3. anonymous

Yes , I want to know How its done

4. anonymous

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

5. anonymous

Um ..substituting sinx ?

6. anonymous

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. $\Large \mathsf{y = \sqrt{\sin{x} + y}}$

7. anonymous

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

8. amistre64

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

9. amistre64

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)*...*

10. amistre64

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

11. anonymous

Okay , i'll just try it

12. amistre64

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

13. anonymous

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