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anonymous
 4 years ago
\[y=\sqrt{sinx+\sqrt{sinx+\sqrt{sinx+\infty}}}\]
anonymous
 4 years ago
\[y=\sqrt{sinx+\sqrt{sinx+\sqrt{sinx+\infty}}}\]

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0What is the value of \(\frac {dy}{dx} \) \[a)\frac{Cosx}{(2y1)}\]\[b) \frac{sinx}{(2y1)}\]\[c) \frac{2y}{(cosx1)}\]\[d) \frac{2y}{(sinx1)}\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Do you really want the answer?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Yes , I want to know How its done

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Okay, but maybe you want to reconsider again. Can you see some substitution that might work here?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Um ..substituting sinx ?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0No, 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. \[\Large \mathsf{y = \sqrt{\sin{x} + y}}\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[\frac{dy}{dx}= \frac{dy}{dx}+\frac{d}{dx}\sqrt{sinx}\]\[\frac{cosx}{2\sqrt{sinx}} + y'(x) \]

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.0this is a murderous application of the "chain rule". its desidned to get you familiar with the process i believe.

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.0given 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)*...*

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.0for example: y = sin(cos(ln(x))) y' = cos(cos(ln(x))) * (sin(ln(x))) * 1/x

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Okay , i'll just try it

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.0to dbl check yourself, you might wanna plug it into the wolframalpha.com

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I tried Wolframalpha but it gave me some other answer which is not in the choices
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