anonymous
  • anonymous
\[y=\sqrt{sinx+\sqrt{sinx+\sqrt{sinx+\infty}}}\]
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
What is the value of \(\frac {dy}{dx} \) \[a)\frac{Cosx}{(2y-1)}\]\[b) \frac{sinx}{(2y-1)}\]\[c) \frac{2y}{(cosx-1)}\]\[d) \frac{2y}{(sinx-1)}\]
anonymous
  • anonymous
Do you really want the answer?
anonymous
  • anonymous
Yes , I want to know How its done

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anonymous
  • anonymous
Okay, but maybe you want to reconsider again. Can you see some substitution that might work here?
anonymous
  • anonymous
Um ..substituting sinx ?
anonymous
  • 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. \[\Large \mathsf{y = \sqrt{\sin{x} + y}}\]
anonymous
  • anonymous
\[\frac{dy}{dx}= \frac{dy}{dx}+\frac{d}{dx}\sqrt{sinx}\]\[\frac{cosx}{2\sqrt{sinx}} + y'(x) \]
amistre64
  • amistre64
this is a murderous application of the "chain rule". its desidned to get you familiar with the process i believe.
amistre64
  • 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)*...*
amistre64
  • amistre64
for example: y = sin(cos(ln(x))) y' = cos(cos(ln(x))) * (-sin(ln(x))) * 1/x
anonymous
  • anonymous
Okay , i'll just try it
amistre64
  • amistre64
to dbl check yourself, you might wanna plug it into the wolframalpha.com
anonymous
  • anonymous
I tried Wolframalpha but it gave me some other answer which is not in the choices

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