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anonymous

  • 4 years ago

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

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  1. anonymous
    • 4 years ago
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    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
    • 4 years ago
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    Do you really want the answer?

  3. anonymous
    • 4 years ago
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    Yes , I want to know How its done

  4. anonymous
    • 4 years ago
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    Okay, but maybe you want to reconsider again. Can you see some substitution that might work here?

  5. anonymous
    • 4 years ago
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    Um ..substituting sinx ?

  6. anonymous
    • 4 years ago
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    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
    • 4 years ago
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    \[\frac{dy}{dx}= \frac{dy}{dx}+\frac{d}{dx}\sqrt{sinx}\]\[\frac{cosx}{2\sqrt{sinx}} + y'(x) \]

  8. amistre64
    • 4 years ago
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    this is a murderous application of the "chain rule". its desidned to get you familiar with the process i believe.

  9. amistre64
    • 4 years ago
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    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
    • 4 years ago
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    for example: y = sin(cos(ln(x))) y' = cos(cos(ln(x))) * (-sin(ln(x))) * 1/x

  11. anonymous
    • 4 years ago
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    Okay , i'll just try it

  12. amistre64
    • 4 years ago
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    to dbl check yourself, you might wanna plug it into the wolframalpha.com

  13. anonymous
    • 4 years ago
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    I tried Wolframalpha but it gave me some other answer which is not in the choices

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