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anonymous

  • 5 years ago

if y=sin^-1(5x), then dy/dx=

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  1. Yuki
    • 5 years ago
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    is it \[\sin^{-1}(5x)\] ?

  2. anonymous
    • 5 years ago
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    yea!

  3. anonymous
    • 5 years ago
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    is that the answer you came up with?

  4. Yuki
    • 5 years ago
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    okay, so you understand that this is arcsin(5x) right ? if you know the formula, you can use that. what I recommend though, is implicit differentiation. are you familiar with it ?

  5. anonymous
    • 5 years ago
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    Erm...I have a pretty concise way of demonstrating the derivation of the derivative of arcsin, let me know if you'd like to see it ayanamarcell. :)

  6. anonymous
    • 5 years ago
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    implicit differeniation?..no im not familar with it.

  7. Yuki
    • 5 years ago
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    you go like this y = arcsin(5x) so using inverse relations, sin(y) = 5x when you differentiate both sides, do you know what happens?

  8. anonymous
    • 5 years ago
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    Yuki, there's a better way! :D (Well, if he/she's not familiar with implicit differentiation that is)

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  9. anonymous
    • 5 years ago
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    @Yuki: no not really

  10. anonymous
    • 5 years ago
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    Sorry Yuki, you were going there. :P But technically, you don't need implicit differentiation for this type of problem (check my file for clarification).

  11. Yuki
    • 5 years ago
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    okay, I don't want to force you down another thing you need to learn, so let's see what QM tells us. I am very interested

  12. anonymous
    • 5 years ago
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    @QM: I cant open your doc

  13. anonymous
    • 5 years ago
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    Ahh. :/ Darn. Then let me put my derivation here:

  14. anonymous
    • 5 years ago
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    thanks Yuki

  15. anonymous
    • 5 years ago
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    \[y = \sin^{-1} x\]\[x = \sin(y)\]\[\frac{dx}{dy} = \cos(y)\] (invert to get dy/dx) \[\frac{dy}{dx} = \frac{1}{\cos(y)}\] Using the main pythagorean identity,\[\frac{dy}{dx} = \frac{1}{\sqrt{1 - \sin^2(y)}}\] And because sin^2(y) = x^2 from our initial conditions,\[\frac{dy}{dx} = \frac{1}{\sqrt{1-x^2}}\]

  16. anonymous
    • 5 years ago
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    Because you only have a 5x inside the operator instead of x, you can multiply everything by the derivative of 5x (chain rule) and you have your derivative!

  17. Yuki
    • 5 years ago
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    Wow that was a GOOD one !

  18. anonymous
    • 5 years ago
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    (And, of course, the x^2 inside the square root will be a (5x)^2, to keep things consistent.) Thanks yuki!

  19. Yuki
    • 5 years ago
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    ayanamarcell, the one with implicit differentiation will be very similar to this proof, so if you understand this one you should not have problem learning it.

  20. anonymous
    • 5 years ago
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    thanks for the help! that was really simple to understand.

  21. anonymous
    • 3 years ago
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    wat about sin^-1 sqr x

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