anonymous
  • anonymous
if y=sin^-1(5x), then dy/dx=
Mathematics
schrodinger
  • schrodinger
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yuki
  • yuki
is it \[\sin^{-1}(5x)\] ?
anonymous
  • anonymous
yea!
anonymous
  • anonymous
is that the answer you came up with?

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yuki
  • yuki
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 ?
anonymous
  • anonymous
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. :)
anonymous
  • anonymous
implicit differeniation?..no im not familar with it.
yuki
  • yuki
you go like this y = arcsin(5x) so using inverse relations, sin(y) = 5x when you differentiate both sides, do you know what happens?
anonymous
  • anonymous
Yuki, there's a better way! :D (Well, if he/she's not familiar with implicit differentiation that is)
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anonymous
  • anonymous
@Yuki: no not really
anonymous
  • anonymous
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).
yuki
  • yuki
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
anonymous
  • anonymous
@QM: I cant open your doc
anonymous
  • anonymous
Ahh. :/ Darn. Then let me put my derivation here:
anonymous
  • anonymous
thanks Yuki
anonymous
  • anonymous
\[y = \sin^{-1} x\]\[x = \sin(y)\]\[\frac{dx}{dy} = \cos(y)\] (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,\[\frac{dy}{dx} = \frac{1}{\sqrt{1-x^2}}\]
anonymous
  • anonymous
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!
yuki
  • yuki
Wow that was a GOOD one !
anonymous
  • anonymous
(And, of course, the x^2 inside the square root will be a (5x)^2, to keep things consistent.) Thanks yuki!
yuki
  • yuki
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.
anonymous
  • anonymous
thanks for the help! that was really simple to understand.
anonymous
  • anonymous
wat about sin^-1 sqr x

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