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anonymous
 5 years ago
if y=sin^1(5x), then dy/dx=
anonymous
 5 years ago
if y=sin^1(5x), then dy/dx=

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0is it \[\sin^{1}(5x)\] ?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0is that the answer you came up with?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0okay, 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
 5 years ago
Best ResponseYou've already chosen the best response.0Erm...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
 5 years ago
Best ResponseYou've already chosen the best response.0implicit differeniation?..no im not familar with it.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0you go like this y = arcsin(5x) so using inverse relations, sin(y) = 5x when you differentiate both sides, do you know what happens?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Yuki, there's a better way! :D (Well, if he/she's not familiar with implicit differentiation that is)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Sorry Yuki, you were going there. :P But technically, you don't need implicit differentiation for this type of problem (check my file for clarification).

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0okay, 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
 5 years ago
Best ResponseYou've already chosen the best response.0@QM: I cant open your doc

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Ahh. :/ Darn. Then let me put my derivation here:

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[y = \sin^{1} x\]\[x = \sin(y)\]\[\frac{dx}{dy} = \cos(y)\] (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,\[\frac{dy}{dx} = \frac{1}{\sqrt{1x^2}}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Because 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!

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Wow that was a GOOD one !

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0(And, of course, the x^2 inside the square root will be a (5x)^2, to keep things consistent.) Thanks yuki!

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ayanamarcell, 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
 5 years ago
Best ResponseYou've already chosen the best response.0thanks for the help! that was really simple to understand.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0wat about sin^1 sqr x
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