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

1. anonymous

is it $\sin^{-1}(5x)$ ?

2. anonymous

yea!

3. anonymous

is that the answer you came up with?

4. anonymous

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

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

implicit differeniation?..no im not familar with it.

7. anonymous

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

Yuki, there's a better way! :D (Well, if he/she's not familiar with implicit differentiation that is)

9. anonymous

@Yuki: no not really

10. 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).

11. anonymous

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

@QM: I cant open your doc

13. anonymous

Ahh. :/ Darn. Then let me put my derivation here:

14. anonymous

thanks Yuki

15. 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,$\frac{dy}{dx} = \frac{1}{\sqrt{1-x^2}}$

16. 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!

17. anonymous

Wow that was a GOOD one !

18. anonymous

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

19. anonymous

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

thanks for the help! that was really simple to understand.

21. anonymous