## calyne 3 years ago Find the derivative of the function y = sin^[-1](2x+1)

1. calyne

i got 2/sqrt(1-(2x+1)^2) but apparently that ain't right

2. lolu

did you try solving out the (2x+1)^2?

3. dpaInc

|dw:1333934423544:dw|

4. dpaInc

expand the radicand and factor out the 4 to get the 2's to cancel... the you'll get y'=1/sqrt(-x^2-x)

5. calyne

uh.. no...? (2x+1)^2 is 4x^2 + 4x + 1. right. so how does that work.

6. calyne

crap

7. trflach

8. calyne

my trig is no good man just show me the way any way just show me so i get it

9. trflach

didnt use trig. just differentiation. treat it as (sin(2x+1))^-1 . bring down the -1, raise whole thing to -2. thats one part of the answer. Then multiply that by the derivative of the inside. The inside (sin(2x+1)) has a derivative of cos(2x+1). Thats the second part. The last part is the derivative of the inside of that. derivative of 2x+1 is 2. Theres the three parts. Multiply them all together. better?

10. calyne

well ok so it's -2 * 1/sin(2x+1)^2 * cos(2x+1)... so it's -2cos(2x+1)/sin(2x+1)^2..... soooooo.........

11. calyne

do i have that right so far? and then what?

12. trflach

thats it i believe. is it not right?

13. calyne

no dpalnc had it it's y'=1/sqrt(-x^2-x) in the textbook

14. calyne

so.... yeah oy

15. trflach

okay, http://www.wolframalpha.com/input/?i=derivative+of++y+%3D+sin%5E-1%282x%2B1%29 i just looked it up. They use trig substitution and turn it into variables. look at the steps.

16. calyne

what. it just gives me the same answer i got. where do i go from there.

17. calyne

it doesn't bring you to the fluttering end solution that it shows it just brings me to 2/sqrt(1-(2x+1)^2 which is what i got anyway

18. calyne

nvm

19. calyne

arccos x just means the inverse cosine function right like cos^-1 (x) ??

20. dpaInc

|dw:1333946189893:dw|

21. calyne

ok thanks but that last q was unrelated

22. dpaInc

take out the 4... so it will be 2 outside.... the radical...