Question

Find dy/dx of
y=sin^(-1)x + cos^(-1)x

Answer 1

\[y=\sin ^{-1}x+\cos^{-1}x\]

Answer 2

use the identity that sin^(-1)x + cos^(-1)x is just pi/2
and dy/dx of y=pi/2 is ?

Answer 3

i got \[\frac{ 1 }{ 2\sqrt{1-x^2} }\]

Answer 4

But its the wrong answer in the book.

Answer 5

wondering what i did wrong.

Answer 6

\(\large \arcsin x +\arccos x =\pi/2\)

Answer 7

How?

Answer 8

its an identity
http://en.wikipedia.org/wiki/Inverse_trigonometric_functions

Answer 9

so you wont use the derivative formula?

Answer 10

if you use derivative formula, you would still get a 0

Answer 11

because i hate memorizing the identities, cant you use \[\sin^{-1}x=\frac{ 1 }{ \sqrt{1-x^2} }\]

Answer 12

1/ sqrt (1-x^2) + (-1/ sqrt (1-x^2)) = 0

Answer 13

Then can u teach me the long way by using the derivative formula?

Answer 14

OHHHHHh ommmggggg that negative in cosine, totally didnt see that. ICIC thanks

Answer 15

welcome ^_^