Question

Differentiate arcsin(sqrt(sinx))

Answer 1

\[\frac{ 1 }{ 1-\sqrt{sinx}^2 }(\frac{ 1 }{ 2 }(sinx^{-1/2}))(cosx)\]

Answer 2

in the denominator it'll be sqrt (1- sinx)

Answer 3

d arc(sinx)/dx= 1/ sqrt(1-x^2)

Answer 4

d [arcsin(sqrt(sinx))] / d [(sqrt(sinx)] = 1/ sqrt( 1- sinx)

Answer 5

\[\frac{ cosx }{ 2(\sqrt{1-(sinx)^2})\sqrt{sinx} }\]

Answer 6

denominator will be 2 sqrt [(1-sinx) sinx]

Answer 7

why is that?

Answer 8

d arc(sinx)/dx= 1/ sqrt(1-x^2)...u know this?

Answer 9

ya

Answer 10

the algebgra on the bottom is messing me up,

Answer 11

\[\sqrt{1-(sinx)^2}\sqrt{sinx}\]

Answer 12

d [arcsin(sqrt(sinx))] / d [(sqrt(sinx)] = 1/ sqrt( 1- sinx)
is it right?

Answer 13

d [arcsin(sqrt(sinx))] / d [(sqrt(sinx)] = 1/ sqrt [1- {sqrt(sinx)} ^2]

Answer 14

yes

Answer 15

im at that point

Answer 16

so why r u writing sqrt ( 1- sin^2 x) in d denominator..it'll be sqrt(1-sinx)

Answer 17

ohh

Answer 18

now u got it....:)

Answer 19

\[\frac{ cosx }{2 \sqrt{1-sinx}\sqrt{sinx} }\]

Answer 20

yes...corect

Answer 21

can that be cleaned up in the denom?

Answer 22

ie combining the sqrts?

Answer 23

oh just all under one radical,i think i got it

Answer 24

u can combine like sqrt(x) sqrt(y) = sqrt(xy), if x and y r not both negative

Answer 25

\[\frac{ cosx }{ 2\sqrt{sinx(1-sinx)} }\]

Answer 26

here u can combine because (1-sinx) will always be be greater than or equal to 0

Answer 27

so sin x and (1-sinx) both cant be negative...so u can combine them in one radical

Answer 28

thanks @akash123 !!!!!!!!!

Answer 29

:)

Answer 30

all set, thank you!!