Question

y=sin^-1(lnx)
differentiate with respect to x.
cant we use the formula sin^-1(x)=1/root(1^2-x^2) straight away?

Answer 1

yes plus chain rule go give
\[\frac{1}{\sqrt{1-\ln^2(x)}}\times \frac{1}{x}\]

Answer 2

but the answer is 1/xcosy

Answer 3

\[y=\sin^{-1}(\ln(x))=>\sin(y)=\ln(x)\]
\[=>\cos(y) \cdot y'=\frac{1}{x}=>y'=\frac{1}{x \cos(y)}\]

Answer 4

\[\cos(y)=\sqrt{1-\sin^2(y)}\]

\[\sqrt{1-\sin^2(y)}=\sqrt{1-\ln(x)^2}\]

since \[\sin(y)=\ln(x)\]

Answer 5

yes why do we move the sin inverse over? can't we use the sin^-1 formula straight?

Answer 6

i mean why we multiply by sinx

Answer 7

why would i remember extra formulas?

Answer 8

but to do questions involving sin^-1, there's this square root formula right

Answer 9

this
\[\frac{d}{dx}\sin^{-1}(x)=\frac{1}{\sqrt{1-x^2}}\]

Answer 10

yes!

Answer 11

we can derive that based on what i did

Answer 12

cant we apply to that question too

Answer 13

that is what satellite73 did

Answer 14

what did he do i don't quite understand..

Answer 15

\[\frac{d}{dx}\sin^{-1}(g(x))=\frac{1}{\sqrt{1-g(x)^2}}g'(x)\]

Answer 16

oh but that isn't the answer in my book...

Answer 17

\[y=\sin^{-1}(x)=>\sin(y)=x=>\cos(y) y'=1=>y'=\frac{1}{\cos(y)}\]
now if sin(y)=x/1=opp/hyp
then we can find the adjacent side to y
using Pythagorean thm
\[adj^2=hyp^2-opp^2=1^2-x^2=>adj=\sqrt{1-x^2}\]
\[y'=\frac{1}{\sqrt{1-x^2}}\]

Answer 18

i just put y in terms of x since we began with x

Answer 19

so is it that both answers are the same?

Answer 20

yes

Answer 21

but y prime has no x 1/x

Answer 22

that is the derivative of the natural log

Answer 23

they are same
\[\cos(y)=\sqrt{1-x^2}\]

Answer 24

well looking at when we have y=sin inverse of x

Answer 25

what was complicated about my way?

Answer 26

i think it is actually easier than what satellite did

Answer 27

can you re explain all over again?

Answer 28

\[y=\sin^{-1}(\ln(x))=>\sin(y)=\ln(x)\]
\[\cos(y)\cdot y'=\frac{1}{x} => y'=\frac{1}{x \cos(y) }\]

Answer 29

we can write cos(y) in terms of x like satellite has his

Answer 30

so there is two answers to this question? the other one we use sin inverse x forumla?

Answer 31

imagine a right triangle
\[\sin(y)=\frac{\ln(x)}{1}=\frac{oppsite}{hypnoteuse} \]
we can find the adjacent side using the Pythagorean thm
\[adj^2=hyp^2-opp^2=1^2-(\ln(x))^2=>adj=\sqrt{1-(\ln(x))^2}\]
so \[\cos(y)=\frac{adjacent}{hypontenues}=\frac{\sqrt{1-(\ln(x))^2}}{1}=\sqrt{1-(\ln(x))^2}\]

Answer 32

satellite and me have same answer see

Answer 33

\[y'=\frac{1}{x \sqrt{1-(\ln(x))^2} }\]
and we don't have to remember extra formulas

Answer 34

y′=1x1−(ln(x))2−−−−−−−−−−√
for this, the root 1-lnx square, why is it one?

Answer 35

i can't read that

Answer 36

y′=1x1−(ln(x))2−−−−−−−−−−√
for this, the root 1-lnx square, why is it one?

Answer 37

it's satellite's answer but im just asking why is the denominator with square root, where did the one come frm?

Answer 38

i showed you above

Answer 39

maybe you can get zarkon to explain it better than me
but that is as good as my explanation will get im sorry