Question

help....
hi can someone pls. explain to me about Inverse trigonometric functions(Differential Calculus).

pls. solve this problem then explain pls.
y=square root Arcsinx

y=xArcsin2x

Answer 1

\[y=\sqrt{Arcsin x}\]

Answer 2

are you looking for the intersection of the two curves?
i don't understand what the question is?
what i'm solving for?

Answer 3

the derivative of the inverse trig. functions

Answer 4

so you have two problems?
this is not one problem?

Answer 5

yes

Answer 6

can you pls. explain?

Answer 7

explanation in the both of pdfs

Answer 8

you have a specific question on what i did let me know

Answer 9

in the part getting the derivative i'm confused

Answer 10

you do know the chain rule right?

Answer 11

quotient rule?

Answer 12

yes

Answer 13

derivative of x with respect to x is 1
derivative of sin(y^2) =(y^2)'cos(y^2)=2yy'cos(y^2)

Answer 14

chain rule
dy/dx=dy/du*du/dx right?

Answer 15

dang i could have sword i just did these!

Answer 16

derivative of 2x=2
derivative of sin(y/x)=(y/x)'cos(y/x)=([y'x-y]/x^2)cos(y/x)

Answer 17

satellite give her your formula

Answer 18

hello myininaya

Answer 19

hey

Answer 20

http://openstudy.com/users/shessa#/users/shessa/updates/4e6026420b8b1f45b4a2374b

Answer 21

i don't think she likes way

Answer 22

all written out, but maybe a step or two is not clear i don't know.

Answer 23

my way*

Answer 24

\[\frac{d}{dx}\sqrt{f(x)}=\frac{f'(x)}{2\sqrt{f(x)}}\] via the chain rule

Answer 25

actually i wrote it out, so maybe it is clear, maybe not. you need two facts
\[\frac{d}{dx}\arcsin(x)=\frac{1}{\sqrt{1-x^2}}\] and the chain rule to finish

Answer 26

of course
\[\frac{d}{dx}\arcsin(f(x))=\frac{f'(x)}{\sqrt{1-f^2(x)}}\] also by the chain rule. gotta love that chain rule

Answer 27

thank you myininaya and satellite73