Question

CALCULUS
derivative of :
f(x)=2Arcsin(x-1)
please help me through the steps

Answer 1

Okay so:
Recall that-
\[\frac{ d }{ dx }\arcsin(x)=\frac{ 1 }{ \sqrt{1-x^2} }\frac{ d }{ dx }\]

Answer 2

yes

Answer 3

This is a chain rule --> So you are going to take the derivative of the outside function first (arcsin) leaving the inside alone and multiplying it by the derivative of the inside...

Answer 4

okay so
\[\frac{ 1 }{ \sqrt{1-x^2} } *2 \] ?

Answer 5

the book gives me that answer and thats what i got but under the square root it gives me 2x-x^2 and idk how they got that 2x

Answer 6

@henryarias14 yep... but what is x?
It's (x-1) right?

Answer 7

yes

Answer 8

wait. so sqr(1-(x-1)) ?

Answer 9

\[f'(x)=\frac{ 2 }{ \sqrt{1-(x-1)^2} }\]

Answer 10

ohhhh

Answer 11

yep:) and we can simplify it further

Answer 12

i get sqr(x^2+2x) but its still not what the book tells me. it says it should be 2x-x^2 unless i did something wrong in my notes. what did you get?

Answer 13

\[f'(x)=\frac{ 2 }{ \sqrt{1-(x^2-2x+1)} }\]

*Expand and cancel -1
\[f'(x)=\frac{ 2 }{ \sqrt{2x-x^2} }\]

Answer 14

oh i get it now. thanks!!!

Answer 15

got a final on thursday

Answer 16

*Multiply by radical
\[f'(x)=\frac{ 2 }{ \sqrt{2x-x^2} }\]
\[f'(x)=\frac{ 2\sqrt{2x-x^2} }{ 2x-x^2 }\]

& @henryarias14 :D good i'm happy!

Answer 17

oh yeah forgot about the radical. thanks! although i think my professor doesn't mind if we leave the radical on the denominator for this type of problems lol

Answer 18

Yeah they usually won't care lol :P