Question

find the derivative of the following function

Answer 1

\[y=arcsinx^{2}+\arcsin(\sqrt{1-x^{4}})\]

Answer 2

Do you remember the derivative of `arcsin x`? :) It will be just a little bit of work beyond that.

Answer 3

\[\frac{ d }{ dx } arcsinu=\frac{ 1 }{ \sqrt{1-u^{2}} }\frac{ du }{ dx }\]

Answer 4

cool c:

Answer 5

Do I just sub it all in?

Answer 6

Yes, just make sure you sub in correctly in the denominator. The entire content of the arcsine gets stuffed into that u.

Answer 7

soooo say \[\frac{ 1 }{ \sqrt{1-x^{4}} } \frac{ d(x^{4}) }{ dx } + \frac{ 1 }{ \sqrt{1-\sqrt{1-x^{4}}} } \frac{ d(\sqrt{1-x^{4}} }{ dx }\]

Answer 8

d(x^2)/dx I mean

Answer 9

yes, and also, it looks like you forgot the square on the second term, in the denominator.

Answer 10

Ohhhh so when I square it the square root goes away!!! okay that was what was confusing me haha thank you!!

Answer 11

hah cool c:

Answer 12

just to double check would the answer be like \[\frac{ 2x }{ \sqrt{1-x^{4}} }+\frac{ 2x^{3} }{ \sqrt{x^{4}}\sqrt{1-x^{4}} }\]

Answer 13

in my notes some of the answers end up with arcsin back in it :S

Answer 14

Hmm you must've done a bunch of simplification :D I'm not sure. Lemme work through a few steps real quick.

\[\large y=\arcsin x^2+\arcsin \sqrt{1-x^4}\]
\[\large y'=\frac{2x}{\sqrt{1-x^4}}+\frac{1}{\sqrt{1+(1-x^4)}}\left(\sqrt{1-x^4}\right)'\]
\[\large y'=\frac{2x}{\sqrt{1-x^4}}+\frac{1}{\sqrt{2-x^4}}\left(\frac{-4x^3}{2\sqrt{1-x^4}}\right)\]

Answer 15

Oh the squareroot should have a negative? lol my bad :) lemme fix that.

Answer 16

\[\large y'=\frac{2x}{\sqrt{1-x^4}}+\frac{1}{\sqrt{x^4}}\left(\frac{-4x^3}{2\sqrt{1-x^4}}\right)\]

Ahhh ok I see what you did ;D hehe. yes very good!

Answer 17

Err maybe you missed the negative when you took the derivative of the inside of the square root, that's the only little boo boo I see.

Answer 18

yea I think I did haha yay thanks!