Question

what is d/dx[tan(arcsin(t))]

Answer 1

if you y=tan(u)

u= arcsin(t)

Answer 2

now

\[\frac{dy}{dx}=\frac{dy}{du}*\frac{du}{dx}\]

Answer 3

wait... are you trying to find dy/dx or dy/dt?

Answer 4

dy/dx =0

Answer 5

since there is no x variable to differentiate

Answer 6

everything else is constant

Answer 7

sorry d/dt

Answer 8

alright so my first post and this would be your chain rule

\[\frac{dy}{dt}=\frac{dy}{du}*\frac{du}{dt}\]

Answer 9

where
\[y=tan(u)\]

and
\[u=arcsin(t)\]

Answer 10

so i get:\[\sec(\arcsin(t))^1 * \frac{ 1 }{ \sqrt{1-x^2} }\]

Answer 11

?

Answer 12

the x^2 should be t^2 but other than that looks good

Answer 13

ty

Answer 14

wait... the derivative of tan(u) is sec^2(u

Answer 15

doesnt the power go n-1?

Answer 16

\[sex^2(u) == \sec(u)^2\]

Answer 17

\[sec^2(u)*\frac{1}{\sqrt{1-t^2}}\]

Answer 18

??

Answer 19

sec^2(u) **

Answer 20

yes but you're finding \[\frac{d}{dx}(tanx)\]

Answer 21

\[\frac{d}{dx}(tanx)=sec^2(x)\]

Answer 22

true.