Question

i want to verify my answer please.. .. find y' if y=sin^-1(e^-2x)

Answer 1

my answer is \[y'=\left(\begin{matrix}1 \\ \sqrt{1+x ^{2}}\end{matrix}\right)\left( e ^{-2x} \right)+\sin^{-1} \left( -2xe ^{-2x} \right)\]

Answer 2

do you think \[y=\arcsin{e^{-2x}}\]

Answer 3

i think is the same

Answer 4

ohhhh

Answer 5

upsss

Answer 6

its is wrong

Answer 7

\[y'=\frac{1}{\sqrt{1-\left(e^{-2x}\right)^2}}\cdot\left(e^{-2x}\right)'\]

Answer 8

ohhhh :)

Answer 9

the chain rule for that will be -2xe^-2x