Question

find an equation of the line tangent to \[y=x ^{\sin x}\] at the point x=1

Answer 1

lny=(sinx)(lnx)

Answer 2

i'm not even sure where you're starting do i have to use chain rule?

Answer 3

\[\frac{ 1 }{y }y'=\frac{ sinx }{ x }+lnx(cosx)\]

Answer 4

implies

Answer 5

\[y'=y(\frac{ sinx }{ x} +lnx(cosx))\]

Answer 6

\[y'=x ^{sinx}(\frac{ sinx }{ x }+lnx(cosx))\]

Answer 7

\[y' at 1 =\sin(1)\]