Question

x^(sinx) = y find y'

Answer 1

idk.

Answer 2

something do natural log of both sides

Answer 3

y = x^(sinx)
lny = SinxLnx

Answer 4

(1/y)(y') = Sinx(1/x) + Lnx(Cosx)

Answer 5

y' = y [ ((sinx)/x)+Cosx Lnx ]

Answer 6

y' = x^(sinx)[ ((sinx)/x)+Cosx Lnx ]

Answer 7

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

Answer 8

haha i got there first.. why do u guys have to use this math input.. people should understand calculator notation..

Answer 9

thank you both