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sinx
cosx
cosx
-sinx
-cos2x
2sin2x
sin^5(x)
5cos^4(x)
e^x
e^x
e^kx
ke^kx
ln(x)
1/x
ln(kx)
1/x
y = kx
kxlnk
proof: log both sides - you know how to differentiate logs