ANSWERS: 3
  • Its really ugly dy/dx = x^cos2x * (ln(cos2x)-[x(4sinxcosx)/(cos2x)]) I made a mistake. This is the better answer
  • y=x^cos2x take the natural log of both sides: lny=ln(x^cos2x) use the power rule for logarithms: lny=cos2x(lnx) use the chain and multiplication rules: (1/y)y'=-sin2x(2)(lnx)+cos2x(1/x) rewrite: y'/y=-2sin2x(lnx)+(cos2x)/x multiply both sides by y: y'=y[-2sin2x(lnx)+(cos2x)/x] replace y with x^cos2x: y'=(x^cos2x)[-2sin2x(lnx)+(cos2x)/x]
  • d/dx y = d/dx x^cos(2x) y' = d/dx x^cos(2x) y' = d/dx e^ln[ x^cos(2x) ] y' = d/dx e^[ cos(2x) ln(x) ] y' = e^[ cos(2x) ln(x) ] d/dx [ cos(2x) ln(x) ] y' = x^cos(2x) [ d/dx cos(2x) ln(x) ] y' = x^cos(2x) [ (d/dx cos(2x)) ln(x) + cos(2x) (d/dx ln(x)) ] y' = x^cos(2x) [ -sin(2x) (d/dx 2x) ln(x) + cos(2x)/x ] y' = x^cos(2x) [-2 sin(2x) ln(x) + cos(2x)/x ]

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