Question #263809

y=ctgx+sin2x3

1
Expert's answer
2021-11-18T15:04:23-0500

y=ctgx+sin2x3y=\ctg x+\sin^2x^3


(ctgx)=1sin2 x(\ctg x)'= -\dfrac1{\sin²\ x}


(sin2x)=sin(2x)a=(sin2x3)=(sinx3×sinx3)u=sinx3;du=3x2cosx3v=sinx3;dv=3x2cosx3da/dx=vdu+udv(sinx3×sinx3)=3x2sinx3cosx3+3x2sinx3cosx3=6x2sinx3cosx3(\sin^2x)' = sin(2x)\\ \therefore a = (\sin^2x^3)' =(\sin x³ × \sin x³)'\\ u = \sin x³ ;\quad du = 3x²\cos x³\\ v = \sin x³ ;\quad dv= 3x²\cos x³\\ da/dx = vdu + udv\\ \therefore(\sin x³ × \sin x³)' = \\ 3x²\sin x³\cos x³ + 3x²\sin x³\cos x³ = 6x²\sin x³\cos x³


dy/dx=1sin2x+6x2sinx3cosx3dy/dx = -\dfrac1{\sin^2 x}+ 6x² \sin x³\cos x³

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