y=ctgx+sin2x3
y=ctgx+sin2x3y=\ctg x+\sin^2x^3y=ctgx+sin2x3
(ctgx)′=−1sin2 x(\ctg x)'= -\dfrac1{\sin²\ x}(ctgx)′=−sin2 x1
(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(sin2x)′=sin(2x)∴a=(sin2x3)′=(sinx3×sinx3)′u=sinx3;du=3x2cosx3v=sinx3;dv=3x2cosx3da/dx=vdu+udv∴(sinx3×sinx3)′=3x2sinx3cosx3+3x2sinx3cosx3=6x2sinx3cosx3
dy/dx=−1sin2x+6x2sinx3cosx3dy/dx = -\dfrac1{\sin^2 x}+ 6x² \sin x³\cos x³dy/dx=−sin2x1+6x2sinx3cosx3
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