\(y=(1-2x)\cdot(3-x+2x^2)\)

setze \(h(x)= 1-2x\) und \(g(x)= 3-x+2x^2\)

\(h'(x)= -2 \\ g'(x)=4x-1\)

\(y'=h'(x)g(x)+h(x)g'(x)= -2 (3-x+2x^2)+(1-2x)(4x-1)=-12x^2+8x-7\)

\(y=(3-2x)\cdot(3+x)\cdot(4-\frac{x}{2})\)

setze

\(h(x)= (3-2x)(3+x)=9-3x-x^2 ,\\ g(x)= 4-\frac x2\)

\(h'(x)=-3-4x \\ g'(x)=-\frac 12\)

\(y'= h'(x)g(x) + h(x)g'(x) = (-3-4x)(4-\frac x2) + (9-3x-2x^2)(-\frac 12)= 3x^2-13x-\frac {33}{2}\)

\(y=3x\sin x\)

setze \(h(x)= 3x , \ g(x)= \sin x\)

\(h'(x)=3\\ g'(x)= \cos x\)

\(y'(x)= h'(x)g(x)+h(x)g'(x)= 3 \sin x + 3x \cos x = 3 (\sin x + x \cos x)\)

\(y=(1-2x+x^2)\cdot\cos x\)

setze \(f(x)= 1-2x+x^2 \) und \(g(x)= \cos x\)

\(f'(x)= 0-2+2x\), \(g'(x)=- \sin x\)

\(y'= (2x-2) \cos x + (1-2x+x^2) (-\sin x)= 2x \cos x - 2\cos x -\sin x +2x \sin x -x^2 \sin x = -(x-1)\cdot\left( (x-1)\sin x-2\cos x\right)\)

\(y=2x\cdot\sin x\cdot\cos x\)

setze \(f(x)= 2x , \ g(x)= \sin x , \ h(x) = \cos x\)

\(f'(x)=2, \ g'(x)= \cos x , \ h'(x)= - \sin x\)

\(y'= (2 \sin x + 2x \cos x ) \cos x + 2x \sin x (- \sin x)= 2 \sin x \cos x + 2x (\cos^2 x -\sin ^2 x)\)

Umformen: \(2 \sin x \cos x = \sin (2x)\) , \(\cos^2 x - \sin ^2 x = \cos (2x)\)

\(y'= \sin(2x) + 2x \cos (2x)\)

\(xy - x + 2y = - 3 \\ y = \frac{-3+x}{x+2} \\ y = (-3+x) \cdot \frac{1}{x+2}\)

\(y'(x) = 1 \cdot \frac{1}{x+2} + (-3+x) \cdot (-1) \cdot \frac{1}{(x+2)^2} \\ y'(x) = \frac{1}{x+2} + \frac{3-x}{(x+2)^2} \\ y'(x) = \frac{x+2+3-x}{(x+2)^2} \\ y'(x) = \frac{5}{(x+2)^2} \)

\(y(x) = \frac{-3+x}{5x+2} \\ y(x) = (-3+x) \cdot \frac{1}{5x+2} \\ y'(x) = 1 \cdot \frac{1}{5x+2} + (-3+x) \cdot (-1) \cdot \frac{1}{(5x+2)^2} \cdot 5 \\\)

\( y'(x) = \frac{1}{5x+2} + \frac{(3-x)\cdot 5}{(5x+2)^2} \\ y'(x) = \frac{5x + 2 + 15 - 5x}{(5x+2)^2} \\ y'(x) = \frac{17}{(5x+2)^2} \)

\(y(x) = x\cdot e^x \\ y'(x) = e^x + x e^x \\ y'(x) = e^x ( 1+x) \)

\(y(x) = x^2 + e^x \\ y'(x) = 2x e^x + e^x x^2\)

\(y'(x) = e^x (2x + x^2) \)

\(y(x) = \frac{e^x}{x} \\ y(x) = e^x \cdot x^{-1} \\ y'(x) = e^x x^{-1} +e^x \cdot (-1) \cdot x^{-2} \\ \)

\( y'(x) = e^x ( x^{-1} - x^{-2}) \\ y'(x) = e^x ( \frac{1}{x} - \frac{1}{x^2}) \\ y'(x) = e^x \frac{x-1}{x^2}\)

\(y(x) = \frac{x^2}{e^x} \\ y(x) = x^2 \cdot e^{-x} \\ y'(x) = 2x \cdot e^{-x} - x^2 \cdot e^{-x} \)

\(y'(x) =\frac{ 2x -x^2}{e^x}\)

\(y(x) = x \ln(x) \\ y'(x) = 1 \cdot \ln(x) + x \cdot \frac{1}{x} \\ y'(x) = \ln(x) + 1\)

\(y(x) = \frac{\ln(x) }{x} = \ln(x) \cdot \frac{1}{x} = \ln(x) \cdot x^{-1} \\\)

\(y'(x) = \frac{1}{x} \cdot x^{-1} + \ln(x) \cdot (-1) \cdot x^{-2} \\ y'(x) = \frac{1}{x} \cdot \frac{1}{x} - \ln(x) \cdot \frac{1}{x^2} \\ y'(x) = \frac{1-\ln(x) }{x^2} \)

\(f(x) = 4 \cdot \frac1{x^{\frac23}} + x = f(x) = 4 \cdot x^{-\frac23} + x \)

\( f(x)' = -4 \frac23 \cdot x^{-\frac53} + 1= f(x)' = - \frac83 \cdot \frac1{x^{\frac53}} + 1= f(x)' = -\frac83 \cdot \frac1{\sqrt[3]{x^5} }+ 1\)

\(y(x) = e^x x^2\)

\(y'(x) = e^x \cdot (x^2 + 2x)\)

\(y''(x) = e^x \cdot ( x^2 + 4x + 2 ) \)

\(y(x) = \frac1{3x^6} = \frac13 x^{-6}\)

\(y'(x) = \frac13 \cdot (-6) \cdot x^{-7} = - \frac2{x^7}\)

\(y(x) = (\sin(x))^2 = \sin(x) \cdot \sin(x)\)

\(y'(x) = \sin(x) \cos(x) + \sin(x) \cos(x) = 2 \sin(x) \cos(x)\)

\(y(x) = (\cos(x))^2 = \cos(x) \cos(x)\)

\(y'(x) = - \sin(x) \cos(x) - \sin(x) \cos(x) = - 2\sin(x) \cos(x)\)

\(f(x) = \frac{e^x}{x^2} = e^x \cdot x^{-2}\)

\(f'(x) = e^x x^{-2} - 2 e^x x^{-3} = e^x ( \frac1{x^2} - \frac2{x^3})\)

\(y(x) = x \cdot \ln(x) \)

\(y'(x) = \ln(x) + x \frac1x = \ln(x) + 1\)

\(y(x) = x \cdot \ln(x) \)

\(y'(x) = \ln(x) + x \frac1x = \ln(x) + 1\)

\(y''(x) = \frac1x\)

\(y(x) = (\sin(x) ) ^2 + (\cos(x))^2 = \sin(x) \sin(x) + \cos(x) \cos(x) \)

\(y'(x) = \sin(x) \cos(x) + \sin(x) \cos(x) - \sin(x) \cos(x) - \sin(x) \cos(x) = 0\)

\(y(x) = \tan(x) = \frac{\sin(x)}{\cos(x)} = \sin(x) \cdot (\cos(x))^{-1}\)

\(y'(x) = \cos(x) \cdot (\cos(x))^{-1} + \sin(x) \cdot (\cos(x))^{-2} \sin(x) = 1 + (\sin(x))^2 \cdot (\cos(x))^{-2} = \)

\(= \frac{(\cos(x))^2 + (\sin(x))^2 }{(\cos(x))^2} = \frac{1 }{(\cos(x))^2} \)