$\displaystyle [(x^3-x)\sin x]'$

$\displaystyle [(x^3-x)\sin x]'= [x^3-x]'\sin x + (x^3-x)[\sin x]'= \bbox[#FFECB3,5px]{(3x^2-1)\sin x + (x^3-x)\cos x}$

$\displaystyle [\sin x\ln x]'$

$\displaystyle [\sin x\ln x]'= [\sin x]'\ln x + \sin x[\ln x]'= \bbox[#FFECB3,5px]{\cos x\ln x + \dfrac{\sin x}{x}}$

$\displaystyle [x^2 2^x]'$

$\displaystyle [x^2 2^x]'= [x^2]'2^x + x^2[2^x]'= 2x\cdot 2^x + x^2 2^x\ln(2)= \bbox[#FFECB3,5px]{(2x+\ln(2)x^2)2^x}$

$\displaystyle \left[\dfrac{3x^3-2x}{x^2+3x-1}\right]'$

$\displaystyle \left[\dfrac{3x^3-2x}{x^2+3x-1}\right]'= \dfrac{[3x^3-2x]'(x^2+3x-1)-(3x^3-2x)[x^2+3x-1]'}{(x^2+3x-1)^2}= \dfrac{(9x^2-2)(x^2+3x-1)-(3x^3-2x)(2x+3)}{(x^2+3x-1)^2}=\\ \qquad\displaystyle \dfrac{(9x^4+27x^3-9x^2-2x^2-6x+2)-(6x^4+9x^3-4x^2-6x)}{(x^2+3x-1)^2}= \bbox[#FFECB3,5px]{\dfrac{3x^4+18x^3-7x^2+2}{(x^2+3x-1)^2}}$

$\displaystyle \left[\dfrac{5x^2-3x}{x\sin x}\right]'$

$\displaystyle \left[\dfrac{5x^2-3x}{x\sin x}\right]'= \left[\dfrac{5x-3}{\sin x}\right]'= \dfrac{[5x-3]'(\sin x)-(5x-3)[\sin x]'}{(\sin x)^2}= \dfrac{(5)(\sin x)-(5x-3)(\cos x)}{\sin^2(x)}= \bbox[#FFECB3,5px]{\dfrac{5}{\sin(x)}+\dfrac{(3-5x)\cos x}{\sin^2(x)}}$