Question 801851
<pre>
{{{expr(d/dx)}}}(x³+2x)(x²-x)|
                |x=2

Use {{{d(u*v)/dx=u*expr(dv/dx)+v*expr(du/dx)}}}

  u = x³+2x      and      v = x²-x
{{{du/dx}}} = 3x²+2              {{{dv/dx}}} = 2x-1

(x³+2x)(2x-1) + (x²-x)(3x²+2)|
                          |x=2

(2³+2·2)(2·2-1) + (2²-2)(3·2²+2)

(8+4)(4-1) + (4-2)(3·4+2)

(12)(3) + (2)(12+2)

36 + (2)(14)

36 + 28

64
Edwin</pre>