Question 594843
<pre>
{{{matrix(2,1,"","f(x)")}}}{{{matrix(2,1,"",""="")}}}{{{matrix(2,1,"",(x + 3)^(2/3))}}}  Use the rule {{{expr(d/(dx))}}}{{{u^n}}} = {{{n*u^(n-1)expr((du)/(dx))}}} where {{{u=x+3}}} and {{{n=2/3}}}

{{{matrix(2,1,"","f'(x)")}}}{{{matrix(2,1,"",""="")}}}{{{matrix(2,1,"",2/3))}}}{{{matrix(2,1,"",(x + 3)^(2/3-1)(1))}}}

{{{matrix(2,1,"","f'(x)")}}}{{{matrix(2,1,"",""="")}}}{{{matrix(2,1,"",2/3))}}}{{{matrix(2,1,"",(x + 3)^(2/3-3/3)(1))}}}

{{{matrix(2,1,"","f'(x)")}}}{{{matrix(2,1,"",""="")}}}{{{matrix(2,1,"",2/3))}}}{{{matrix(2,1,"",(x + 3)^(-1/3))}}}

Use the rule again for {{{"f''(x)"}}} with {{{n=-1/3}}}

{{{matrix(2,1,"","f''(x)")}}}{{{matrix(2,1,"",""="")}}}{{{matrix(2,1,"",expr(-1/3)expr(2/3)))}}}{{{matrix(2,1,"",(x + 3)^(-1/3-1)(1))}}}

{{{matrix(2,1,"","f''(x)")}}}{{{matrix(2,1,"",""="")}}}{{{matrix(2,1,"",expr(-1/3)expr(2/3)))}}}{{{matrix(2,1,"",(x + 3)^(-1/3-3/3)(1))}}}

{{{matrix(2,1,"","f''(x)")}}}{{{matrix(2,1,"",""="")}}}{{{matrix(2,1,"",-2/9))}}}{{{matrix(2,1,"",(x + 3)^(-4/3))}}}

Edwin</pre>