Question 1203615
<pre>
h(x) = {{{ (3x-12)(5x^4-sqrt(x))(e^x) }}}

The product rule can be applied:

If  h(x) = A(x)B(x) then 
 
    h'(x) = A(x)B'(x) + B(x)A'(x)

Choosing A(x) = (3x-12) and B(x) = {{{ (5x^4-sqrt(x))(e^x) }}}

 A'(x) = 3

Unfortunately, B'(x) will be messy:
 B'(x) = {{{ (5x^4 - sqrt(x))*(e^x) + (e^x)*(20x^3-(1/(2sqrt(x))))  }}} 


Now you have:
   h'(x) = {{{ (3x-12)*(5x^4 - sqrt(x))*(e^x) + (e^x)*(20x^3-(1/(2sqrt(x))))  }}}  + {{{ (5x^4-sqrt(x))(e^x)*3 }}}


Expand the first term, combine like-terms, and pull out {{{e^x}}}, and you should get:

   h'(x) = {{{e^x * (15x^5+15x^4-240x^3-3x*sqrt(x)+(15/2)sqrt(x)+6/sqrt(x)) }}}

[ Sometimes writing {{{sqrt(x) }}} as x^(1/2) makes the workout easier. ]

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EDIT 9/14/23:  tutor greenestamps answer is missing a factor of {{{e^x}}}