<PRE><B>Question 1203615</B>
&lt;br&gt;
A basic application of the product rule for derivatives tells us&lt;br&gt;
(d/dx)((f(x)e^x)) = f(x)e^x+f&#39;(x)e^x&lt;br&gt;
So in this problem probably the easiest way to find h&#39;(x) is to write f(x)=(3x-12)(x^4-x^(1/2)) and find the derivative using that pattern.&lt;br&gt;
f(x) = (3x-12)(5x^4-x^(1/2)) = 15x^5-60x^4-3x^(3/2)+12x^(1/2)&lt;br&gt;
f&#39;(x) = 75x^4-240x^3-(9/2)x^(1/2)+6x^(-1/2)&lt;br&gt;
f(x)+f&#39;(x) = 15x^5+15x^4-240x^3-3x^(3/2)+(15/2)x^(1/2)+6x^(-1/2)&lt;br&gt;
And so&lt;br&gt;
ANSWER: h&#39;(x) = (15x^5+15x^4-240x^3-3x^(3/2)+(15/2)x^(1/2)+6x^(-1/2))&lt;br&gt;
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