<s>[INST] Simplest form of $h'(y)$ given $h(y)= (1-3y^2)^5 \cdot ( y^2 + 2)^6$ Find $h'(y)$ in the simplest form if the $$h(y)= (1-3y^2)^5 \cdot ( y^2 + 2)^6$$ 
My answer was: $$-30y(1-3y^2)^4 \cdot (y^2+2)^6 + 12y(y^2+2)^5 \cdot (1-3y^2)^5$$ 
But according to wolfram alpha the answer was $$ -30y(1-3y^2)^4 \cdot (y^2+2)^6 $$ only. which makes it weird that they did not differentiate the second term my answer was different. 
I posted it again because I did not get an answer. 
 [/INST] You differentiated correctly.
Since you have been asked to write the derivative in its simplest form:
you can factor out common factors in each term of your sum: 
$$-30y(1-3y^2)^4 \cdot (y^2+2)^6 + 12y(y^2+2)^5(1-3y^2)^5$$ $$ = 6y(1-3y^2)^4(y^2 +2)^5\Big((-5)(y^2+2) + 2(1-3y^2)\Big)$$ $$ = 6y(1-3y^2)^4(y^2 +2)^5\Big(-5y^2-10 +2 -6y^2\Big) $$ $$=6y(1-3y^2)^4(y^2 +2)^5(-11y^2 - 8)$$
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