# \ #

# "Yes, you are exactly right: it can be correctly thought of as" #

# "a product rule and chain rule problem." #

# "Let's compute" \ y' ":" #

# \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad y \ = \ x (1 - x )^4. #

# \qquad \qquad \qquad \qquad \quad :. \qquad \qquad \qquad \quad y \ = \ [ x ] \cdot [ (1 - x )^4 ]. #

# "Product Rule:" \qquad \qquad \quad \ y' \ = \ x [ (1 - x )^4 ]' + [ x ]' [ (1 - x )^4 ]. #

# "Chain Rule for first differentiated quantity:" #

# \qquad \qquad \quad \quad \ y' \ = \ x [ 4 (1 - x )^3 (1 - x )' ] + 1 \cdot [ (1 - x )^4 ]. #

# "Continuing (notice the -1 that appears -- important):" #

# \qquad \qquad \quad \quad \ y' \ = \ x [ 4 (1 - x )^3 ( -1 ) ] + (1 - x )^4. #

# "Simplifying -- factor out lowest powers of same quantities:" #

# \qquad \qquad \quad \quad \ y' \ = \ (1 - x )^3 [x ( 4 ) ( -1 ) + (1 - x )^1 ] #

# \qquad \qquad \qquad \qquad \ \ = \ (1 - x )^3 [x ( -4 ) + (1 - x ) ] #

# \qquad \qquad \qquad \qquad \ \ = \ (1 - x )^3 [ -4 x + 1 - x ] #

# \qquad \qquad \qquad \qquad \ \ = \ (1 - x )^3 ( 1 - 5 x ). #

# \qquad \qquad \qquad \qquad \ \ = \ ( 1 - 5 x ) (1 - x )^3. #

# "Thus:" #

# \qquad \qquad \qquad \qquad \qquad \qquad \qquad \ \ y' \ = \ ( 1 - 5 x ) (1 - x )^3. #

# \ #

# "Now we can calculate the critical points." #

# "Recall that the critical points are the points where the derivative" #

# "vanishes or is undefined." #

# "So we solve:" #

# \qquad \qquad \qquad \qquad \qquad \qquad \qquad y' = 0 \quad \quad "and" \quad \quad y' = "undefined" #

# "So we solve:" #

# \quad ( 1 - 5 x ) (1 - x )^3 = 0 \qquad "and" \qquad ( 1 - 5 x ) (1 - x )^3 = "undefined" #

# "In the second part of the previous, we note that" #

# ( 1 - 5 x ) (1 - x )^3 \ \ "is defined everywhere, so the second part" #

# "yields no solutions. So we continue solving only the first part:" #

# \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad ( 1 - 5 x ) (1 - x )^3 = 0 #

# \qquad \qquad \qquad \qquad \qquad \qquad 1 - 5 x = 0 \qquad \qquad (1 - x )^3 = 0 #

# \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad \ 5 x = 1 \qquad \qquad 1 - x = 0 #

# \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad x = 1/5 \qquad \qquad x = 1 #

# "These are our critical points." #

# \ #

# "Summarizing:" #

# "The critical points of" \ \ y = x (1 - x )^4 \ \ "are:" \qquad x = 1/5, \ 1. #