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Answer 1 · 2017-11-29T22:11:38

#dy/dx = (3)/(2-x)^2#

#dy/dx = d/dx (u/v) = (v (du)/dx - u (dv)/dx )/v^2#

Use #u = 1+x# and #v=2-x#:

#dy/dx = ( (2-x) d/dx (1+x) - [(1+x) d/dx (2-x)] ) / (2-x)^2#

Use the Power Rule to differentiate:

#dy/dx = ((2-x)(0+1) - [(1+x)(0-1)] ) / (2-x)^2#

# = ( (2-x) - (-1-x) ) / (2-x)^2#

# = ( 2-x + 1 + x ) / (2-x)^2 #

# = 3/(2-x)^2#

Thus, #dy/dx = (3)/(2-x)^2#.