How to differentiate #y=(x+2)^2(3x-1)^3# , find #dy/dx# as a product of its factor ?

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Answer 1 · 2018-07-17T14:18:22

#dy/dx=(x+2)(15x+16)(3x-1)^2#

Given that

#y=(x+2)^2(3x-1)^3#

differentiating w.r.t. #x# as follows

#dy/dx=d/dx((x+2)^2(3x-1)^3)#

#=(x+2)^2d/dx(3x-1)^3+(3x-1)^3d/dx(x+2)^2#

#=(x+2)^2(3(3x-1)^2(3))+(3x-1)^3(2(x+2))#

#=9(x+2)^2(3x-1)^2+2(x+2)(3x-1)^3#

#=(x+2)(3x-1)^2(9x+18+6x-2)#

#=(x+2)(3x-1)^2(15x+16)#

#=(x+2)(15x+16)(3x-1)^2#

Answer 2 · 2018-07-17T14:23:21

#dy/dx=2(x+2)(3x-1)^3+9(x+2)^2(3x-1)^2#

#(uv)'=u'v+uv'#

#(f(g(x)))'=f'(g(x))*g'(x)# we get

#dy/dx=2(x+2)(3x-1)^3+(x+2)^2*3(3x-1)^2*3#
this is equal to

#dy/dx=2(x+2)(3x-1)^3+9(x+2)^2(3x-1)^2#