How do you differentiate #y=(x+5)(2x-3)(3x^2+4)#?

2 Answers
Apr 24, 2018

Answer:

#y'=(2x-3)(3x^2+4)+2(x+5)(3x^2+4)+6x(2x-3)(x+5)#

#y'=24x^3+63x^2-74x+28#

Explanation:

If #y=uvw#, where #u#, #v#, and #w# are all functions of #x#, then:
#y'=uvw'+uv'w+u'vw# (This can be found by doing a chain rule with two functions substitued as one, i.e. making #uv=z#)

#u=x+5#
#u'=1#

#v=2x-3#
#v'=2#

#w=3x^2+4#
#w'=6x#

#y'=(2x-3)(3x^2+4)+2(x+5)(3x^2+4)+6x(2x-3)(x+5)#

#y'=6x^3+8x-9x^2-12+6x^3+8x+30x^2+40+12x^3+60x^2-18x^2-90x#

#y'=24x^3+63x^2-74x+28#

Apr 24, 2018

Answer:

#dy/dx=24x^3+63x^2-74x+28#

Explanation:

#"expand the factors and differentiate using the "color(blue)"power rule"#

#•color(white)(x)d/dx(ax^n)=nax^(n-1)#

#y=(x+5)(2x-3)(3x^2+4)#

#color(white)(y)=6x^4+21x^3-37x^2+28x-60#

#rArrdy/dx=24x^3+63x^2-74x+28#