How do i solve -4(2x^3+5x^2)^3 using chain rule?

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Answer 1 · 2018-03-23T11:18:43

#(dy)/(dx)=24x^5(2x+5)^2(3x+5)#

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

We take,

#y=4u^3,........#where #u=2x^3+5x^2#

#=>(dy)/(dx)=4*3u^(3-1)=12u^2, and (du)/(dx)=6x^2+10x#

Applying Chain Rule:

#color(red)((dy)/(dx)=(dy)/(du)*(du)/(dx)#

#(dy)/(dx)=(12u^2)(6x^2+10x)#

#(dy)/(dx)=12(2x^3+5x^2)^2(6x^2+10x)#

#(dy)/(dx)=12x^4(2x+5)^2x(3x+5)#

#(dy)/(dx)=24x^5(2x+5)^2(3x+5)#