#y# is a quotient in the form of #color(blue)(y=(u(x))/(v(x)))#
The deferentiation of the quotient is as follows :
#color(blue)(y'=((u(x))'v(x)-(v(x))'u(x))/(v(x))^2)#
Let us find #(u(x))'# and #(v(x))'#
#color(green)((u(x))'=?)#
#u(x)# is a composite of two functions #f(x)# and #g(x)# where:
#f(x)=x^5# and #g(x)=x^3+4#
We have to use chain rule to find #color(green)((u(x))')#
#u(x)=f(g(x))# then
#color(green)((u(x))'=f'(g(x))*g'(x))#
#f'(x)=5x^4# then
#f'(g(x))=5(g(x))^4#
#color(green)(f'(g(x))=5(x^3+4)^4)#
#color(green)((g(x))'=3x^2)#
So,#(u(x))'=5(x^3+4)^4*3x^2#
#color(green)((u(x))'=15x^2(x^3+4)^4)#
#color(red)((v(x))'=?)#
#v(x)=3x^4-2#
#color(red)((v(x))'=12x^3)#
Now,let us substitute #color(green)((u(x))'# and #color(red)((v(x))'# in #color(blue)y'#
#color(blue)(y'=((u(x))'v(x)-(v(x))'u(x))/(v(x))^2)#
#y'=(color(green)(15x^2(x^3+4)^4)*(3x^4-2)-color(red)(12x^3)(x^3+4)^5)/(3x^4-2)^2#
#y'=((x^3+4)^4[15x^2(3x^4-2)-12x^3(x^3+4)])/(3x^4-2)^2#
#y'=((x^3+4)^4[45x^6-30x^2-12x^6-48x^3])/(3x^4-2)^2#
#y'=((x^3+4)^4(45x^6-12x^6-48x^3-30x^2))/(3x^4-2)^2#
Therefore,
#color(blue)(y'=((x^3+4)^4(33x^6-48x^3-30x^2))/(3x^4-2)^2)#
