How do you differentiate the equation?

Question

#(x^3 + 2x)^2 = f(x)#

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Answer 1 · 2018-05-04T14:18:33

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

#f'(x)#
#=2(x^3+2x)^(2-1)(3x^2+2)#(Chain Rule)
#=2(x^3+2x)^1(3x^2+2)#

Chain Rule:
In general,
#d/dx[f(x)]^n=n[f(x)]^(n-1)xxf'(x)#

Answer 2 · 2018-05-04T14:37:16

#dy/dx=f'(x)=2(x^3+2x)(3x^2+2)#

Given: #f(x)=(x^3+2x)^2#

Set: #y=(x^3+2x)^2#

Let #u=x^3+2x => (du)/dx=3x^2+2#

Set #y=u^2 => dy/(du)=2u#

We need #dy/dx# and this is obtained by:

#dy/dx =color(white)("dd") dy/(du)color(white)("ddd")xxcolor(white)("dd")(du)/dx#

#dy/dx =color(white)("dd") (2u)color(white)("ddd")xx(3x^2+2)#

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

#dy/dx=f'(x)=2(x^3+2x)(3x^2+2)#