How do you differentiate the equation?

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

2 Answers
May 4, 2018

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

Explanation:

#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)#

May 4, 2018

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

Explanation:

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)#