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)