How do you find the derivative of #f(x)=(x-2)^3#?

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Answer 1 · 2016-12-20T13:38:42

It depends on whether you have learned the chain rule yet or not.

With the chain rule:

For #u# a function of #x# and #n# an integer,

#d/dx(u^n) = n u^(n-1) (du)/dx#.

So, #f'(x) = 3(x-2)^2 * d/dx(x-2)#

# = 3(x-2)^2 * 1 = 3(x-2)^2#.

Without the chain rule

Expand,

#f(x) = (x+2)^3 = x^3+6x^2+12x+8#.

So, #f'(x) = 3x^2+12x+12#