How do you find f'(x) using the definition of a derivative #f(x) =(x^2 + 2)^2#?

1 Answer
Nov 13, 2015

#f'(x) = 2(x^2 + 2)*2x#

Explanation:

There are two forms of the limit definition of the derivative.

#f'(x) = lim_(a->x)(f(x)-f(a))/(x-a)#
and
#f'(x) = lim_(h->0)(f(x+h)-f(x))/(h)#
(Showing that these are equivalent may be done through the substitution #x = a + h# and then a change in variables)

For this problem, the first form is easier to use.

#f'(x) = lim_(a->x)((x^2+2)^2 - (a^2+2)^2)/(x-a)#

#=>f'(x) = lim_(a->x)(x^4+4x^2+4 - a^4-4a^2-4)/(x-a)#

#=>f'(x) = lim_(a->x)((x^4 - a^4)+4(x^2-a^2))/(x-a)#

#=>f'(x) = lim_(a->x)((x^2+a^2)(x^2-a^2)+4(x^2-a^2))/(x-a)#

#=>f'(x) = lim_(a->x)((x^2-a^2)(x^2 + a^2 + 4))/(x-a)#

#=>f'(x) = lim_(a->x)((x+a)(x-a)(x^2 + a^2 + 4))/(x-a)#

#=>f'(x) = lim_(a->x)(x+a)(x^2 + a^2 + 4)#

#=>f'(x) = lim_(a->x)(x+a)(x^2 + a^2 + 4)#

#=>f'(x) =(x+x)(x^2 + x^2 + 4) = 2x(2x^2 + 4)#

#=>f'(x) = 2(x^2 + 2)*2x#

Which is the same result obtained from later techniques such as the power rule and chain rule.