Using the limit definition, how do you find the derivative of #f (x) = 3x^5 + 4x #?

1 Answer
Jun 1, 2016

Apply the limit definition and use some algebra to simplify to find that

#f'(x) = 15x^4+4#

Explanation:

There are two equivalent definitions commonly used for the derivative of a function at a point:

#f'(a) = lim_(h->0)(f(a+h)-f(a))/h#

and

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

Note that we can show the second as being equivalent to the first by making the substitution #h=x-a# into the first. We will be using the second for this problem:


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

#=lim_(x->a)(3x^5+4x-3a^5+4a)/(x-a)#

#=lim_(x->a)(3*(x^5-a^5)/(x-a)+4*(x-a)/(x-a))#

#=lim_(x->a)(3(x^4+x^3a+x^2a^2+xa^3+a^4)+4)#

#=3(a^4+a^4+a^4+a^4+a^4)+4#

#=3(5a^4)+4#

#=15a^4+4#

Thus, we have #f'(x) = 15x^4+4#