Using the limit definition, how do you find the derivative of #f(x) = 3x^2 + 8x + 4 #?

1 Answer
Mar 25, 2016

#f'(x)=6x+8#; see below for explanation

Explanation:

The limit definition of the derivative states that for a function #f(x)# its derivative equals

#f'(x)=lim_(hrarr0)(f(x+h)-f(x))/h#

So, when #f(x)=3x^2+8x+4#, we see that #f(x+h)=3(x+h)^2+8(x+h)+4#.

Applying the limit definition, we obtain

#f'(x)=lim_(hrarr0)(3(x+h)^2+8(x+h)+4-(3x^2+8x+4))/h#

Find #(x+h)^2#. Distribute the negative into #-(3x^2+4x+8)#.

#f'(x)=lim_(hrarr0)(3(x^2+2hx+h^2)+8(x+h)+4-3x^2-8x-4)/h#

Distribute the #3# and the #8#.

#f'(x)=lim_(hrarr0)(3x^2+6hx+3h^2+8x+8h+4-3x^2-8x-4)/h#

Cancel all like terms.

#f'(x)=lim_(hrarr0)(color(red)(cancel(color(black)(3x^2)))+6hx+3h^2color(blue)(cancel(color(black)(+8x)))+8hcolor(green)(cancel(color(black)(+4)))color(red)(cancel(color(black)(-3x^2)))color(blue)(cancel(color(black)(-8x)))color(green)(cancel(color(black)(-4))))/h#

#f'(x)=lim_(hrarr0)(6hx+3h^2+8h)/h#

Divide #h# from each term.

#f'(x)=lim_(hrarr0)6x+3h+8#

To evaluate the limit, plug in #0# for #h#.

#f'(x)=6x+3(0)+8#

#f'(x)=6x+8#