How do you find the derivative using limits of #f(x)=9-1/2x#?

2 Answers
Oct 13, 2017

#f'(x) = -1/2#
See below for details.

Explanation:

We will use the limit definition, #f'(x) = lim_(h->0) (f(x + h) - f(x))/h#.

#f'(x) = lim_(h->0) (9 - 1/2(x + h) - (9 - 1/2x))/h#

#f'(x) = lim_(h->0) (9- 1/2x - 1/2h - 9 + 1/2x)/h#

#f'(x) = lim_(h->0) (-1/2h)/h#

#f'(x) = lim_(h->0) -1/2#

#f'(x) = -1/2#

Hopefully this helps!

Oct 13, 2017

This is the formula for finding the derivative using limits:

#d/dxf(x) = lim_(h->0)(f(x-h)-f(x))/h#

Plug in #(x-h)# wherever you see an #x# in your #f(x)# for the first part of your numerator and then just the function itself for the second part of the numerator.

Then, its just basic arithmetic after that:

#=lim_(h->0)((9-(x-h)/2)-(9-x/2))/h#

#=lim_(h->0)((18/2-(x-h)/2)-(18/2-x/2))/h#

#=lim_(h->0)(((18-x+h)/2)-((18-x)/2))/h#

#=lim_(h->0)((18-x+h)/2-(18+x)/2)/h#

#=lim_(h->0)((cancel18cancel-x+cancelh)/2-(cancel18+cancelx)/2)/cancelh#

#=lim_(h->0)-1/2#

#=-1/2#