February 1st, 2018, 05:04 PM  #1 
Newbie Joined: Jan 2018 From: MI Posts: 11 Thanks: 1  Calc help
I am not sure what the expression is asking for, I've tried a whole bunch of different solutions but I just don't get it.

February 1st, 2018, 05:08 PM  #2 
Senior Member Joined: Sep 2016 From: USA Posts: 535 Thanks: 306 Math Focus: Dynamical systems, analytic function theory, numerics 
The derivative of a function at $10$ is the following number if it exists: \[ \lim\limits_{h \to 0} \frac{f(10 + h)  f(10))}{h} .\] Evaluate this in your example. 
February 1st, 2018, 06:13 PM  #3 
Newbie Joined: Jan 2018 From: MI Posts: 11 Thanks: 1 
I did that and got 8h154. I'm not sure what I'm supposed to do next.
Last edited by skipjack; February 1st, 2018 at 07:48 PM. 
February 1st, 2018, 08:01 PM  #4 
Global Moderator Joined: Dec 2006 Posts: 20,116 Thanks: 1912 
You should instead evaluate $\displaystyle \lim_{h \to 0} \frac{f(x + h)  f(x)}{h}$ and you initially get $\displaystyle \frac{f(x + h)  f(x)}{h} = 8h + 16x + 6$ which has a limit of $16x + 6$ as $h$ tends to 0. That is the expression that is asked for, and is the derivative of $f(x)$. Its limit as $x$ tends to 10 can be found by substituting 10 for $x$. Note: it's a rather poor question. 
February 2nd, 2018, 09:27 AM  #5 
Math Team Joined: Jul 2011 From: Texas Posts: 2,805 Thanks: 1449  $\displaystyle f'(a) = \lim_{x \to a} \dfrac{f(x)f(a)}{x  a}$ $\displaystyle f'(10) = \lim_{x \to 10} \dfrac{f(x)f(10)}{x  (10)}$ $\displaystyle f'(10) = \lim_{x \to 10} \dfrac{8x^2+6x+11  751}{x +10}$ $\displaystyle f'(10) = \lim_{x \to 10} \dfrac{8x^2+6x740}{x +10}$ $\displaystyle f'(10) = \lim_{x \to 10} \dfrac{2(4x37)(\cancel{x+10})}{\cancel{x +10}} = 2(77) = 154$ 

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