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 September 20th, 2015, 10:09 AM #1 Member   Joined: Jun 2014 From: Alberta Posts: 56 Thanks: 2 Use the Definition of a Derivative for f(x) = kx + b I can't figure out how to do this. f'(x) = lim(as h goes to 0) (k(x+h) + b - (k(x) + b))/h I tried using a conjugate, and I really don't know what other trick to try without getting an indeterminate answer.
September 20th, 2015, 11:00 AM   #2
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Quote:
 Originally Posted by Mathbound I can't figure out how to do this. f'(x) = lim(as h goes to 0) (k(x+h) + b - (k(x) + b))/h I tried using a conjugate, and I really don't know what other trick to try without getting an indeterminate answer.
You are thinking this too hard. You don't need to use the conjugate for this one.

$\displaystyle f'(x) = \lim_{h \to 0} \frac{(k(x + h) + b) - (kx + b)}{h}$

$\displaystyle f'(x) = \lim_{h \to 0} \frac{(kx + kh + b) - (kx + b)}{h}$

$\displaystyle f'(x) = \lim_{h \to 0} \frac{kx + kh + b - kx - b}{h}$

Can you finish from here?

-Dan

September 20th, 2015, 06:30 PM   #3
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Quote:
 Originally Posted by topsquark You are thinking this too hard. You don't need to use the conjugate for this one. $\displaystyle f'(x) = \lim_{h \to 0} \frac{(k(x + h) + b) - (kx + b)}{h}$ $\displaystyle f'(x) = \lim_{h \to 0} \frac{(kx + kh + b) - (kx + b)}{h}$ $\displaystyle f'(x) = \lim_{h \to 0} \frac{kx + kh + b - kx - b}{h}$ Can you finish from here? -Dan
Oh yeah! Thanks!

### f(x)=kx b

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