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$y= 3x+5$; $x_0=-1$

I know the answer is $3$ but I don't know how to solve it. Can you please help me?

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4  
$\frac{dy}{dx}=3$, a constant. So its value does not depend on $x_0$. –  André Nicolas Feb 28 '13 at 23:47
1  
So, that's it? Thank you so much! –  Joanna Feb 28 '13 at 23:52
    
You are welcome. Sometimes when a problem is too simple, it can be puzzling. You might have had an easier time with $y=x^3+5x$. –  André Nicolas Feb 28 '13 at 23:59

1 Answer 1

$$ {dy \over dx} \equiv \lim_{h \to 0} {f(x+h)-f(x) \over h} \\ $$

Where $y = f(x)$.

$$ {dy \over dx} = \lim_{h \to 0} {3 x + 3 h + 5 - 3 x - 5 \over h} = 3 $$

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That is the formula that we have been using in class and I didn't understand it. However, you have made it so much clear! Thank you! –  Joanna Feb 28 '13 at 23:59
    
@SeanHaugh Are you sure you mean $\equiv$, instead of $=$? –  Jeel Shah Feb 28 '13 at 23:59
    
@gekkostate, I'm not Sean Haugh, but the $\equiv$ sign is also used for definitions or identities. –  George V. Williams Mar 1 '13 at 0:07
    
@GeorgeV.Williams oh! I was not aware of that. Thank you for the clarification. –  Jeel Shah Mar 1 '13 at 0:43

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