# Difference quotient

Where am I going wrong?

Find the difference quotient for: $f(x)=2-x-3x^2$

$$\frac{[ 2-(x+h)-3(x+h)^2 ] - [ 2-x-3x^2 ]}{h}$$

$$\frac{2-x-h-3x^2-6hx-3h^2-2+x+3x^2}{h}$$

$$\frac{-3h^2-6hx-h}{h}$$

$$-3h-6x-1$$

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## 1 Answer

Well, nowhere. Everything is OK.

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But shouldn't the answer be -6x-1? –  AFerrara May 30 '12 at 2:33
@AFerraro: That is when $h \to 0$ which happens to be the derivative :) –  user9413 May 30 '12 at 2:33
Yes, but you need to take the limit, and what you have found is the quotient. To find the derivative you have to let $h \to 0$ –  Pedro Tamaroff May 30 '12 at 2:34
@AFerrara It will be $-6x-1$ only when you take the limit as $h \rightarrow 0$. –  user17762 May 30 '12 at 2:34
Ok, I thought they were the same, Thanks –  AFerrara May 30 '12 at 2:35