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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Well, nowhere. Everything is OK.
Pedro
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But shouldn't the answer be -6x-1? – AFerrara May 30 '12 at 02:33
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@AFerraro: That is when $h \to 0$ which happens to be the derivative :) – May 30 '12 at 02:33
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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 May 30 '12 at 02:34
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1@AFerrara It will be $-6x-1$ only when you take the limit as $h \rightarrow 0$. – May 30 '12 at 02:34
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Ok, I thought they were the same, Thanks – AFerrara May 30 '12 at 02:35