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I'm attempting to find the derivative of the function: $$f(x) = 4x^2+3x+5$$ Using the alternative formula: $$\frac{f(z)-f(x)}{z-x}$$

Here are my steps so far: $$\frac{4z^2+3z+5-(4x^2+3x+5)}{z-x}$$ $$\frac{4(z^2-x^2)+3(z-x)}{z-x}$$

I have no idea where to go from this point. I've tried several different things to come up with the correct answer - which I know is $8x+3$. Can someone please guide me through this problem? I'm completely stuck. Also, sorry about the formatting. I'm using this editor and don't have it completely figured out yet.

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The formatting was good. The formulas were all legible and correct, so I'd say that whatever you've been doing, keep doing it. – MJD Feb 10 '13 at 14:40
@MJD Somebody fixed it for me. There was a red /[ before and after each equation. And when I tried to delete those, it would break my formatting. – Scott Feb 10 '13 at 14:43
I fixed the red brackets. I just deleted them, and nothing broke. But even with the red brackets, everything else was perfectly clear, so it was easy to fix, and that's good enough. – MJD Feb 10 '13 at 14:45
this is not a simple alternative formula.this is called lagaranges theorem to check differentiability in a closed interval.There are some conditions before applying this formula – iostream007 May 10 '13 at 13:08
You must know that $z^2-x^2 = (z-x)(z+x)$ – Steven Gregory Mar 21 at 12:54
up vote 5 down vote accepted

Use the difference of two squares:

$$z^2-x^2 = (z-x)(z+x)$$

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You must calculate the limit too: $$\lim_{z\to x}\frac{f\left(z\right)-f\left(x\right)}{z-x}$$

In your case it's $$\lim_{z\to x}\frac{4(z^{2}-x^{2})+3(z-x)}{z-x}$$

Just simplify the fraction to $4\left( z+x \right)+3$ and you will see that the limit is equal to $8x+3$. Then you have done it.

These steps (if you need them more explicit): $$\frac{4\left(z^{2}-x^{2}\right)+3\left(z-x\right)}{z-x}=\frac{4\left(z+x\right)\left(z-x\right)+3\left(z-x\right)}{z-x}=4\left(z+x\right)+3$$ $$\lim_{z\to x}\left(4\left(z+x\right)+3\right)=4\left(x+x\right)+3=8x+3$$

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Good point. I should have included the lim z->x on every step. – Scott Feb 10 '13 at 14:58
To use less space you can simplify fraction and only then use lim z->x on simplified expression. In such simple examples where only polynomials are used it is useful. – zaarcis Feb 10 '13 at 15:12

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