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.
$\endgroup$ 52 Answers
$\begingroup$Use the difference of two squares:
$$z^2-x^2 = (z-x)(z+x)$$
$\endgroup$ $\begingroup$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):
$$\frac{4\left(z^{2}-x^{2}\right)+3\left(z-x\right)}{z-x}=\frac{4\left(z+x\right)\left(z-x\right)+3\left(z-x\right)}{z-x}=4\left(z+x\right)+3$$
$$\lim_{z\to x}\left(4\left(z+x\right)+3\right)=4\left(x+x\right)+3=8x+3$$
