Hello bobsanchez Originally Posted by

**bobsanchez** So basically add h to whatever the quantity is and then subtract that quantity from it?

Yes - more or less.

An alternative notation to $\displaystyle \delta x,\;y,\;\delta y$ is to use function notation - $\displaystyle y=f(x)$ - so that P is the point $\displaystyle \big(x,f(x)\big)$ and Q is $\displaystyle \big(x+h,f(x+h)\big)$. The gradient of PQ is then

$\displaystyle \dfrac{f(x+h)-f(x)}{h}$

This looks quite different from what I wrote in my first reply, but is simply an alternative way of saying the same thing.

In the example I gave you before, then, $\displaystyle f(x) = x^2$, and so the gradient of PQ is

$\displaystyle \dfrac{f(x+h)-f(x)}{h}=\dfrac{(x+h)^2-x^2}{h}$

$\displaystyle =\dfrac{x^2+2xh+h^2-x^2}{h}$

$\displaystyle =\dfrac{2xh+h^2}{h}$

$\displaystyle =2x+h$

So, as before, as $\displaystyle h\to 0$ the limit of this gradient - which is therefore the gradient of the tangent at P - is $\displaystyle 2x$.

Grandad