Proof of Product Rule

I don't understand the bit I circled in the proof. Why as small delta X tends towards 0 does small delta y/small delta X tend toward dy/dx etc. ?


Posted from TSR Mobile

The definition of $\frac{dy}{dx}$ is as the limit of $\frac{\delta y}{\delta x}$ as $\delta x$ tends to 0. This is because it is an "instantaneous gradient".

Oh I see. And what exactly does limit mean?


Posted from TSR Mobile

I don't understand the bit I circled in the proof. Why as small delta X tends towards 0 does small delta y/small delta X tend toward dy/dx etc. ?


Posted from TSR Mobile

Basically, thinking of a graph of $y = f(x)$, to find the average gradient of the graph, you'd do $\frac{\Delta y}{\Delta x} = \frac{ f(x+\delta x) - f(x)}{(x+\delta x) - x} = \frac{f(x+\delta x) - f(x)}{\delta x}$
Typically, to find the gradient between 2 points, $\delta x$ is rather big, but if we're trying to find the tangent, we need to make the distance between the two points as small as possible, so we make the tangent at that point to be the gradient of x at the 'limit' as $\delta x$ approaches 0

If we tried y = x^2
$\frac{dy}{dx} = \displaystyle\lim_{\delta x \rightarrow 0} \frac{\Delta y}{\Delta x} = \displaystyle\lim_{\delta x \rightarrow 0} \frac{(x+\delta x)^2 -x^2}{\delta x} = \displaystyle\lim_{\delta x \rightarrow 0} \frac{2x\delta x + (\delta x)^2}{\delta x} = \displaystyle\lim_{\delta x \rightarrow 0} 2x + \delta x$
As the change in x approaches 0, the gradient approaches 2x

In the book, instead of saying $f(x+\delta x)$ it's saying $y + \delta y$ or $v + \delta v$

So as the change in X approaches 0, the gradient of the line approaches the gradient of the tangent? Is that basically what it means?


Posted from TSR Mobile