derivative

Let's say we want to know the rate of change of the \(f(x) = x^{2}\). Because this is not linear (the is a parabola and is therefore curved), we can only take the rate of change by finding the tangent line to a point on the curve. In other words, for any point \((x_{0}, y_{0})\), we want to find the straight line that touches this point, and no other points on the parabola.

The gameplan will look like this: first, we find the slope between two points \((x_{0}, y_{0})\) and \((x_{0} + dx, y_{0} + dy)\), where \(dx\) and \(dy\) are the difference in x and y respectively, or in other words, just a small change in x and change in y. Then, we make these points infinitely close to each other and see what the slope is. This resulting line will be infinitely close to the tangent line.

Now we want to find the equation for the slope between these two points:

\begin{align*} m = \frac{y_{1} - y_{0}}{x_{1} - x_{0}} \\ m = \frac{y_{0} + dy - y_{0}}{x_{0} + dx - x_{0}} \\ m = \frac{dy}{dx} \end{align*}

Because both of these points need to satisfy the \(f(x) = x^{2}\):

\begin{align*} y_{0} = x_{0}^{2} \\ y_{0} + dy = (x_{0} + dx)^{2} \\ dy = (x_{0} + dx)^{2} - y_{0} \\ dy = (x_{0} + dx)^{2} - x_{0}^{2} \\ m = \frac{(x_{0} + dx)^{2} - x_{0}^{2}}{dx} \end{align*}

Then we use a binomial expansion:

\begin{align*} m = \frac{x_{0}^{2} + 2x_{0}dx + dx^{2} - x_{0}^{2}}{dx} \\ = \frac{2x_{0}dx + dx^{2}}{dx} \\ = 2x_{0} + dx \end{align*}

Now we see that since \(dx\) is infinitely close to zero (but not zero because otherwise that would be dividing by zero), we can say that the tangent line of \(x^{2}\) at this point is:

\begin{align*} 2x_{0} \end{align*}

And since this works for all points over \(f(x)\), we can simply say:

\begin{align*} \frac{dy}{dx} = 2x \\ f'(x) = 2x \end{align*}

These two notations are both valid. The first is called Leibniz notation, and the second is called Lagrange notation.

Note that you can easily show that the process we did for \(x^{2}\) works for most functions, and is defined as follows:

\begin{align*} \frac{d}{dx}f(x) = \lim_{h\to0}\frac{f(x + h) - f(x)}{h} \end{align*}

This \(lim_{h\to0}\) notation is a limit. It broadly dictates that \(h\) is going infinitely close to zero but is not exactly zero. You will also see \(\frac{d}{dx}\) used as an operator on a much of the time, which also means you're taking a derivative with whatever \(\frac{d}{dx}\) is multiplied with.

Usually, instead of using the definition in order to calculate derivatives, we use some simpler rules to do so. We derive many of them here.

The equation of a circle centered at the origin is:

\begin{align*} x^{2} + y^{2} = r^{2} \end{align*}

This is clearly dependent on \(y\), and no, we don't need to do algebra to isolate the y (yet, we can do that later). instead, we can simply take the derivative of both sides:

\begin{align*} \frac{d(x^{2} + y^{2})}{dx} = \frac{d(r^{2})}{dx} \end{align*}

the right hand side is obviously going to reduce to zero because it is a constant inside a derivative. Because we consider \(y = y(x)\), taking the derivative of \(y\) in terms of \(x\) means we have to apply the chain rule.

\begin{align*} 2x + 2y(x) * y'(x) = 0 \end{align*}

Remember, the we are taking the derivative of here is \((y(x))^{2}\), which is why the \(y'(x)\) term appears; you're doing the chain rule on an inner that you don't know the value of but that you can represent nonetheless.

Now, we move everything to the other side in order to find \(y'(x)\):

\begin{align*} y'(x) = -\frac{x}{y(x)} \end{align*}

and then we finally find \(y(x)\) and substitute it in:

\begin{align*} y(x) = (r^{2} - x^{2})^{\frac{1}{2}} \\ y'(x) = -\frac{x}{(r^{2} - x^{2})^{\frac{1}{2}}} \end{align*}

The benefit of this strategy is that you can find the derivative of a circle (or as we will see later, many other curves) in terms of \(y\), which is useful for converting coordinate systems. Implicit differentiation is also useful for some other things, like: