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Notion of a Derivative

April 9th, 2009 |

Author: Merrin

Our first application of limits will be to calculate the slope of tangent lines to a function at a given coordinate. This slope function, the slope of the tangent line as a function of x, is called the derivative. There are several notations for representing the derivative.

$\cfrac{d}{dx} f(x)=\cfrac{df}{dx} =f'(x)=\dot{f}(x)$

The notation on the left is preferred, and it is due to Leibniz, one of the inventors of calculus. The notation of the dot over the function is due to Newton and is called a fluxion. Despite its cool name, compact notation, and Newton being the other inventor of calculus, fluxions didn’t quite catch on. Dots however live on in physics and more advanced formulations of calculus. It is not as important which symbol to use for a derivative as long as we know what the derivative means.

Leibniz notation is suggestive of a change in the numerator divided by a change in the denominator which hints at how to actually calculate the derivative.

In life before limits, a comparison of rates would be written as

$\cfrac{\Delta y}{\Delta x} =\cfrac{y_{2} -y_{1} }{x_{2} -x_{1} }$

But the instantaneous rates of change or derivative is defined as

$\cfrac{d}{dx} f(x)=\mathop{\lim }\limits_{\Delta x\to 0} \cfrac{f(x+\Delta x)-f(x)}{\Delta x} =\mathop{\lim }\limits_{\Delta x\to 0} \cfrac{\Delta y}{\Delta x}$

The old style average rate of change is recognizable as the exact equation for the slope of a line. In calculating the slope of a line with the limit as Delta x tended to zero, then Delta y would also respond and tend towards zero also. The ratio, the slope m, would be the same because it’s for a line. We can repeat the same procedure for a more general function which we call a curve.

Figure 3-1: As the point x_2 moves toward the left to x_1 , the limiting process produces a tangent line at x_1.

There are two ways to calculate the slope of the tangent line using a limit process. Make a point at x + Delta x approach x from the right, or make a point at x minus Delta x approach x from the left. Both limits must exist and be the same for the derivative to exist. Traditionally most people are thinking of the approach from the right, but there is plenty of room for left handers.

$\cfrac{d}{dx} f(x)=\mathop{\lim }\limits_{\Delta x\to 0} \cfrac{f(x+\Delta x)-f(x)}{(x+\Delta x)-x}=\mathop{\lim }\limits_{\Delta x\to 0}\cfrac{f(x)-f(x-\Delta x)}{x-(x-\Delta x)}$

For a line f(x)=mx+b we know that the derivative is just the slope,

$\cfrac{d}{dx} f(x)=m$

We can calculate this from the definition of the derivative.

$\mathop{\lim }\limits_{\Delta x\to 0} \cfrac{(mx+m\Delta x+b-mx-b)}{x+\Delta x-x} =m\mathop{\lim }\limits_{\Delta x\to 0} \cfrac{\Delta x}{\Delta x} =m$

Example 1. Show that the left hand derivative is also valid from the definition of the right hand derivative.

$\cfrac{d}{dx} f(x)=\mathop{\lim }\limits_{\Delta x\to 0} \cfrac{f(x)-f(x-\Delta x)}{\Delta x}$

Solution 1. This can be shown several ways. One way is start with the original definition of the derivative and reverse the sign of Delta x.

${\cfrac{d}{dx} f(x)} = {\mathop{\lim }\limits_{\Delta x\to 0} \cfrac{f(x+\Delta x)-f(x)}{\Delta x} } \\ ={\mathop{\lim }\limits_{\Delta x\to 0} \cfrac{f(x)-f(x-(-\Delta x))}{-\Delta x} } \\ = {\mathop{\lim }\limits_{\varepsilon \to 0} \cfrac{f(x)-f(x-\varepsilon )}{\varepsilon } } \\ = {\mathop{\lim }\limits_{\Delta x\to 0} \cfrac{f(x)-f(x-\Delta x)}{\Delta x} }$

Example 2. There are other definitions to calculate derivatives. Show this alternate definition of the derivative, \textit{provided the derivative at x exists}.

$\cfrac{d}{dx} f(x)=\mathop{\lim }\limits_{\Delta x\to 0} \cfrac{f(x+\Delta x/2)-f(x-\Delta x/2)}{\Delta x}$

Solution 2. This definition of the derivative is more balanced with respect to the point of differentiation. Half of the limit is on the left and half of it is on the right, but no part appears on x. This definition might lead us to trouble if there is a singularity or a removable discontinuity at x, read the fine print. The derivative must exist at x. To prove this identity, we recall that sometimes it is helpful to add and subtract the same thing in an equation. We write

${f'}={\mathop{\lim }\limits_{\Delta x\to 0} \cfrac{f(x+\Delta x/2)-f(x)+f(x)-f(x-\Delta x/2)}{\Delta x} } \\ ={\mathop{\lim }\limits_{\Delta x\to 0} \cfrac{f(x+\Delta x/2)-f(x)}{\Delta x} +\mathop{\lim }\limits_{\Delta x\to 0} \cfrac{f(x)-f(x-\Delta x/2)}{\Delta x} } \\ = {\cfrac{1}{2} \mathop{\lim }\limits_{\Delta x\to 0} \cfrac{f(x+\Delta x/2)-f(x)}{\Delta x/2} +\cfrac{1}{2} \mathop{\lim }\limits_{\Delta x\to 0} \cfrac{f(x-\Delta x/2)-f(x)}{-\Delta x/2} } \\ ={\cfrac{1}{2} f'(x)+\cfrac{1}{2} f'(x)=f'(x)}$

We used the fact that the derivative at x from the normal definition must exist.

Example 3. Show that the derivative at x can also be written in a different form which implies bringing two points x and z toward each other to find the slope.

$\cfrac{d}{dx} f(x)=\mathop{\lim }\limits_{z\to x} \cfrac{f(z)-f(x)}{z-x}$

Solution 3. The connection to the previous formula is z=x+Delta x. This form shows directly how z moves towards x as a limit in the slope calculation.

One entire branch of calculus can be said to be related to differentiation the other integration. We have explored several ways to define the derivative

which all agree. The rest of this chapter is devoted to proving and understanding the rules of differentiation for the elementary functions. When we finish it won’t be necessary to take any limits to calculate the derivatives of any elementary functions and their combinations. We will use the definition however to build these rules. It is important to know the foundations of where calculus comes from.

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