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- Derivative is Slope of the Tangent
- Sloppy Picture of a Tangent
- A Better Picture of Secant
- To the Tangent

The geometric interpretation of a derivative of a function at a point is the slope of the tangent line of the graph of the function at that point.

A tangent line of circle, which is where most people first encounter the term *tangent,* necessary touches a circle at exactly one point. The circle, however, is a special case. In the more general case, a tangent line may touch the graph of a function at more than one point.

Finding the slope of the tangent line at a point, if there is one) is tantamount to finding the tangent line since given a point and a slope, the line can be found with the point-slope formula for a line:

{$$ y - y_0 = m( x - x_0) $$}

so the slope is really all we need to find, even if we have no idea what the tangent would be, and the derivative is how we can find the slope.

If we have two non-coincidental points on the graph of a function, **P** and **Q**, the line they determine is called a *secant.* If we imagine that **Q** moves closer and closer to **P**, and the secant rotates so it still contains **Q** and **P**, eventually the secant will have a slope nearer and near that of the *tangent* at **P**. That is the hope, anyway; for some functions this never happens.

Obviously, the point **Q**, which we imagine as being moveable along the graph of the function, could be somewhere on the graph of the function where the slope of the secant line does not resemble the slope of the tangent very much at all, and in **Q**'s travels to get near **P** the slope of the secant may be all over the place. So the claim that the slope of the secant gets near the slope of the tangent is only valid when **Q** arbitrarily close to **P**.

Now the problem with this simple illustration is, the tangent line has been drawn in. I have assumed it is there and have drawn what I think it would look like. But in fact, sometimes for some points on some functions, the tangent line is not there. Moreover. comparing the slope of the secant to the slope of the tangent line as I think it should be, is exactly backwards. It is only when the slope of the secant line approaches a value as a limit that we know there is a tangent.

Figure 3 is part of the graph of **y=1/x**. There is a symmetrical part in the third column. This function never crosses or touches either the **X** or **Y** axis. (An overview of the function is in Fig. 3a in the sidebar.)

The figure has been overloaded with labels to illustrate the variety of notations commonly used, and often used interchangeably in the same calculation by many authors and lectures.

The function {$ y = 1/x $} is often expressed as {$ f(x) = 1/x $}. Only {$ y $} and {$ x $} appear in the first form, and the first form is solved for {$ y $}, so it is not completely clear in such a simple example why the {$ f(x) $} notation would ever be used, but its virtues become apparent in more complicate function in which there may be more unknowns or variables.

The co-ordinates of {$ P $} and {$ Q $} appear in the figure. These have been precalculate, of course, to ensure they really are on the graph. Although points are often represented if the form {$ (x_0, y_0) $}, they often appear as {$ (x_0, f(x_0) ) $}.

This emphasizes that the point really is on the graph of the function rather than being just any point on the co-ordinate plane. In this case it represents what really happened. I picked the values of {$ x $} and computed the values of {$ y $} using the function.

Because we are only discussing the secant {$ \overline{PQ} $} and all the values are given, there is no calculus problem here. But we can find the equation of the secant.

Given two points:

{$$ P = (x_0, y_0) \text{ and } Q = (x_1, y_1) $$}

the line containing them is given by

{$$ y - y_0 = \frac{y_0 - y_1}{x_0 - x_1}(x - x_0) $$}

provided {$ x_0 \ne x_1 $}.

The slope of the line, conventionally called {$ m $}, given by

{$$ m = \frac{y_0 - y_1}{x_0 - x_1} $$}

It does not matter which point is which, and in particular which component is subtracted from which in computing the slope, so long they are taken in the same order in the numerator and denominator. The order can be chosen to simplify computations or picked arbitrarily.

To compute the slope of the secant {$ \overline{PQ} $}:

{$$ \begin{align} m &= \frac{y_0 - y_1}{x_0 - x_1} \cr &= \frac{3/2 - 1/3}{2/3 - 3} \cr &= \frac{9/6 - 2/6}{4/6 - 18/6} \cr &= \frac{7/6}{- 14/6} \cr &= \frac{7}{- 14} \cr m &= -1/2 \end{align}$$}

Although it is slightly beside the point here, the equation of the secant line can now be found:

{$$ \begin{align} y - y_0 &= m(x - x_0) \cr y - 3/2 &= -1/2(x + 2/3) \cr y &= -1/2x + (1/2)(2/3) + 3/2 \cr y &= -1/2x + 1\frac{5}{6} \end{align} $$}

which I leave in y-intercept form to make comparison with the figure easier.

The numerator of the slope is also represented by {$ \Delta y $}, and naturally the denominator is also represent by {$ \Delta x $}. These are also read "the change in y" and "the change in x" respectively. In a (Cartesian co-ordinate) figure, the change in y is a vertical change, also called "rise" whether up or down, and the change in x is a horizontal change, also called "run." This is from a mnemonic for slope: slope is the ratio of rise to run, which is useful to remember which is the numerator and the denominator in the slope formula (but does not help if the subtractions are not taken in the same order).

With {$ \Delta y $} and {$ \Delta x $} the length of the secant segment can be computed which is occasionally useful. More pertinent at this point is that {$ \Delta y $} is the change in y (the rise) while {$ \Delta x $} is the change in x (the run}, so

{$$ m = \frac{y_0 - y_1}{x_0 - x_1} = \frac{\Delta y}{\Delta x} $$}

With trivial exceptions, geometry cannot find the slope of a line tangent to a function at a point. But it can make the processes of calculus seem much more reasonable.

Figure 4 is meant to illustrate what happens if we imagine that the point {$ Q $} (not labeled) moves closer and closer to the point {$ P $} on the function {$ f(x) = 1/x $}.

In effect, the secant line {$ \overline{PQ} $} rotates. If {$ Q $} ever reached {$ P $} exactly, there would not be a secant line because {$ Q $} would be the same point as {$P $}, and one point does not define a line.

This makes sense because the slope of the secant is m where

{$$ m = \frac{\Delta y}{\Delta x} $$}

and if {$ Q $} becomes one point with {$ P $}, {$ \Delta y $} and {$ \Delta x $} would both be 0 and {$ m $} would be undefined. The secant which would not exist would have a slope which does not exist.

Nonetheless, looking at Figure 4, we have the feeling there ought to be a line at the point {$ P $} when it is also known as {$ Q $}. They are now only one point, but if we had a slope, we could use the point-slope formula to find the line that we feel ought to be there. Fortunately we have limits, which helps us deal with things that get near other things regardless of whether the things become identical.

This brings us to the definition of derivative (geometrical version). Given a function {$ f(x) $}, its derivative is {$ f'(x) $} where

{$$ f^\prime (x) = \lim_{ \Delta x \rightarrow 0} {m} = \lim_{ \Delta x \rightarrow 0} {{\Delta y} \over { \Delta x}}$$}

or in the specific case

{$$ f^\prime (x_0) = \lim_{ x \rightarrow x_0} {{f(x_0) -f(x)} \over { x_0 - x}} $$}

Now to find the slope of the tangent of

{$$ \begin{gather} y = 1/x \cr f(x) = y = 1/x \end{gather} $$}

at {$ P = (x_0, y_0) = (x_0, f(x_0)) = (2/3, 3/2) $}

{$$ \begin{align} f^\prime (2/3) &= \lim_{ x \rightarrow x_0} {{f(x_0) -f(x)} \over { x_0 - x}} \cr &= \lim_{ x \rightarrow x_0} {{f(x_0) -f(x)} \over { x_0 - x}} \cr &= \lim_{ x \rightarrow 2/3} {{3/2 - 1/x} \over { 2/3 - x}} \end{align} $$}

Notice we are taking the limit as {$ x \rightarrow 2/3 $}, so we cannot merely substitute because that would result in {$ 0/0 $}. This is always the case in derivatives no matter how they are set up or what notation is used. Getting rid of the {$ 0 $} in the denominator is most of what differential calculus is about.

{$$ \begin{align} f^\prime (2/3) &= \lim_{ x \rightarrow 2/3} {{3/2 - 1/x} \over { 2/3 - x}} \cr &= \lim_{ x \rightarrow 2/3} {{3/2 - 1/x} \over {-(x - 2/3)}} \tag{i.} \cr &= \lim_{ x \rightarrow 2/3} {{ (3x - 2)/2x } \over {-(x - 2/3)}} \tag{ii.} \cr &= \lim_{ x \rightarrow 2/3} {{ (x - 2/3)(3/2x) } \over {-(x - 2/3)}} \tag{iii.} \cr &= \lim_{ x \rightarrow 2/3} {{ (3/2x) } \over { ( - 1 )}} \cr &= \lim_{ x \rightarrow 2/3} {{ -3 } \over { 2x }} \end{align} $$}

Now the limit can be solved by substitution because the denominator will not be {$ 0 $}.

{$$ \begin{align} f^\prime (2/3) &= \lim_{ x \rightarrow 2/3} {{ -3 } \over { 2x }} \cr &= {{ -3 } \over { 2(2/3) }} \cr &= {{ -3 } \over { 4/3 }} \cr &= {{ -9 } \over { 4 }} \end{align} $$}

At this point the derivative has been found, but for comparison to the general solution for this function, the form of the solution is changed:

{$$ \begin{align} f^\prime (2/3) &= {{ -9 } \over { 4 }} \cr &= (-1) \left( {{ 1 } \over { 4/9 }} \right) \cr \therefore f^\prime (2/3) &= (-1) \left( {{ 1 } \over { (2/3)^2 }} \right) \end{align} $$}

Hairy algebra is very common in derivatives because it is always necessary to get the zero out of the denominator.

i. -1 is factored in the denominator. ii. The fractions in the numerator expressed with a common denominator. iii. 3 is factored in the numerator to make it clear that there is a factor that cancels with a factor in the denominator.

*Sources:*

- FooPlot: Online graphing calculator and function plotter
- MIT OpenCourseWare http://ocw.mit.edu "18.01SC Single Variable Calculus: Fall 2010" Creative Commons License Terms. Secants and Tangents (problem set)
- FooPlot: Online graphing calculator and function plotter
- FooPlot: Online graphing calculator and function plotter

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