# Derivative of a function is the equation of the tangent line?

So what exactly is a derivative? Is that the EQUATION of the line tangent to any point on a curve? So there are 2 equations? One for the actual curve, the other for the line tangent to some point on the curve? How can the equation of the tangent line be the same equation throughout the curve?

The derivative of a function at a point is the slope of the tangent line at that point, not the line's equation. The function $f'(x)$ tells you this slope for each point in the domain of $f$ where a non-vertical tangent line exists.

The derivative is the instantaneous rate of change of a function $f(x)$, usually denoted $f'(x)$ or $\frac{d}{dx}\begin{pmatrix}f(x)\end{pmatrix}$. What does instantaneous rate of change really mean? The rate of change of a line is it's slope. The slope of a line can be calculated by using two points on the line, $(x_1,f(x_1))$ and $(x_2,f(x_2))$. You probably know the formula:

$$\frac{f(x_2)-f(x_1)}{x_2-x_1}.$$

Now consider finding the rate of change of something other than a line. If you use the above formula, you will only be calculating the average rate of change of the function over that interval $[x_1,x_2]$. This is because the behavior of this non-linear function may change drastically over the interval. For example, if the average rate of change is positive on the interval $[x_1,x_2]$, nothing is preventing the function from increasing rapidly to $y>f(x_2)$ and then decreasing to $f(x_2)$ as $x$ approaches $x_2$ (This is merely an example of how a non-linear function could behave).

So given that slope breaks down for curve other than a line, it might be instructive to calculate the rate of change in another way. This is why the derivative is used. It is the instantaneous rate of change because it calculates the slope as the points $x_1$ and $x_2$ become arbitrarily close together. This leads to the definition of the limit

$$f'(x)=lim_{\Delta x\to0}\frac{f(x+\Delta x)-f(x)}{(x+\Delta x)-x}=lim_{\Delta x\to0}\frac{f(x+\Delta x)-f(x)}{\Delta x}$$

You can easily see that you are taking the average rate of change between the points $(x,f(x))$ and $(x+\Delta x,f(x+ \Delta x))$, and tending $\Delta x$ towards zero, to calculate the instantaneous rate of change. You could loosely think about this as the "slope" of the curve at that point.

Take a look at this picture to see how you move from a so-called secant line to tangent line as $\Delta x$ vanishes:

(You can easily find many more examples and visual aid)

You should note the the nice derivative formulas you can find in tables are found by applying the limit definition of a derivative above to a function, such $\sin x$, $e^x$, etc.

Now, given that you can think about the derivative at a point as the "slope" of the curve at that point, you can calculate the equation of the line tangent to the curve at a point. So suppose we have a curve, $f(x)$, and we want to find the equation of the tangent line at $x=a$. Recall that to calculate the equation of a line we need a slope and a point. Using the derivative of $f$ we have all the information we need. To find the slope of the curve at a point, or the slope of the line tangent to that point, we find $f'(a)$, which is the derivative evaluated at $a$. Then, we simply compute $f(a)$ to get our point, $(a,f(a))$. Now we find the tangent line as

$$y-f(a)=f'(a)(x-a)$$

as it is standard to define a line.

This is all in one dimension, but it can be extended into $n$-dimensional space.

Hint

Suppose that you run a car and that you registered the distance as a function of time; this means that you have a function $$\text{distance}=f(\text{time})$$ The derivative is the change of the distance over a small period of time; this is the speed $$\text{speed}=f'(\text{time})$$ When considering a curve, the derivative of the function at a given value of $x$ gives the slope of the tangent line to the curve at this specific point.

So what exactly is a derivative?

The derivative is instantaneous (i.e. at any given precise moment in time) the rate of change of a dependent variable (usually $y$) with respect to the independent variable (usually $x$). For straight lines, the derivative is simply the slope or gradient of the line.

For curves, whose gradient (slope) is constantly changing, the derivative gives the instantaneous rate of change at a given point; i.e. the slope of the tangent at that point.

There is one and only one tangent (that's what tangent means in Latin- "touching"to a curve at any given point. The tangent is a straight line that just touches the curve.

So there are 2 equations? One for the actual curve, the other for the line tangent to some point on the curve?

In a sense, yes. There's the equation of the curve, $y=f(x)$, which could be any function ($x^2, e^x, \ln(x), \tan(x),$ etc.), and then there's the equation of the tangent line at a given point.

If you choose a different point on the curve, not only will you change the gradient of the tangent at the point, but also the equation of the tangent.

How can the equation of the tangent line be the same equation throughout the curve?

This is not true.

I'll show you with a sketch of an arbitrary function $y=f(x).$

I've just picked 3 random points on the curve (denoted by purple dots). At each of these points, I've drawn the tangents to the curve at that specific point. As you can see, these are clearly different lines, so they can't possibly have the same equation.

• I see from your example that the tangent lines should not have the same equation BUT, they are equal to the derivative of y = f(x). So the equation we get as a result of taking the derivative is the equation of the tangent line right? there is only one answer if we take the derivative and that results in ONE equation: f'(x). According to my understanding that should be the equation of all the tangent lines on the graph? So it doesnt make sense how the tangent lines should have different equations... – TazMan Jul 2 '14 at 14:55

Just to give some completion here, if the function $y=f(x)$ passes through the point $(x_0,y_0)$, and has derivative $f'(x_0)$ at that point, the equation of the tangent at $(x_0,y_0)$ is:$$y-y_0=f'(x_0)(x-x_0)$$ This is the line which passes through $(x_0,y_0)$ and has gradient $f'(x_0)$.

This relates to the fact that $$f'(x_0)=\lim _{x\to x_0}\frac {f(x)-f(x_0)}{(x-x_0)}=\lim _{x\to x_0}\frac {y-y_0}{(x-x_0)}$$