Frozen ice water partially melted illustrating what derivatives represent

Calculus

Derivative Basics - Representations Explained

08.11.2022 • 10 min read

Rachel McLean

Subject Matter Expert

In this article, we’ll discuss the definition of the derivative, and learn what the derivative represents. We’ll examine three different ways to conceptualize derivatives, including instantaneous rates of change, the slope of the tangent line, and velocity.

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In This Article

  1. What Does the Derivative Represent?

  2. 3 Steps on How To Calculate Derivatives

  3. Derivatives and Rates of Change

  4. Derivatives and Slopes of Tangent Lines

  5. Derivatives and Velocity

What Does the Derivative Represent?

Derivatives measure rates of change. More specifically, derivatives are concerned with instantaneous rates of change. The instantaneous rate of change of a function at x=ax = a is equal to the slope of the tangent line at x=ax = a.

What is a derivative exactly? We can formally define the derivative of a function using limits:

f(x)=limΔx0f(x+Δx)f(x)Δx=Lf’(x) = \mathop{\lim }\limits_{\Delta{x} \to 0} \frac{{f\left( {x + \Delta{x} } \right) - f\left( x\right)}}{\Delta{x} }=L

The limit given above defines a function f(x)f’(x), which is called the derivative. This function f(x)f’(x) allows us to assign some value f(a)f’(a) to each number aa in the domain of y=f(x)y = f(x):

f(a)=limΔx0f(a+Δx)f(a)Δx=Lf’(a) = \mathop{\lim }\limits_{\Delta{x} \to 0} \frac{{f\left( {a + \Delta{x} } \right) - f\left( a\right)}}{\Delta{x} }=L

If this limit exists at x=ax = a, then LL is the derivative of the function at x=ax = a.

If this limit does not exist at x=ax = a, then there is no tangent, and so the function is not differentiable at aa.

If a function is differentiable at every point in its domain, then we consider it a differentiable function.

3 Steps on How To Calculate Derivatives

Below are 3 steps to calculating derivatives using the limit definition of a derivative:

1. Substitute

Substitute your function into the limit definition of a derivative formula. To do this, replace xx with the expression (x+Δx)(x + \Delta{x}) wherever xx appears in f(x)f(x). Be cautious here and remember that f(x+h)f(x)+f(h)f(x + h) \neq f(x) + f(h).

2. Simplify

3. Evaluate

Evaluate the resulting limit.

The symbol Δx\Delta{x} represents a small value that xx changes by. By evaluating the limit as Δx\Delta{x} approaches zero, we obtain an instantaneous rate of change.

Alternatively, you might also see f(x)f’(x) expressed by using the substitution b=x+hb = x + h:

f(x)=limbxf(b)f(x)bxf’(x) = \mathop{\lim }\limits_{b \to x} \frac{{f\left( {b } \right) - f\left( x\right)}}{b - x}

Similarly, f(a)f’(a) will use the substitution b=a+hb = a + h:

f(a)=limbaf(b)f(a)baf’(a) = \mathop{\lim }\limits_{b \to a} \frac{{f\left( {b } \right) - f\left( a\right)}}{b - a}

The derivative f(x)f’(x) represents two things. First, f(x)f’(x) represents the instantaneous rate of change of y=f(x)y = f(x) with respect to xx, for all values of xx where the limit exists. Second, f(x)f’(x) represents the slope of the curve of f(x)f(x) at any point in its domain.

In the next sections, we’ll go into more detail about the meaning of rates of change and tangent lines, and their relationship with derivatives.

Derivatives and Rates of Change

Rates of change describe the change that occurs in one variable as another variable changes. More specifically, rates of change represent the change in the dependent variable as the independent variable changes.

For any function ff, the average rate of change over the interval [a,b][a, b] is given by:

Average Rate of Change=ΔyΔx=y2y1x2x1=f(b)f(a)ba\text{Average Rate of Change} = \frac{\Delta{y}}{\Delta{x}} = \frac{y_2 - y_1}{x_2-x_1} = \frac{f(b)-f(a)}{b-a}

This value represents the average rate of change of ff as xx changes from aa to bb. We also refer to this value as the difference quotient.

To better understand the connection between the average rate of change and the derivative, we can change the notation of our interval from [a,b][a, b] to [a,a+Δx][a, a +\Delta{x}]. In this notation, Δx\Delta{x} represents the distance between two values of xx, namely aa and a+Δxa + \Delta{x}.

Now, we express the average rate of change of ff as aa changes from aa to a+Δxa +\Delta{x} as:

Average Rate of Change=f(a+Δx)f(a)Δx\text{Average Rate of Change} = \frac{f(a + \Delta{x})-f(a)}{\Delta{x}}

This value gives us the average rate of change over an interval, but not an exact rate of change at a point. To find the exact rate of change of ff at x=ax = a, we need to make Δx\Delta{x} as small as we can.

We can do this by taking the limit of the difference quotient as Δx\Delta{x} approaches zero. The resulting value is called the instantaneous rate of change of ff at x=ax = a:

Instantaneous Rate of Change=limΔx0f(a+Δx)f(a)Δx\text{Instantaneous Rate of Change}= \mathop{\lim }\limits_{\Delta{x} \to 0} \frac{{f\left( {a + \Delta{x} } \right) - f\left( a\right)}}{\Delta{x} }

Often, we use the substitution h=Δxh =\Delta{x} to simplify our notation:

limΔx0f(a+Δx)f(a)Δx=limh0f(a+h)f(a)h\mathop{\lim }\limits_{\Delta{x} \to 0} \frac{{f\left( {a + \Delta{x} } \right) - f\left( a\right)}}{\Delta{x} } = \mathop{\lim }\limits_{h \to 0} \frac{{f\left( {a + h } \right) - f\left( a\right)}}{h }

Alternatively, we can express the instantaneous rate of change of ff at x=ax = a as:

Instantaneous Rate of Change=limbaf(b)f(a)ba\text{Instantaneous Rate of Change}= \mathop{\lim }\limits_{b \to a} \frac{{f\left( {b } \right) - f\left( a\right)}}b - a

Notice that both ‌definitions of the instantaneous rate of change are equal to the limit definitions of f(a)f’(a) given earlier!

Rate of Change Example

Let’s do one example together. We’ll find the instantaneous rate of change of f(x)=1xf(x) = \frac{1}{x} at x=5x = 5. Using the definition of instantaneous rate of change, we have:

Instantaneous Rate of Change=limΔx0f(a+Δx)f(a)Δx\text{Instantaneous Rate of Change}= \mathop{\lim }\limits_{\Delta{x} \to 0} \frac{{f\left( {a + \Delta{x} } \right) - f\left( a\right)}}{\Delta{x} }

=limΔx0f(5+Δx)f(5)Δx=\mathop{\lim }\limits_{\Delta{x} \to 0} \frac{{f\left( {5 + \Delta{x} } \right) - f\left( 5\right)}}{\Delta{x} }
=limΔx015+Δx15Δx= \mathop {\lim }\limits_{\Delta{x} \to 0} \frac{\frac{1}{5 + \Delta{x}} - \frac{1}{5}}{\Delta{x}}
=limΔx055(5+Δx)(5+Δx)5(5+Δx)Δx= \mathop {\lim }\limits_{\Delta{x} \to 0} \frac{\frac{5}{5(5 + \Delta{x})} - \frac{(5 + \Delta{x})}{5(5+ \Delta{x})}}{\Delta{x}}
=limΔx055Δx5(5+Δx)Δx= \mathop {\lim }\limits_{\Delta{x} \to 0} \frac{\frac{5-5-\Delta{x}}{5(5+\Delta{x})}}{\Delta{x}}
=limΔx0Δx5(5+Δx)1Δx= \mathop {\lim }\limits_{\Delta{x} \to 0} \frac{-\Delta{x}}{5(5+ \Delta{x})} \cdot \frac{1}{\Delta{x}}
=limΔx015(5+Δx)= \mathop {\lim }\limits_{\Delta{x} \to 0} \frac{-1}{5(5+\Delta{x})}
=15(5+0)= \frac{-1}{5(5+0)}
=155= \frac{-1}{5 \cdot 5}
=152= -\frac{1}{5^2}
=125= -\frac{1}{25}

Thus, the instantaneous rate of change of the function f(x)=1xf(x) = \frac{1}{x} at x=5x = 5 at x=5x = 5 is 125-\frac{1}{25}.

Derivatives and Slopes of Tangent Lines

So far, we’ve learned that we can formally define a derivative using limits and that this limit represents the instantaneous rate of change for all xx values where the limit exists:

f(x)=limΔx0f(x+Δx)f(x)Δx=Lf’(x) = \mathop{\lim }\limits_{\Delta{x} \to 0} \frac{{f\left( {x + \Delta{x} } \right) - f\left( x\right)}}{\Delta{x} }=L

You might be wondering how to geometrically interpret this limit. Let’s review some important definitions.

A secant line is a straight line that passes through two points on a curve. For example, the below graph of a function illustrates the secant line through x=1x = 1 and x=2x = 2.

Graph showing a secant line is a straight line that passes through two points on a curve.

The slope of the secant line is equal to the quotient of the change in yy and the change in xx.

Slope of the Secant Line=ΔyΔx=y2y1x2x1=f(x+Δx)f(x)Δx\text{Slope of the Secant Line} = \frac{\Delta{y}}{\Delta{x}} = \frac{y_2 - y_1}{x_2-x_1} = \frac{{f\left( {x + \Delta{x} } \right) - f\left( x\right)}}{\Delta{x} }

Both the numerator and denominator should look very familiar—the equation for the slope of the secant line is equal to the average rate of change over [x,x+Δx][x, x + \Delta{x}]!

So, the slope of the secant line gives us the slope between two points on a curve. But what if we want to find the slope of the curve at a single point, instead of between two points?

To find the slope of a curve at a single point, we must find the slope of the tangent line at that point. A tangent line touches the curve at just one point and also matches the direction of the curve at that point.

Consider again the secant line on the graph, as well as Δx\Delta{x} on the x-axis. Imagine making Δx\Delta{x} smaller and smaller until it nearly reaches zero. Visualize how the secant line would change at each alteration so that it matches the direction of the curve at x=1x = 1 more and more closely each time.

As Δx\Delta{x} approaches zero, our approximation of the slope of the tangent line becomes more and more accurate. So, to find the precise slope of the curve at x=ax = a, we can take the limit of the slope of the secant line through (a,f(a))(a, f(a)) and (a+Δx,f(a+Δx))(a + \Delta{x}, f(a +\Delta{x})) as Δx\Delta{x} approaches zero.

Thus, the slope of the tangent line at the point (a,f(a))(a, f(a)) is given by:

Slope of the Tangent Line=limΔx0f(a+Δx)f(a)Δx\text{Slope of the Tangent Line}= \mathop{\lim }\limits_{\Delta{x} \to 0} \frac{{f\left( {a + \Delta{x} } \right) - f\left( a\right)}}{\Delta{x} }

Notice that this formula is equal to the definition of f(a)f’(a) given earlier, as well as the definition of the instantaneous rate of change given in the previous section!

Let’s try an example. We’ll find the slope of the tangent line of the parabola f(x)=x2+3f(x) = x^2 + 3 at x=1x = 1. The graph of the function is below.

Graph of a function where we'll find the slope of the tangent line of the parabola

Using the definition of the slope of the tangent line, we have:

Slope of the Tangent Line=limΔx0f(a+Δx)f(a)Δx\text{Slope of the Tangent Line}= \mathop{\lim }\limits_{\Delta{x} \to 0} \frac{{f\left( {a + \Delta{x} } \right) - f\left( a\right)}}{\Delta{x} }
=limΔx0f(1+Δx)f(1)Δx= \mathop{\lim }\limits_{\Delta{x} \to 0} \frac{{f\left( {1 + \Delta{x} } \right) - f\left( 1\right)}}{\Delta{x} }
=limΔx0[(1+Δx)2+3][12+3]Δx= \mathop{\lim }\limits_{\Delta{x} \to 0} \frac{[(1 + \Delta{x})^2 + 3] - [1^2 + 3]}{\Delta{x}}
=limΔx0[(Δx)2+2Δx+1+3][1+3]Δx= \mathop{\lim }\limits_{\Delta{x} \to 0} \frac{[(\Delta{x})^2 + 2\Delta{x} + 1 + 3] - [1 + 3]}{\Delta{x}}
=limΔx0(Δx)2+2ΔxΔx= \mathop{\lim }\limits_{\Delta{x} \to 0} \frac{(\Delta{x})^2 + 2\Delta{x} }{\Delta{x}}
=limΔx0Δx(Δx+2)Δx= \mathop{\lim }\limits_{\Delta{x} \to 0} \frac{\Delta{x}(\Delta{x} + 2)}{\Delta{x}}
=limΔx0(Δx+2)= \mathop{\lim }\limits_{\Delta{x} \to 0} (\Delta{x} + 2)
=2=2

So, the slope of the tangent line is 2 at the given point x=1x = 1. After determining the slope of the line, we can find the rest of the equation for the tangent line using the point-slope formula yy1=m(xx1)y - y_1 = m(x - x_1).

Derivatives and Velocity

We can think about velocity as an extension of the “rates of change” interpretation of the derivative. Earlier, we noted that rates of change describe how much one variable changes in response to a change in another variable.

Velocity is one kind of rate of change. Velocity is the rate of change of distance over elapsed time and measures the rate at which an object changes its position regarding time.

Velocity is a vector, meaning that it has both a magnitude and a direction. The magnitude of a velocity vector is speed. The direction of a velocity vector indicates the direction in which an object is being displaced.

For example, define the polynomial s(t)=t2t2s(t) = t^2 - t - 2 as the distance in feet of an object from its starting position, where tt is in seconds. We’ll find the instantaneous velocity of the object at t=2t = 2 seconds.

Using the definition of instantaneous rate of change, we have:

Instantaneous Velocity=limΔt0f(a+Δt)f(a)Δt\text{Instantaneous Velocity}= \mathop{\lim }\limits_{\Delta{t} \to 0} \frac{{f\left( {a + \Delta{t} } \right) - f\left( a\right)}}{\Delta{t} }
=limΔt0f(2+Δt)f(2)Δt=\mathop{\lim }\limits_{\Delta{t} \to 0} \frac{{f\left( {2 + \Delta{t} } \right) - f\left( 2\right)}}{\Delta{t} }
=limΔt0[(2+Δt)2(2+Δt)2][2222]Δt=\mathop{\lim }\limits_{\Delta{t} \to 0} \frac{{[(2 + \Delta{t})^2 - (2 + \Delta{t}) - 2] - [2^2 - 2 - 2]}}{\Delta{t} }
=limΔt0[4+4Δt+(Δt)22Δt2][44]Δt=\mathop{\lim }\limits_{\Delta{t} \to 0} \frac{{[4 + 4\Delta{t} + (\Delta{t})^2 - 2 - \Delta{t} - 2] - [4-4]}}{\Delta{t} }
=limΔt0(Δt)2+3ΔtΔt=\mathop{\lim }\limits_{\Delta{t} \to 0} \frac{{(\Delta{t})^2 + 3\Delta{t}}}{\Delta{t} }
=limΔt0[Δt+3]=\mathop{\lim }\limits_{\Delta{t} \to 0} [\Delta{t} + 3]
=3=3

Thus the velocity of the given function at t=2t = 2 is 3 feet per second.

We’ve been using LaGrange’s notation for the derivative, which is f(x)f’(x). We often read aloud this as “f prime of x.” Another common notation is Leibniz’s notation, which is dydx\frac{dy}{dx}.“ We read this notation aloud as “dy dx.”

For velocity and other physics applications, we sometimes use Newton’s notation instead. Newton’s notation, or dot notation, looks like x˙ \dot{x}, and places a dot over the dependent variable.

Acceleration is the rate of change of velocity. We can think about acceleration as the second derivative of a function. In particular, acceleration is the second derivative of position with respect to time. Second derivatives are an example of higher-level derivatives and can be found by taking the derivative of the first derivative. Higher-level derivatives give us interesting information about the behavior of a function, such as concavity.

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