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AP Calc AB · Mathematics
Derivatives · Derivatives

The Derivative as a Function

On a road trip across Texas, the odometer tells you where you are, and the speedometer tells you how fast the odometer is changing at this very instant. The speedometer reading is a new function built from the odometer function, one value for every moment of the trip. That is the derivative function $f'(x)$. By the end of this page you will know how to build it from a limit, from rules, and from curves that are not even solved for $y$.

The Derivative Defined by a Limit

The slope of a line through two points on a curve is a difference quotient. Slide the second point closer and closer to the first, and the secant line turns into the tangent line.

A parabola with a secant line through points P and Q and the tangent line at P; as Q slides toward P the secant approaches the tangent. P Q secant tangent slope of secant = [f(x+h) − f(x)] / h As Q → P, the secant becomes the tangent.

Why a limit? The difference quotient is an average rate of change over an interval of width $h$. Letting $h\to 0$ squeezes the interval to a single instant and gives the instantaneous rate of change, the slope of the tangent line.

Direct substitution always gives $\tfrac{0}{0}$, so the algebra must cancel the $h$ in the denominator first.

To get the derivative at one point, such as $f'(3)$ for $f(x)=x^2$, you can plug into $f'(x)=2x$, or use the point form directly: $\displaystyle\lim_{x\to3}\frac{x^2-9}{x-3}=\lim_{x\to3}(x+3)=6$.

With a square root, the trick is to multiply by the conjugate.

Practice★☆☆Not started
A skater moves along a track following a trajectory whose position (in feet) is given by the function $s(t) = t^2 + 3t$, where $t$ is the time in seconds. Calculate,…

A skater moves along a track following a trajectory whose position (in feet) is given by the function $s(t) = t^2 + 3t$, where $t$ is the time in seconds. Calculate, using the definition of the derivative as a limit, the instantaneous velocity of the skater at the instant $t = 2$ seconds.

t [s] s(t) [ft] 2 2+h 10 s(2+h) Δt = h Δs s(t) = t² + 3t Secant Tangent (v) h→0
Solve
Practice★★★Not started
Making exclusive use of the definition of the derivative as a limit (difference quotient), calculate the derivative function of $f(x) = \sqrt{2x^2 + 1}$. Subsequently,…

Making exclusive use of the definition of the derivative as a limit (difference quotient), calculate the derivative function of $f(x) = \sqrt{2x^2 + 1}$. Subsequently, use the result obtained to find the equation of the tangent line to the curve at the point with x-coordinate $x = 2$.

x y 0 2 3 P(2, 3) f(x) y = (4/3)x + 1/3 Δx Δy
Solve

Differentiation Rules

Taking a limit every time is slow. These rules, each of which can be proved from the limit definition, do the work for you.

$f(x)$ $f'(x)$
$c$ (constant) $0$
$x^n$ (any real $n$) $nx^{n-1}$
$e^x$ $e^x$
$a^x$ $a^x\ln a$
$\ln x$ $\dfrac1x$
$\sin x$ $\cos x$
$\cos x$ $-\sin x$
$\tan x$ $\sec^2 x$

Constants come out front and sums split: $(cf\pm g)'=cf'\pm g'$. Rewrite roots and reciprocals as powers first: $\sqrt{x}=x^{1/2}$ and $\dfrac1{x^3}=x^{-3}$.

For example, $(x^2\sin x)'=2x\sin x+x^2\cos x$. The graph shows $f$ and its derivative $f'$; notice that $f'$ is zero exactly where $f$ has a horizontal tangent.

Practice★☆☆Not started
Calculate the derivative function of the following elementary functions using the rules of differentiation: a) $f(x) = 5x^4 - 3x^2 + 7x - 1$ b)…

Calculate the derivative function of the following elementary functions using the rules of differentiation:

  1. a) $f(x) = 5x^4 - 3x^2 + 7x - 1$
  1. b) $g(x) = \frac{1}{x^3} + \sqrt{x}$
  1. c) $h(x) = 3e^x - 2\cos(x)$
Solve
Practice★☆☆Not started
Given the function $f(x) = x \cdot \ln(x)$, calculate its derivative $f'(x)$ applying the product rule and determine the value of the slope of the curve at the point…

Given the function $f(x) = x \cdot \ln(x)$, calculate its derivative $f'(x)$ applying the product rule and determine the value of the slope of the curve at the point where $x = e$.

Solve
Practice★☆☆Not started
An access ramp for a skatepark is being designed. The side profile of the first section follows the function $f(x) = \frac{x^2+4}{x+1}$. Calculate the derivative…

An access ramp for a skatepark is being designed. The side profile of the first section follows the function $f(x) = \frac{x^2+4}{x+1}$. Calculate the derivative function $f'(x)$ to find out how the incline of the ramp varies at any point in that section.

Solve

The Chain Rule

Most functions you meet are layers: an outside function applied to an inside function, like $\sin(5x+2)$ or $e^{x^2}$.

A chain of two functions: x goes into the inner function g to give u, which goes into the outer function f to give y. x u = g(x) y = f(u) g f dy/dx = f′(u) · g′(x)

A very useful special case: $\dfrac{d}{dx}\ln u=\dfrac{u'}{u}$.

When a logarithm contains a quotient or product, simplify with log properties first; it is usually much easier.

Practice★☆☆Not started
Apply the chain rule to find the derivative of the following composite functions: a) $f(x) = (3x^2 - 5)^4$ b) $g(x) = \sin(5x + 2)$ c) $h(x) = e^{x^2}$

Apply the chain rule to find the derivative of the following composite functions:

  1. a) $f(x) = (3x^2 - 5)^4$
  1. b) $g(x) = \sin(5x + 2)$
  1. c) $h(x) = e^{x^2}$
Solve
Practice★★☆Not started
Calculate the derivative function of the following composite function, simplifying the result as much as possible:…

Calculate the derivative function of the following composite function, simplifying the result as much as possible:

$f(x) = \ln\left(\frac{1 + \sin(x)}{1 - \sin(x)}\right)$

Solve

Tangent Lines and Normal Lines

Since $f'(a)$ is the slope of the tangent line at $x=a$, you can write that line with point-slope form. The normal line is perpendicular to the tangent line at the same point.

A curve with its tangent line at point P and the normal line perpendicular to it at P. P tangent normal normal slope = −1 / f′(a)
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Find the equation of the tangent line to the graph of the function $f(x) = x^2 - 4x + 1$ at the point with x-coordinate $x = 3$.

Find the equation of the tangent line to the graph of the function $f(x) = x^2 - 4x + 1$ at the point with x-coordinate $x = 3$.

x y 0 f(x) = x² - 4x + 1 V(2, −3) y = 2x - 8 Δx = 1 Δy = 2 3 −2 P(3, −2)
Solve
Practice★★☆Not started
Find the equation of the tangent line to the curve $f(x) = \frac{x^2 - 1}{x + 2}$ at the point where the function intersects the y-axis ($OY$).

Find the equation of the tangent line to the curve $f(x) = \frac{x^2 - 1}{x + 2}$ at the point where the function intersects the y-axis ($OY$).

x = −2 X Y 1 2 3 −1 1 2 3 4 −1 0 f(x) Tangent line P(0, −0.5) y-intercept
Solve

Higher-Order Derivatives and Patterns

You can differentiate a derivative. The result is the second derivative, and so on.

For functions like $\dfrac1x$, $e^{2x}$, or $\ln x$, the derivatives follow a pattern, and you can write $f^{(n)}(x)$ without differentiating $n$ times.

The first derivatives of 1/x written as powers, with the general pattern for the nth derivative. f(x) = x⁻¹ f′(x) = −1 · x⁻² f″(x) = 2 · x⁻³ f‴(x) = −6 · x⁻⁴ f⁽ⁿ⁾(x) = (−1)ⁿ · n! · x⁻⁽ⁿ⁺¹⁾

The chain rule adds one factor each time. For a linear inside, $\dfrac{d^n}{dx^n}\,(ax+b)^{-1}=\dfrac{(-1)^n\,n!\,a^n}{(ax+b)^{n+1}}$: every derivative picks up one more factor $a$.

Rational functions with a factored denominator become easy after partial fractions.

How do you know a pattern is true for every $n$, not only for the first few? Use mathematical induction on the pattern for $(x-c)^{-1}$:

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Calculate the second derivative of the function $f(x) = x^3 - 6x^2 + 9x - 2$. For what value of $x$ is this second derivative zero? (This point is important for studying…

Calculate the second derivative of the function $f(x) = x^3 - 6x^2 + 9x - 2$. For what value of $x$ is this second derivative zero? (This point is important for studying concavity, which you will see later.)

Solve
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Given the function $f(x) = \ln(2x - 3)$, find the general expression for its n-th derivative $f^{(n)}(x)$ and use it to calculate the value of the fourth derivative at…

Given the function $f(x) = \ln(2x - 3)$, find the general expression for its n-th derivative $f^{(n)}(x)$ and use it to calculate the value of the fourth derivative at the point $x = 2$.

Solve
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Consider the function $f(x) = \frac{1}{x^2 - 3x + 2}$. a) Decompose the function into partial fractions. b) Using the previous decomposition, find a general formula for…

Consider the function $f(x) = \frac{1}{x^2 - 3x + 2}$.

  1. a) Decompose the function into partial fractions.
  2. b) Using the previous decomposition, find a general formula for the $n$-th derivative, $f^{(n)}(x)$, for any $n \in \mathbb{N}$.
  3. c) Prove the validity of your formula for every $n \in \mathbb{N}$ by mathematical induction, explicitly checking the first three orders of differentiation.
Solve

Velocity and Acceleration

If $s(t)$ is the position of an object moving along a line, then:

  • velocity $v(t)=s'(t)$ (negative means moving in the negative direction),

  • acceleration $a(t)=v'(t)=s''(t)$,

  • speed $=|v(t)|$.

The object is momentarily at rest when $v(t)=0$. It changes direction only when $v$ changes sign. The speed is increasing when $v$ and $a$ have the same sign and decreasing when they have opposite signs.

In the graph above, $x$ plays the role of $t$: the first curve is $s(t)$ and the second is $v(t)$.

With exponentials, factor out the exponential before analyzing signs.

Practice★★☆Not started
A surveillance drone follows a trajectory whose height in feet is given by the function $h(t) = \frac{1}{3}t^3 - 4t^2 + 11t + 5$, where $t$ is the time in seconds (…

A surveillance drone follows a trajectory whose height in feet is given by the function $h(t) = \frac{1}{3}t^3 - 4t^2 + 11t + 5$, where $t$ is the time in seconds ($0 \leq t \leq 10$).

  • a) Calculate the velocity function $v(t)$ and the acceleration function $a(t)$.

  • b) At what instants of time is the drone momentarily stationary (velocity equal to zero)?

Solve
Practice★★★Not started
A surveillance drone follows a flight path along a straight axis. Its position (in feet) with respect to the control center is given by the function…

A surveillance drone follows a flight path along a straight axis. Its position (in feet) with respect to the control center is given by the function $s(t) = t^2 \cdot e^{-t/4}$, where $t$ is the time in minutes since take-off ($t \ge 0$).

  1. a) Determine the velocity function $v(t)$ and the acceleration function $a(t)$ of the drone.
  2. b) At what instant does the drone change its direction of motion?
  3. c) Calculate the exact moment when the acceleration is zero. What physical interpretation does this point have with respect to the drone's velocity?
4 6 10 12 16 18 20 2 4 6 8 10 t (min) s(t) (ft) s(t) = t²·e⁻ᵗ/⁴ 8 8.66 v(t) = 0 Change of direction 2.34 a(t) = 0 Max velocity 13.65 a(t) = 0 Min velocity
Solve

Reading $f'$ and $f''$

The derivative tells the story of the original function.

The graph above shows $f(x)=x^3-3x$ and $f'(x)=3x^2-3$. Where does $f'$ cross the $x$-axis? Where does $f'$ have its minimum?

A different kind of problem gives you conditions about $f$, $f'$, and $f''$ and asks you to find the function. Translate each condition into an equation.

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The following graph corresponds to the derivative function, $f'(x)$, of a function $f(x)$ which is continuous throughout its domain $\mathbb{R}$. Based on the graph of…

The following graph corresponds to the derivative function, $f'(x)$, of a function $f(x)$ which is continuous throughout its domain $\mathbb{R}$.

x y −2 2 2 f'(x) - ↘ + ↗ - ↘

Based on the graph of $f'(x)$, answer the following questions about the original function $f(x)$, giving reasons:

  1. a) State the intervals of increase and decrease of $f(x)$.
  2. b) Determine the x-coordinates of the relative extrema of $f(x)$ and classify them as maxima or minima.
  3. c) At what value of $x$ does $f(x)$ have a point of inflection? Justify your answer based on the relationship between $f''(x)$ and the slope of $f'(x)$.
Solve
Practice★★★Not started
Consider the function $f(x) = x^4 + ax^3 + bx^2 + cx + d$. Determine the coefficients $a, b, c$ and $d$ knowing that the following conditions are simultaneously met: 1.…

Consider the function $f(x) = x^4 + ax^3 + bx^2 + cx + d$. Determine the coefficients $a, b, c$ and $d$ knowing that the following conditions are simultaneously met:

  1. The graph of $f(x)$ passes through the origin.

  2. The function has a point of inflection at $x = 1$.

  3. The tangent line to the curve at the point with x-coordinate $x = 0$ is parallel to the line $y = -2x + 5$.

  4. The value of the second derivative at $x = 0$ is $f''(0) = 6$.

Solve

Differentiability and Continuity

A function is differentiable at $a$ if $f'(a)$ exists, that is, if the two one-sided limits of the difference quotient exist and are equal:

$$\lim_{h\to0^-}\frac{f(a+h)-f(a)}{h}=\lim_{h\to0^+}\frac{f(a+h)-f(a)}{h}.$$

So a function can fail to be differentiable at a point in four common ways.

Four graphs where a derivative fails to exist: a corner, a cusp, a vertical tangent, and a jump discontinuity. corner cusp vertical tangent jump

At a corner, the left and right slopes are different numbers. At a cusp or vertical tangent, the slope is $\pm\infty$. At a jump, the function is not even continuous. For example, $f(x)=\sqrt[3]{x}$ at $0$ has $\dfrac{f(h)-f(0)}{h}=h^{-2/3}\to\infty$, a vertical tangent.

Absolute value always makes corners at its zeros. Here is a shortcut with a reason.

For example, take $f(x)=|x^2-4|$ at $x=2$. Here $g(x)=x^2-4$ has $g(2)=0$ and $g'(2)=4\neq0$, so the one-sided derivatives are $-4$ and $4$, and there is a corner at $x=2$ (but $f$ is continuous there).

Practice★☆☆Not started
Determine the differentiability of the following function at the point $x = 2$, first checking if it is continuous at that point: $$f(x) = \begin{cases} x^2 & \text{if }…

Determine the differentiability of the following function at the point $x = 2$, first checking if it is continuous at that point:

$$f(x) = \begin{cases} x^2 & \text{if } x \leq 2 \\ 4x - 4 & \text{if } x > 2 \end{cases}$$ x y 0 x ≤ 2 x > 2 2 4 y = x² y = 4x - 4 P(2,4)
Solve
Practice★★☆Not started
Study the differentiability of the function $f(x) = |x^2 - 4x + 3|$ at the points $x = 1$ and $x = 3$. State whether one-sided derivatives exist at these points and…

Study the differentiability of the function $f(x) = |x^2 - 4x + 3|$ at the points $x = 1$ and $x = 3$. State whether one-sided derivatives exist at these points and whether the function is continuous.

X Y 0 1 2 3 4 1 2 3 y = x² - 4x + 3 f(x) f'(1⁻) = −2 f'(1⁺) = 2 f'(3⁻) = −2 f'(3⁺) = 2 x=1 x=3
Solve

Making a Piecewise Function Differentiable

When a piecewise function has unknown constants, you choose them so the two pieces join smoothly. A smooth join needs two things at the junction $x=c$.

Two pieces joined with a corner on the left and joined smoothly on the right. corner: slopes differ smooth: slopes match

Once the one-sided derivatives agree at $c$, you can ask a follow-up: is $f'$ itself continuous at $c$? If both pieces are built from ordinary functions (polynomials, exponentials, logs on their domains), their derivative formulas are continuous. So the left limit and right limit of $f'$ at $c$ equal the matching one-sided derivatives, and $f'$ is continuous at $c$. In the example, $4x+3$ and $3x^2+4$ both equal $7$ at $x=1$.

Practice★★☆Not started
Determine the values of the parameters $a$ and $b$ so that the following function is differentiable throughout its domain, paying special attention to the joining point…

Determine the values of the parameters $a$ and $b$ so that the following function is differentiable throughout its domain, paying special attention to the joining point $x = 0$:

$f(x) = \begin{cases} e^{ax} & \text{if } x \leq 0 \\ x^2 + bx + 1 & \text{if } x > 0 \end{cases}$

Solve
Practice★★★Not started
Consider the piecewise function $f: \mathbb{R} \to \mathbb{R}$ given by the following expression: $$f(x) = \begin{cases} a \cdot e^{x-2} + b \cdot x & \text{if } x \le 2…

Consider the piecewise function $f: \mathbb{R} \to \mathbb{R}$ given by the following expression:

$$f(x) = \begin{cases} a \cdot e^{x-2} + b \cdot x & \text{if } x \le 2 \\ \ln(x^2 - 3) + x^2 & \text{if } x > 2 \end{cases}$$

Determine the values of the parameters $a, b \in \mathbb{R}$ so that the function is differentiable throughout its domain. Once found, discuss whether the derivative function $f'(x)$ is, in turn, continuous at $x = 2$.

Solve

Implicit Differentiation

Some curves are not written as $y=f(x)$, such as the circle $x^2+y^2=25$ or the Folium of Descartes $x^3+y^3-3xy=0$. They still have a tangent slope at each point. Near a point, $y$ is secretly a function of $x$, so differentiate both sides with respect to $x$ and treat $y$ as $y(x)$.

The circle x squared plus y squared equals 25 with the tangent line at the point (3, 4), which has slope negative three fourths. (3, 4) tangent slope = −3/4

For the circle: $2x+2y\,y'=0$, so $y'=-\dfrac{x}{y}$. At $(3,4)$ the slope is $-\dfrac34$, which matches the figure (the tangent is perpendicular to the radius).

Notice the pattern: horizontal tangent means numerator $=0$; vertical tangent means denominator $=0$ (provided the numerator is not also zero).

Practice★★☆Not started
Given the equation of the implicitly defined curve $x^3 + y^3 - 3xy = 0$ (known as the Folium of Descartes), calculate the value of the slope of the tangent line to the…

Given the equation of the implicitly defined curve $x^3 + y^3 - 3xy = 0$ (known as the Folium of Descartes), calculate the value of the slope of the tangent line to the curve at the point $( \frac{3}{2}, \frac{3}{2} )$.

O x y 3/2 3/2 P(3/2, 3/2) m = ? x³ + y³ - 3xy = 0
Solve
Practice★★★Not started
Given the curve defined by the implicit equation $x^3 + y^3 - 3xy = 0$ (known as the Folium of Descartes): a) Obtain the expression for the derivative $y'$ in terms of…

Given the curve defined by the implicit equation $x^3 + y^3 - 3xy = 0$ (known as the Folium of Descartes):

  1. a) Obtain the expression for the derivative $y'$ in terms of $x$ and $y$ using implicit differentiation.
  2. b) Find the points on the curve where the tangent line is horizontal.
  3. c) Determine the equation of the normal line to the curve at the point $( \frac{3}{2}, \frac{3}{2} )$.
Solve

Logarithmic Differentiation

How do you differentiate $x^x$? The power rule needs a constant exponent, and the exponential rule needs a constant base. When both the base and the exponent vary, take the natural log first.

Two practical details:

  • Domain. For a real function $u^v$ with a variable exponent, you need $u>0$ (and $v$ defined). Find the domain of $f$ by requiring that, and the domain of $f'$ from the final formula.

  • Zero slope. Since $y=u^v>0$, the factor outside the bracket is never zero. So $y'=0$ exactly when the bracket equals $0$. To show that the bracket has a zero, check its sign at two test points and use the Intermediate Value Theorem.

Practice★★☆Not started
Use the technique of logarithmic differentiation to find the derivative of the following power-exponential function: $f(x) = (x^2 + 1)^{\cos(x)}$

Use the technique of logarithmic differentiation to find the derivative of the following power-exponential function:

$f(x) = (x^2 + 1)^{\cos(x)}$

Solve
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Use the technique of logarithmic differentiation to calculate the derivative function of: $$f(x) = \left( \cos(x) \right)^{\ln(x)}$$ State, with justification, the…

Use the technique of logarithmic differentiation to calculate the derivative function of:

$$f(x) = \left( \cos(x) \right)^{\ln(x)}$$

State, with justification, the domain of definition of the function $f(x)$ and that of its derivative function $f'(x)$. Is there any point in the interval $(0, \pi/2)$ where the slope of the tangent line is zero?

Solve

Summary

Back to the start: the speedometer is the derivative of the odometer. $f'(x)$ is a new function that gives the slope, and the instantaneous rate of change, at every $x$. You can build it from the limit definition or from the rules.

See all the exercises in this section