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Introduction
1. Limits
2. Derivative basics
2.1 The derivative
2.2 Tangent lines & slopes
2.3 Power rule
2.4 Product & quotient rules
2.5 Special derivatives
2.6 Differentiability & continuity
3. Advanced differentiation
4. Contextual uses
5. Analytical uses
6. Integration
7. Differential equations
8. Applications of integrals
Wrapping up
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2.5 Special derivatives
Achievable AP Calculus AB
2. Derivative basics
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Special derivatives

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What you’ll learn

  • Derivatives of exponential, logarithmic, and trigonometric functions

So far, we’ve worked with derivatives of polynomials and rational functions. This section introduces the derivatives of non-algebraic functions.

Exponential functions

Exponential functions model rapid growth and decay and show up in population models, radioactive decay, and compound interest.

The most important exponential function is the natural exponential function ex, whose rate of change is equal to itself.

Derivative of the natural exponential function:

dxd​(ex)=ex

More generally, for an exponential function of the form f(x)=bx (where b>0 and b=1), the derivative is

Derivative of a general exponential function:

dxd​(bx)=bx⋅ln(b)

In other words, rewrite the original function and multiply by ln(b).

Sidenote
Power rule vs. derivatives of exponentials

Be careful not to mix these up. Here, x is a variable and b is a constant.

  1. Power rule if f(x)=xb:

    • e.g. dxd​(x3)=3x2
  2. Rule for exponentials if f(x)=bx:

    • e.g. dxd​(3x)=3xln(3)

Examples

  1. At what value(s) of x does the graph of

f(x)=x2ex

have horizontal tangent lines?

(spoiler)

Horizontal tangent lines occur where f′(x)=0.

Differentiating f using the product rule,

f′(x)​=x2⋅dxd​(ex)+ex⋅dxd​(x2)=x2ex+2xex​

Set equal to 0 and solve.

x2ex+2xex=0

Avoid dividing both sides by xex right away, since that can remove solutions. Instead, factor:

xex(x+2)=0

Because ex=0 for any real x, the solutions come from x=0 or x=−2. At these values of x, the tangent lines to f have a slope of 0.

  1. At what value of x does the graph of

f(x)=7xx​

have a horizontal tangent line?

(spoiler)

A horizontal tangent line occurs where f′(x)=0. Differentiating using the quotient rule,

f′(x)​=(7x)27x⋅dxd​(x)−x⋅dxd​(7x)​=72x7x−x(7x⋅ln(7))​=72x7x(1−xln(7))​=7x1−xln(7)​​

This equals 0 when the numerator equals 0. Solving,

1−xln(7)−xln(7)x​=0=−1=ln(7)1​​

Logarithmic functions

Derivative of the natural logarithmic function:

dxd​(lnx)=x1​

More generally, any logarithmic function f(x)=logb​x, where the base b is a positive number and b=1, can be rewritten using the change-of-base rule:

f(x)​=ln(b)lnx​=ln(b)1​⋅lnx​

Then, since lnb1​ is a constant, it stays out front when differentiating.

Derivative of a general logarithmic function:

dxd​logb​x=xln(b)1​

Sidenote
The domain of the derivative

A derivative formula is only valid where the original function is defined. For logarithms, the argument must be >0.

Examples

  1. Determine the point at which the tangent line to f(x)=xln(x) is parallel to the line y=3x−1.
(spoiler)

Parallel lines have the same slope, so find where f′(x)=3.

Differentiating using the product rule,

f′(x)​=x⋅x1​+(lnx)(1)=1+lnx​

Set equal to 3 and solve:

1+lnxlnxx​=3=2=e2​

Then the corresponding y-coordinate is

f(e2)=e2ln(e2)=2e2

At the point (e2,2e2), the tangent line to f is parallel to y=3x−1.

  1. Let f(x)=logk​x, where k is a positive real number. Find the value of k such that the slope of the tangent line at x=1 equals 2.
(spoiler)

Differentiating,

f′(x)=xln(k)1​

“The slope of the tangent line at x=1 equals 2” means that f′(1)=2. So substitute 1 for x and set f′(x) equal to 2 to solve for k:

22ln(k)ln(k)k​=1⋅ln(k)1​=1=21​=e1/2​

Trigonometric functions

Begin with the basic trig functions sinx and cosx.

dxd​(sinx)dxd​(cosx)​=cosx=−sinx​

From there, the other trig functions can be differentiated by first rewriting in terms of sinx and cosx.

For example, f(x)=tanx can be rewritten as the quotient

tanx=cosxsinx​

Then differentiating using the quotient rule,

f′(x)​=cos2(x)cosx⋅cosx−sinx⋅(−sinx)​=cos2xcos2x+sin2x​​

Using the Pythagorean identity sin2x+cos2x=1,

f′(x)=cos2x1​=sec2x​

The same can be done to f(x)=cotx.

Find the derivative of f(x)=cotx by rewriting it as a quotient.

(spoiler)

Using the quotient rule,

f′(x)​=sin2(x)sinx⋅(−sinx)−cosx⋅cosx​=sin2x−sin2x−cos2x​=sin2x−(sin2x+cos2x)​=sin2x−1​=−csc2x​​

Now do the same for the remaining two reciprocal functions:

Find the derivatives of

a) f(x)=secx

b) f(x)=cscx

Solutions

a) Differentiate f(x)=secx

(spoiler)

Since secx=cosx1​,

f′(x)​=cos2x(cosx)(0)−1(−sinx)​=cos2xsinx​=cosx1​⋅cosxsinx​=secxtanx​​

b) Differentiate f(x)=cscx

(spoiler)

Since csc(x)=sinx1​,

f′(x)​=sin2x(sinx)(0)−(1)(cosx)​=sin2x−cosx​=sinx−1​⋅sinxcosx​=−cscxcotx​​

Memorizing the trig derivatives saves time, but it’s also useful to know how to derive them from the derivatives of sinx and cosx.

Higher-order derivatives

Because a derivative is just another function, you can differentiate it again to find a second derivative, and repeat the process indefinitely through repeated differentiation. For example, if f(x)=x2, then its first 3 derivatives are

f′(x)f′′(x)f′′′(x)​=2x=2=0​

While polynomials eventually differentiate down to zero, other functions exhibit different patterns. For example, trigonometric functions like sinx and cosx repeat in a four-step cycle, allowing them to be differentiated indefinitely.

These higher-order derivatives are often denoted f(n)(x). where n indicates the number of times the function has been differentiated.

Note: In AP Calculus AB, applications are generally limited to the 1st and 2nd derivatives.

Exponential functions

  • dxd​(ex)=ex

  • dxd​(bx)=bxln(b)

  • Distinguish between power rule (xb) and exponential rule (bx)

Logarithmic functions

  • dxd​(lnx)=x1​
  • dxd​(logb​x)=xln(b)1​
  • Domain: only defined for x>0

Trigonometric functions

  • dxd​(sinx)=cosx

  • dxd​(cosx)=−sinx

  • dxd​(tanx)=sec2x

  • dxd​(cotx)=−csc2x

  • dxd​(secx)=secxtanx

  • dxd​(cscx)=−cscxcotx

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Special derivatives

What you’ll learn

  • Derivatives of exponential, logarithmic, and trigonometric functions

So far, we’ve worked with derivatives of polynomials and rational functions. This section introduces the derivatives of non-algebraic functions.

Exponential functions

Exponential functions model rapid growth and decay and show up in population models, radioactive decay, and compound interest.

The most important exponential function is the natural exponential function ex, whose rate of change is equal to itself.

Derivative of the natural exponential function:

dxd​(ex)=ex

More generally, for an exponential function of the form f(x)=bx (where b>0 and b=1), the derivative is

Derivative of a general exponential function:

dxd​(bx)=bx⋅ln(b)

In other words, rewrite the original function and multiply by ln(b).

Sidenote
Power rule vs. derivatives of exponentials

Be careful not to mix these up. Here, x is a variable and b is a constant.

  1. Power rule if f(x)=xb:

    • e.g. dxd​(x3)=3x2
  2. Rule for exponentials if f(x)=bx:

    • e.g. dxd​(3x)=3xln(3)

Examples

  1. At what value(s) of x does the graph of

f(x)=x2ex

have horizontal tangent lines?

(spoiler)

Horizontal tangent lines occur where f′(x)=0.

Differentiating f using the product rule,

f′(x)​=x2⋅dxd​(ex)+ex⋅dxd​(x2)=x2ex+2xex​

Set equal to 0 and solve.

x2ex+2xex=0

Avoid dividing both sides by xex right away, since that can remove solutions. Instead, factor:

xex(x+2)=0

Because ex=0 for any real x, the solutions come from x=0 or x=−2. At these values of x, the tangent lines to f have a slope of 0.

  1. At what value of x does the graph of

f(x)=7xx​

have a horizontal tangent line?

(spoiler)

A horizontal tangent line occurs where f′(x)=0. Differentiating using the quotient rule,

f′(x)​=(7x)27x⋅dxd​(x)−x⋅dxd​(7x)​=72x7x−x(7x⋅ln(7))​=72x7x(1−xln(7))​=7x1−xln(7)​​

This equals 0 when the numerator equals 0. Solving,

1−xln(7)−xln(7)x​=0=−1=ln(7)1​​

Logarithmic functions

Derivative of the natural logarithmic function:

dxd​(lnx)=x1​

More generally, any logarithmic function f(x)=logb​x, where the base b is a positive number and b=1, can be rewritten using the change-of-base rule:

f(x)​=ln(b)lnx​=ln(b)1​⋅lnx​

Then, since lnb1​ is a constant, it stays out front when differentiating.

Derivative of a general logarithmic function:

dxd​logb​x=xln(b)1​

Sidenote
The domain of the derivative

A derivative formula is only valid where the original function is defined. For logarithms, the argument must be >0.

Examples

  1. Determine the point at which the tangent line to f(x)=xln(x) is parallel to the line y=3x−1.
(spoiler)

Parallel lines have the same slope, so find where f′(x)=3.

Differentiating using the product rule,

f′(x)​=x⋅x1​+(lnx)(1)=1+lnx​

Set equal to 3 and solve:

1+lnxlnxx​=3=2=e2​

Then the corresponding y-coordinate is

f(e2)=e2ln(e2)=2e2

At the point (e2,2e2), the tangent line to f is parallel to y=3x−1.

  1. Let f(x)=logk​x, where k is a positive real number. Find the value of k such that the slope of the tangent line at x=1 equals 2.
(spoiler)

Differentiating,

f′(x)=xln(k)1​

“The slope of the tangent line at x=1 equals 2” means that f′(1)=2. So substitute 1 for x and set f′(x) equal to 2 to solve for k:

22ln(k)ln(k)k​=1⋅ln(k)1​=1=21​=e1/2​

Trigonometric functions

Begin with the basic trig functions sinx and cosx.

dxd​(sinx)dxd​(cosx)​=cosx=−sinx​

From there, the other trig functions can be differentiated by first rewriting in terms of sinx and cosx.

For example, f(x)=tanx can be rewritten as the quotient

tanx=cosxsinx​

Then differentiating using the quotient rule,

f′(x)​=cos2(x)cosx⋅cosx−sinx⋅(−sinx)​=cos2xcos2x+sin2x​​

Using the Pythagorean identity sin2x+cos2x=1,

f′(x)=cos2x1​=sec2x​

The same can be done to f(x)=cotx.

Find the derivative of f(x)=cotx by rewriting it as a quotient.

(spoiler)

Using the quotient rule,

f′(x)​=sin2(x)sinx⋅(−sinx)−cosx⋅cosx​=sin2x−sin2x−cos2x​=sin2x−(sin2x+cos2x)​=sin2x−1​=−csc2x​​

Now do the same for the remaining two reciprocal functions:

Find the derivatives of

a) f(x)=secx

b) f(x)=cscx

Solutions

a) Differentiate f(x)=secx

(spoiler)

Since secx=cosx1​,

f′(x)​=cos2x(cosx)(0)−1(−sinx)​=cos2xsinx​=cosx1​⋅cosxsinx​=secxtanx​​

b) Differentiate f(x)=cscx

(spoiler)

Since csc(x)=sinx1​,

f′(x)​=sin2x(sinx)(0)−(1)(cosx)​=sin2x−cosx​=sinx−1​⋅sinxcosx​=−cscxcotx​​

Memorizing the trig derivatives saves time, but it’s also useful to know how to derive them from the derivatives of sinx and cosx.

Higher-order derivatives

Because a derivative is just another function, you can differentiate it again to find a second derivative, and repeat the process indefinitely through repeated differentiation. For example, if f(x)=x2, then its first 3 derivatives are

f′(x)f′′(x)f′′′(x)​=2x=2=0​

While polynomials eventually differentiate down to zero, other functions exhibit different patterns. For example, trigonometric functions like sinx and cosx repeat in a four-step cycle, allowing them to be differentiated indefinitely.

These higher-order derivatives are often denoted f(n)(x). where n indicates the number of times the function has been differentiated.

Note: In AP Calculus AB, applications are generally limited to the 1st and 2nd derivatives.

Key points

Exponential functions

  • dxd​(ex)=ex

  • dxd​(bx)=bxln(b)

  • Distinguish between power rule (xb) and exponential rule (bx)

Logarithmic functions

  • dxd​(lnx)=x1​
  • dxd​(logb​x)=xln(b)1​
  • Domain: only defined for x>0

Trigonometric functions

  • dxd​(sinx)=cosx

  • dxd​(cosx)=−sinx

  • dxd​(tanx)=sec2x

  • dxd​(cotx)=−csc2x

  • dxd​(secx)=secxtanx

  • dxd​(cscx)=−cscxcotx

More from Derivative basics

  • Tangent lines & slopes
  • Power rule
  • Product & quotient rules
  • Differentiability & continuity