AP Calculus AB
Class Notes
3.7 Implicit Differentiation
pp 157 - 164
Objectives:
Take the derivative implicitly. Use all previous rules (sum & difference rule, constant multiple rule, power
rule, product rule, quotient rule, chain rule, derivatives of trig functions) in your practice of implicit
differentiation.
Outline
Determine the derivative of a relation by
differentiating implicitly.
Find the equations of lines tangent to and normal to
an implicit curve.
Find higher order derivatives using implicit
differentiation.
Vocabulary
implicit differentiation
normal to a surface
Power Rule for Rational Powers of x
Omissions
none
Day 1
Introduction to Implicit Differentiation
Implicit differentiation is a trick that allows us to differentiate relations. As you may recall, when we have
graph that violates the vertical line test, it is not a function, but rather it is a relation. An example of a relation
is
𝑥 3 + 𝑦 3 − 9𝑥𝑦 = 0
The graph of this relation looks like the following:
y
As you can see, this relation clearly
violates the vertical line test; it is not a function. Furthermore, it cannot
easily be broken down to two functions. Nonetheless, we can see that the relation has a tangent everywhere
except where the tangent is horizontal.

Implicit differentiation allows us to find the derivative
without putting the relation into function form.


How do we differentiate a relation that is dependent
on x and y? It’s kind of a trick. We apply the chain
rule to a function of y. For example:



x













𝑑
𝑑𝑦
𝑦=
𝑑𝑥
𝑑𝑥

I know that that sounds simplistic. It is.


Let’s try the simple differentiation:
𝑑 3
𝑦
𝑑𝑥

To perform this, we treat y as a function, and use the
chain rule: we differentiate the outside (the cube), leave the inside alone, and multiply by the derivative of the
inside (y). The derivative of the function 𝑦 3 is
𝑑 3
𝑑𝑦
𝑦 𝑏𝑒𝑐𝑜𝑚𝑒𝑠 3𝑦 2
𝑑𝑥
𝑑𝑥
Class Notes 3.4
Page 1 of 19
Let’s practice taking the derivative of several different terms:
Term
𝒅
𝒅𝒙
y
𝑦2
√𝑦
cos 𝑦
csc 𝑦
𝑥𝑦
𝑥2𝑦
𝑥𝑦 2
𝑥3𝑦2
𝑥
𝑦
𝑦
𝑥
Class Notes 3.4
Page 2 of 19


Find
𝑑𝑦
𝑑𝑥
for the relation 𝑥 3 + 𝑦 3 − 9𝑥𝑦 = 0





x















End of 3.7 Day 1
3.7 Day 2 Finding the Second Derivative Implicitly
To find the second derivative of a function implicitly, you find the first derivative, and then the second. When
you take the second derivative, you’ll have some terms expressed as 𝑦′. Substitute-in the first derivative at
these places.
𝑑2 𝑦
Example: Find 𝑑𝑥 2 of the following: 3𝑥 2 − 4𝑦 2 = 7
𝑑
[3𝑥 2 − 4𝑦 2 = 7]
𝑑𝑥
6𝑥 − 8𝑦𝑦 ′ = 0
−8𝑦𝑦 ′ = −6𝑥
3𝑥
𝑦′ =
4𝑦
𝑑
Now, take the 𝑑𝑥 of each side again:
𝑑 ′ 3𝑥
[𝑦 =
]
𝑑𝑥
4𝑦
3 𝑦 ∙ 1 − 𝑥𝑦′
𝑦 ′′ =
4
𝑦2
Now, subsititue the first derivative into the second derivative’s equation above.
3𝑥
𝑦 − 𝑥 [4 𝑦 ]
3
𝑦 ′′ =
4
𝑦2
Simplify.
3𝑥2
3𝑥
′′
𝑦 =
3 𝑦−𝑥[4𝑦]
4
𝑦2
=
4𝑦2 3𝑥2
3 𝑦− 4𝑦 3 4𝑦 − 4𝑦
=4 𝑦 2
4 𝑦2
2
2
=
3
4
4𝑦2 −3𝑥2
4𝑦
𝑦2
3 4𝑦 2 −3𝑥 2
4𝑦 3
=4
Now, recall that the original equation was 3𝑥 − 4𝑦 = 7. We can therefore, subsitute this into the numerator
and get:
3 4𝑦 2 − 3𝑥 2 3 −1(3𝑥 2 − 4𝑦 2 ) 3 −7
−21
=
=
=
3
3
3
4
4𝑦
4
4𝑦
4 4𝑦
16𝑦 3
Class Notes 3.4
Page 3 of 19


Let’s try several more:
𝑑2 𝑦
Find 𝑑𝑥 2
𝑥3 + 𝑦3 = 1
𝑥3𝑦3 − 4 = 0
𝑥𝑐𝑜𝑠 𝑦 = 𝑦
End of 3.7 Day 2
Class Notes 3.4
Page 4 of 19
3.7 Day 3
Applications of Implicit Differentiation
The main application of implicit differentiation is finding the tangent and normal lines, and variations on these.
Here’s an example:
Find the point(s) (if any) of horizontal tangent lines: 𝑥 2 + 𝑥𝑦 + 𝑦 2 = 6 and write the equation of the tangent
lines at thos points.
Find 𝑦′ if 𝑥 3 + 𝑦 3 = 6𝑥𝑦.
Find the tangen to the folium of Descartes 𝑥 3 + 𝑦 3 = 6𝑥𝑦 at the point (3,3).
At what point in the first quadrant is the tangen t line horizontal.
Class Notes 3.4
Page 5 of 19
Use implicit differentiation to find an equation of the tangent line to the curve at the given point.
𝜋 𝜋
𝑦 sin 2𝑥 = 𝑥 cos 2𝑦, ( , )
2 4
At what point(s) is the tangent line to the curve 𝑦 2 = 2𝑥 3 perpendicular to the line 4𝑥 − 3𝑦 + 1 = 0 ?
End of 3.7 Day 3
Class Notes 3.4
Page 6 of 19
AP Calculus AB
Class Notes
3.8 Derivatives of Inverse Functions
pp 165 - 171
Objectives:
In this section we’ll find the value of the derivative of the inverse of a function without finding the analytical
derivative of the inverse function or even knowing the inverse function itself. We’ll set this up graphically, and
then memorize the analytical formula. Next, we’ll memorize and apply the derivatives of the three main trig
functions: arcsin x, arccos x, and arctan x.
Outline
Draw the graphs that illustrate the relationship
between the slope of the tangent of a function with
the slope of its inverse.
Use formulae to find the derivative of arcsin, arccos,
and arctan.
Vocabulary
implicit differentiation
normal to a surface
Power Rule for Rational Powers of x
Omissions
none
3.8 Day 1 The Derivative of an Inverse Function
We learned in earlier classes that we can find the inverse of many functions by reversing the x and y variables
and solving for y. But in many cases, you cannot analytically find the inverse function. For example, the
function:
𝑦 = 3𝑥 2 − 2𝑥 + 5
does not have an inverse as a function of x. However, we can use what we know about inverse functions to
find the value of the derivative without knowing the inverse function.
We’ll start with a property of the behavior of the derivative of the inverse of a function.
Graphs of inverses have reciprocal slopes.
𝑓(𝑥) = 𝑥 2
𝑓(𝑥) = 𝑥 3
3
𝑓 −1 (𝑥) = √𝑥
𝑓 −1 (𝑥) = √𝑥
𝑓(𝑥) = 𝑒 𝑥
𝑓
= ln 𝑥
−1 (𝑥)
𝑓(𝑥) = cos 𝑥
𝑓 −1 (𝑥) = cos−1 𝑥
In the next example, we’ll see how the slopes of inverse functions are reciprocals.
Class Notes 3.4
Page 7 of 19
For the following functions and inverses, find the domains of 𝑓 and 𝑓 −1 . Find the ranges. Graph each. That the
slopes of the graphs of 𝑓 and 𝑓 −1 are reciprocals at the given points.
1 1
𝑓(𝑥) = 3 − 4𝑥 (2 , 8)
𝑓 −1 (𝑥) =
3−𝑥
4
1 1
(8 , 2 )
𝑓 domain__________range______________
𝑓
−1
domain__________range______________
𝑓(𝑥) = √𝑥 − 4 (5,1)
𝑓 −1 (𝑥) = 𝑥 2 + 4
(1,5)
𝑓 domain__________range______________
𝑓 −1 domain__________range______________
𝑓(5) =________________________
1
2
𝑓 ( ) =________________________
𝑓 −1 (1) =________________________
1
𝑓 −1 (8) =________________________
Here’s a closer look at the phenomenon that slopes of inverses are reciprocal.
For the sake of this discussion, let’s start with a generic function, 𝑓(𝑥) and it’s inverse, 𝑔(𝑥). Follow the steps.
 Draw the arbitrary 𝑓(𝑥).
 Reflect 𝑓(𝑥) in the line 𝑦 = 𝑥 to find its inverse.
 Label a point on 𝑓 with the point (𝑥, 𝑓(𝑥)).
 Draw the tangent to f through (𝑥, 𝑓(𝑥)). The slope of this
line is m.
 Now, find the inverse reflection, (𝑓(𝑥), 𝑥) on the function
g.
 Draw the tangent to g through (𝑓(𝑥), 𝑥). The slope of this
1
line is 𝑚
 Since the two derivatives are inverses, we can say that the
1
derivative of 𝑓(𝑥) = ′
. This you must memorize.
𝑔 (𝑓(𝑥))
Class Notes 3.4
Page 8 of 19
Let’s do the same thing again, this time with g being our the function g.
Type equation here.
𝑓 ′ (𝑥) =
𝑔′ (𝑥) =
1
𝑓 ′ (𝑔(𝑥))
1
𝑔′ (𝑓(𝑥))
Memorize This!
Now, let’s use it. Recall the function
𝑓(𝑥) = 3𝑥 2 − 2𝑥 + 5
−1
Use this function to find (𝑓 )′(6).
Throughout this unit, we’ll label 𝑓 −1 (𝑥) as 𝑔(𝑥).
Find 𝑔′(6).
From the formula, 𝑔(𝑥) =
1
.
𝑓 ′ (𝑔(6))
First of all, what is 𝑔(6)? Since 𝑓(1) = 6,
𝑔(6) = 1.
Since we know what 𝑓 is, we can find 𝑓 ′ .
𝑓 ′ (𝑥) = 6𝑥 − 2
𝑓 ′ (1) = 4
𝑔′ (6) =
Class Notes 3.4
1
4
Page 9 of 19
√3
Example: If 𝑓(𝑥) = sin 𝑥, and f and g are inverse functions, find 𝑔′ ( 2 ).
1
𝑔′ (𝑥) = ′
𝑓 (𝑔(𝑥))
√3
The expression 𝑔′ ( 2 ) asks the question what value of x has 𝑓(𝑥) =
and trig, we know that sin 𝜋/3=
√3
.
2
From our knowledge of the unit circle
√3
.
2
𝑔′ (
1
1
√3
= ′
)=
2
𝑓 ′ (𝑔(√3⁄2)) 𝑓 (𝜋/3)
1
Since 𝑓(𝑥) = sin 𝑥, 𝑓 ′ (𝑥) = cos 𝑥. Therefore, 𝑓 ′ (𝜋/3) = cos(𝜋⁄3) = 2.
That means that
𝑔′ (
1
1
1
1
√3
= ′
=
=
=2
)=
2
𝑓 ′ (𝑔(√3⁄2)) 𝑓 (𝜋/3) cos(𝜋⁄3) 1/2
Find 𝑔′(4) when 𝑓(𝑥) = 𝑥 3 − 4 and g is the inverse of f.
Find 𝑔′ (2) when 𝑓(𝑥) = 2𝑥 2 − 7𝑥 + 2
Find 𝑔′ (1) when 𝑓(𝑥) = √𝑥 − 3
End of 3.8 Day 1
Class Notes 3.4
Page 10 of 19
3.8 Day 2 Inverse Trig Functions
Finding the Derivative of the Inverse Numerically
Before we go onto the formulas for the inverse trig functions, let’s revisit yesterday’s work.
Using the following table, please find 𝑔′ (5) when f and g are inverse functions.
x
𝑓(𝑥)
𝑓 ′ (𝑥)
-3
2
-1
-1
3
1
0
-2
3
1
0
12
2
5
-2
3
8
0
Find 𝑔′ (8)
Find 𝑔′ (8)
Yesterday was about finding the value of the inverse of the derivative of an inverse function. We saw that we
could find that value regardless of whether or not we could name the inverse function. Today, we’re going to
Now, we’re going to learn the derivatives of the inverse trig functions. You must memorize these.Tomorrow,
we’ll derive each one (or you can do the derivation in the book).
𝑑
𝑢′
arcsin 𝑢 =
𝑑𝑥
√1 − 𝑢2
𝑑
𝑢′
arccos 𝑢 = −
𝑑𝑥
√1 − 𝑢2
𝑑
𝑢′
arctan 𝑢 =
𝑑𝑥
1 + 𝑢2
We’ll apply these as we’d do for any other derivative, including the use of our rules (constant multiple, sum
and difference, product, quotient, chain, and trig formulae).
Differentiate each of the following Ready. Set. Go.
arcsin(2𝑥)
Class Notes 3.4
arctan(3𝑥)
arcsin √𝑥
Page 11 of 19
2 arcsin(x − 1)
arcsin t 2
arcsin(3𝑥)
𝑥
x 2 arctan(x)
Class Notes 3.4
3arccos
𝑥
2
arcsin √𝑥
Page 12 of 19
AP Calculus AB
Class Notes
3.9
Derivatives of Exponential and Logarithmic Functions
pp 172 - 180
Objectives:
Find the derivatives of exponential and logarithmic functions. Use all previous rules of differentiation
(constant multiple, power, product, quotient, chain, implicit, trig, inverse) in conjunction with the new rules
for logarithms and exponential functions.
Outline
Differentiate exponential functions.
Differentiate logarithmic functions.
Vocabulary
Omissions
Logarithmic differentiation is not part of the AP Calculus AB curriculum.
3.9 Day 1 Exponential Differentiation
The derivative of the exponential function is truly the most bizzare in all of calculus.
𝑑 𝑥
𝑒 = 𝑒𝑥
𝑑𝑥
What does this mean? The slope of the tangent to the function 𝑒 𝑥 at the point x is 𝑒 𝑥 !
Take a look at the table:
x
0
1
2
⋮
n
𝑓(𝑥)
𝑒 0 or 1
𝑒 0 or 𝑒
𝑒2
⋮
𝑒𝑛
𝑓 ′ (𝑥)
1
𝑒
𝑒2
⋮
𝑒𝑛
tangent line
𝑦 = 1(𝑥 − 0) + 1
𝑦 = 𝑒(𝑥 − 1) + 𝑒
𝑦 = 𝑒 2 (𝑥 − 1) + 𝑒 2
⋮
𝑦 = 𝑒 𝑛 (𝑥 − 𝑛) + 𝑒 𝑛
y





x

Class Notes 3.4


Page 13 of 19
If the argument of 𝑒 𝑥 is a function, for example, 𝑒 sin 𝑥 , then the derivative of 𝑒 𝑢 is given by
𝑑 𝑢
𝑒 = 𝑢′ 𝑒 𝑢
𝑑𝑥
𝑑
Let’s give it a try. Find 𝑑𝑥 of the following expressions.
𝑒 7𝑥
𝑒 −5𝑥
𝑥3 𝑒𝑥
𝑒 1/𝑥
𝑒 𝑥 − 𝑒 −𝑥
𝑒 𝑥 + 𝑒 −𝑥
sin(𝑒 𝑥 )
𝑒 𝑥𝑡𝑎𝑛 𝑥
𝑒 (𝑥+𝑥
Class Notes 3.4
2)
Page 14 of 19
There is a more obscure exponential function that shows up from time to time. It’s the exponential function in
the form of 𝑎𝑢 as in 𝑓(𝑥) = 8𝑥 . There are two way to look at this.
1) Write the function as a function as an exponential to the base e.
Original function.
𝑦 = 8𝑥
Take the natural logarithm of the function.
Exponentiate to the base e to undo previous step.
Differentiate both sides wrt x.
Derivative of the exponential function.
Bring down the x with the exponent of the argument of a
logarithm rule.
Bring out the constant with the constant multiple rule.
𝑑
𝑥=1
𝑑𝑥
Done!
Now, let’s create the general form: we do this by using the definition of the logarithm.
Original function.
𝑦 = 𝑎𝑢
Take the natural logarithm of the function.
Exponentiate to the base e to undo previous step.
Differentiate both sides wrt x.
Derivative of the exponential function.
Bring down the x with the exponent of the argument of a logarithm rule.
Bring out the constant with the constant multiple rule.
𝑑
𝑥 = 𝑢′
𝑑𝑥
Exponent of the logarithm rule. Done!
This is shown on page 173 of your book with the general form of the equation:
𝑑 𝑢
𝑑𝑢
(𝑎 ) = 𝑎𝑢 ln 𝑎
𝑑𝑥
𝑑𝑥
Seen next to it’s cousin, it starts to make sense:
Class Notes 3.4
Page 15 of 19
𝑑 𝑢
𝑒
𝑑𝑥
Try some.
5𝑥
𝑑𝑢
𝑑
(𝑎𝑢 )
𝑑𝑥
= 𝑒 𝑢 𝑑𝑥
𝑑𝑢
= 𝑎𝑢 ln 𝑎 𝑑𝑥
𝑑
Find 𝑑𝑥 of the following expressions.
3−𝑥
𝜋 𝑥 tan 𝑥
𝜋 sin 𝑥
End of 3.9 Day 1
3.9 Day 2 Logarithmic Differentiation
Since we know that the inverse of the logatrithmic function is the exponential function , and we know how to
differentiate the exponential function, we can use our knowledge to find the derivative of ln 𝑢.
1.
𝑦 = ln 𝑢
Original equation.
2.
Exponentiate both sides to the base e.
3.
Since 𝑒 ln 𝑢 = 𝑢 …
4.
Reflexive property.
5.
Take derivative of both sides wrt x.
6.
𝑑𝑢
Rename 𝑑𝑥 .
7.
Derivative of exponent rule.
8.
From step 4: 𝑒 𝑦 = 𝑢
9.
Reflexive property.
10.
Solving for 𝑦′ we have….
Thus, we have the formula:
Let’s try some:
ln 2
Class Notes 3.4
𝑑
1 𝑑𝑢
ln 𝑢 =
𝑑𝑥
𝑢 𝑑𝑥
ln(𝑥 3 )
Page 16 of 19
(ln 𝑥)2
ln (sin 𝑥)
ln(2 + √𝑥)
ln (
ln √𝑥
√1 + ln2 𝑥
x
)
1 + x2
Logarithims other than base e
To find the derivative of log 𝑏 𝑢, we’ll use the rule for finding the logarithm with an arbitrary base
ln 𝑢
log 𝑏 𝑢 =
.
ln 𝑏
Class Notes 3.4
Page 17 of 19
Original problem.
𝑦 = log 𝑏 𝑢
Rewrite with the expansion of a logarithm with arbitrary base, b.
Take derivative of each side wrt x.
Since ln 𝑏 is a constant, we can take it out with the constant multiple
rule.
𝑑
Recalling from the previous derivation that 𝑑𝑥 ln 𝑢 =
𝑢′
,
𝑢
we get…
Our two formulas for the derivative of a logarithm are therefore
𝑑
1 𝑑𝑢
ln 𝑢 = 𝑢 𝑑𝑥
𝑑𝑥
𝑑
Once again, let’s try this on.
log 3 𝑥
log10 cos 𝑥
log 5 √𝑥 2 − 1
log 2
Class Notes 3.4
1
log 𝑏 𝑢 = ln 𝑏 ∙
𝑑𝑥
and
𝑢′
𝑢
𝑥2
𝑥−1
Page 18 of 19
Assignments
simple harmonic motion
jerk
(p. 143)
(p. 144)
Chain Rule
Power Chain Rule
orthogonal curves
implicit differentiation
normal to the surface
Power Rule for Rational Powers of x
inverse function–inverse cofunction identities
Power Rule for Arbitrary Real Powers
logarithmic differentiation
orthogonal families
derivative of f at a
differentiable function
(p. 149)
(p. 151)
(p. 154)
(p. 157)
(p. 159)
(p. 161)
(p. 168)
(p. 176)
(p. 177)
(p. 180)
(p. 99)
(p. 99)
Class Notes 3.4
Page 19 of 19