As you will quickly see, Calculus is not really a subject of AP Physics; it is a tool that we use from the first day.
However, many of you are now just starting Calculus and a few of you are not taking and have not taken
Calculus. For that reason, we have this high speed mini-course to get you able to use the Calculus tools.
To let the cat out of the bag, you all have already done Calculus in first year Physics – multiple times – without knowing it. Every time you used a Tangent Line to find a slope you were doing Differential Calculus. Every time you found an area under a curve you were doing Integral Calculus. The only differences were that we usually used real data points and approximation whereas now in AP Physics C (and in Calculus) we will use equations fitted to real data to find the slopes (Differential Calculus) or the areas (Integral Calculus).
Since we are finding Tangent line slopes without looking at real data, we need to find a way to sneak up on the slopes. Tangent lines to a curve touch the curve “locally” at only one point (the point of tangency). Since we need two points to compute a slope we will use the Secant Line slope method of finding approximate slopes and then driving the secant line to a tangent line by moving one end of the secant line toward the other end.
Suppose we have the following graph:
Using basic algebra, the slope of the line through the two points
(x, f(x)) and (x+
 x, f(x+
 x)) = slope of the secant line which is (recalling that slope is rise over run) 𝑠𝑙𝑜𝑝𝑒 = 𝑓(𝑥+∆𝑥)−𝑓(𝑥)
(𝑥+∆𝑥)−𝑥
= 𝑓(𝑥+∆𝑥)−𝑓(𝑥)
∆𝑥
(1)
Now what we really want is the slope as
 x approaches zero (moving the right hand point closer and closer to the left hand point) and the secant line starts to look like the tangent line touching the function f(x) at only one point “near” x. 𝑠𝑙𝑜𝑝𝑒 𝑜𝑓 [𝑓(𝑥)𝑎𝑡 𝑥] = d dx f(x) = f ′ (x) = lim
∆x→0
( f(x+∆x)−f(x)
∆x
) (2)
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Example 1
Find the slope of the graph of f(x) = kx 2 at any x.
Use equation (1) above to solve this problem by first finding the slope of the secant line: 𝑠𝑙𝑜𝑝𝑒 = 𝑓(𝑥 + ∆𝑥) − 𝑓(𝑥)
∆𝑥
= 𝑘(𝑥 + ∆𝑥) 2
∆𝑥
− 𝑘𝑥 2
As those of you taking (or already completed) Calculus know, a large chunk of Calculus involves sneaky algebraic manipulations which often start out making things messy and then proceed to reduce to something easy.
There are THREE MAJOR TRICKS IN MATHEMATICS that one can do to an expression that does not change the numeric value of the expression:
1.
Expand any expression that can be expanded.
2.
Add ZERO to the expression
3.
Multiply the entire expression by ONE.
We will now use method 1 (also called, MAKE IT WORSE AND SEE WHAT HAPPENS.”) f ( x
  x )
 x
 f ( x )

 k ( x kx
2 
  x )
2
 x
2 kx
 x

 kx
2
 x k
 x
2
 k ( x
 kx
2
2 

2 x
 x
  x
2
2 kx

 x x
 k
 x
2
)

 x kx
2
 kx

2 kx
 k
 x
2 
Thus the slope of the secant line is:
2 kx
 x

 x k
 x
2  kx
2
2kx+k
 x
To find the slope of graph at x (or the slope of a tangent line at x) we use equation (2): 𝑠𝑙𝑜𝑝𝑒 𝑜𝑓 [𝑓(𝑥)𝑎𝑡 𝑥] = d dx f(x) = f ′ (x) = lim
∆x→0
( f(x + ∆x) − f(x)
) = lim
∆x ∆𝑥→0
(2𝑘𝑥 + 𝑘∆𝑥) = 2𝑘𝑥
So: 𝑑 𝑑𝑥 𝑘𝑥 2 = 2𝑘𝑥
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If you go through the same grueling algebraic process you will find that: 𝑑 𝑑𝑥 𝑑 𝑘𝑥 3 𝑘𝑥 4 𝑑𝑥
= 3𝑘𝑥
= 4𝑘𝑥
2
3
And in general: 𝐝 𝐝𝐱 𝐤𝐱 𝐧 = 𝐧𝐤𝐱 𝐧−𝟏
Example 2: Examples from Physics
Suppose you are moving so that your position is given by x = 3t
4
, then your speed (velocity) is the slope of the position curve and can be calculated using the general rule above: 𝑣 = 𝑑 𝑑𝑡
(3𝑡 4 ) = (4)3𝑡 4−1 = 12𝑡 3
Now your acceleration (which is the slope of the velocity) must be 𝑎 = 𝑑 𝑑𝑡
(12𝑡 3 ) = (3)12𝑡 3−1 = 36𝑡 2
If you wish, you can even find the jerk (change in acceleration = slope of acceleration curve) which is 𝑗 = 𝑑 𝑑𝑡
(36𝑡 2 ) = (2)36𝑡 2−1 = 72𝑡
You can even find the snap (change in jerk = slope of jerk curve) which is 𝑠 = 𝑑 𝑑𝑡
(72𝑡) = (1)72𝑡 1−1 = 72𝑡 0 = 72
Going berserk one can find the crackle (snap-crackle-pop = I made it up) which is 𝑐 = 𝑑 𝑑𝑡
(72) = 𝑑 𝑑𝑡
(72𝑡 0 ) = (0)72𝑡 0−1 = 0
The last derivative is better stated by the rule that the derivative of a constant is 0.
There are some other rules you have to deal with when taking derivatives, which can all be proven using equations (1) & (2), but I will leave this to Mr. Decker.
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Sum (or Difference) Rule:
If you have f(x) = g(x) ± h(x) , then f’(x) = g’(x) ± h’(x).
Example 3
Find the slope of the function f(x) = 4x
5
+ 1.5x
2.5 – 7x -3
, when x = 2.
First find f’(x) {forget about the x =2 for now}: 𝑓 ′ (𝑥) = 𝑑 𝑑𝑥
(4𝑥 5 ) + 𝑑 𝑑𝑥
(1.5𝑥 2.5
) −
= 20𝑥 4 𝑑 𝑑𝑥
(7𝑥 −3 ) = 20𝑥 4
+ 3.75𝑥 1.5
+
21 𝑥 4
Now substitute x = 2 in to find the slope of f(x): 𝑓 ′ (2) = 20(2) 4 + 3.75(2) 1.5
+
+ 3.75𝑥
21
(2) 4
= 329.3
1.5
− (−21𝑥 −4 )
Derivatives of Basic Functions: 𝒅 𝒅𝒙 𝐬𝐢𝐧 𝒙 = 𝐜𝐨𝐬 𝒙 𝒅 𝐜𝐨𝐬 𝒙 = − 𝐬𝐢𝐧 𝒙 𝒅𝒙 𝒅 𝒆 𝒙 𝒅𝒙
= 𝒆 𝒙 𝒅 𝒅𝒙 𝐥𝐧 𝒙 =
𝟏 𝒙
There are other derivatives we may need, but we will learn them as we come across them.
Product Rule:
Suppose we have a function which is the product of two functions: 𝒇(𝒙) = 𝒈(𝒙)𝒉(𝒙)
Then the derivative of the function 𝑓(𝑥) is given by: 𝒇 ′ (𝒙) = 𝒈(𝒙)𝒉 ′ (𝒙) + 𝒈 ′ (𝒙)𝒉(𝒙)
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Example 4
Find the derivative of:
(i) 𝑓(𝑥) = 𝑥 2 sin 𝑥 𝑓 ′ (𝑥) = (𝑥 2 𝑑
) ( 𝑑𝑥
(ii) ℎ(𝑡) = 𝑡 ln 𝑡 sin 𝑥) + ( ℎ ′ (𝑡) = (𝑡) ( 𝑑 𝑑𝑡 𝑑 𝑑𝑥 𝑥 2 ) (sin 𝑥) = 𝑥 ln 𝑡) + ( 𝑑 𝑑𝑡
2 (cos 𝑥) + (2𝑥)(sin 𝑥) = 𝑥 2 cos 𝑥 + 2𝑥 sin 𝑥 𝑡) (ln 𝑡) = (𝑡) (
1 𝑡
) + (1)(ln 𝑡) = 1 + ln 𝑡
Quotient Rule:
Suppose you have a function that is the quotient of two functions: 𝒇(𝒙) = 𝒈(𝒙) 𝒉(𝒙)
Then the derivative of 𝑓(𝑥) is: 𝒇 ′ (𝒙) = 𝒉(𝒙)𝒈 ′ (𝒙) − 𝒈(𝒙)𝒉′(𝒙)
(𝒉(𝒙)) 𝟐
Sometimes cleverly stated as “Ho DHi minus Hi DHo over Ho squared”.
Example 5
Find the derivative of : ℎ(𝑧) = 𝑧 3 cos 𝑧 ℎ ′ (𝑧) =
(cos 𝑧) ( 𝑑 𝑑𝑧 𝑧 3 ) − (𝑧 3 )( 𝑑 𝑑𝑧
(𝑧 3 ) 2 cos 𝑧)
=
(cos 𝑧)(3𝑧 2 ) − (𝑧 3 )(− sin 𝑧) 𝑧 6
=
3𝑧 2 cos 𝑧 + 𝑧 𝑧 6
3 sin 𝑧
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Chain Rule:
Suppose you have a function of a function. In other words, compute a function then use the result of that computation to compute a second function. It sounds uncommon but it is very common. Here are some examples:
y = ln(2x)
The 2x must be computed first then the ln computed. y = (x
2
+1)
3
Normally, one would compute the x 2 +1 first then cube the result though it could be expanded.
Mathematically, we call these types of functions composite functions.
Now, suppose you have a function defined as a composite function: 𝒇(𝒙) = 𝒈(𝒉(𝒙))
Then the derivative of 𝑓(𝑥) is: 𝒇 ′ (𝒙) = 𝒈 ′ (𝒉(𝒙))𝒉′(𝒙)
Example 6
Find the derivative of the following:
(i) 𝑦 = ln 2𝑥 𝑑𝑦
= 𝑑𝑥 𝑑 𝑑𝑥 ln 2𝑥 =
1
2𝑥
(2) =
1 𝑥
(ii) 𝑦 = (𝑥 2 + 1) 3 𝑑𝑦 𝑑𝑥
= 𝑑 𝑑𝑥
(𝑥 2 + 1) 3 = 3(𝑥 2 + 1) 2 (2𝑥) = 6𝑥(𝑥 2 + 1) 2
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