MT112
Calculus - I
Week#06
Lecture#01
Derivative of Trigonometric Functions
𝒅
1. 𝒅𝒙
𝒔𝒊𝒏𝒙 = 𝒄𝒐𝒔𝒙
𝒅
2. 𝒅𝒙
𝒄𝒐𝒔𝒙 − −𝒔𝒊𝒏𝒙
𝒅
3. 𝒕𝒂𝒏𝒙 = 𝐬𝐞𝐜 𝟐 𝒙
𝒅𝒙
𝒅
4. 𝒅𝒙
𝒔𝒆𝒄 = 𝒔𝒆𝒄𝒙 𝒕𝒂𝒏𝒙
𝒅
5. 𝒅𝒙
𝒄𝒔𝒄𝒙 = −𝒄𝒔𝒄𝒙 𝒄𝒐𝒕𝒙
𝒅
6. 𝒄𝒐𝒕𝒙 = −𝒄𝐬𝐜 𝟐 𝒙
𝒅𝒙
Derivative of Trigonometric Functions
Example:
Find 𝑓 ′′ (𝜋/4) if 𝑓(𝑥) = sec 𝑥.
Solution:
𝑓 ′ (𝑥) = sec 𝑥 tan 𝑥
Using product rule
𝑑 𝑑𝑦
𝑓 ′′ 𝑥 = sec 𝑥 . [tan 𝑥] + tan 𝑥 · [sec 𝑥]
𝑑𝑥 𝑑𝑥
= sec 𝑥 · sec 2 𝑥 + tan 𝑥 · sec 𝑥 tan 𝑥
= sec 3 𝑥 + sec 𝑥 tan2 𝑥
𝜋 𝜋 𝜋
𝑓 ′′ = sec 3 + sec tan2 (𝜋/4)
4 4 4
= (√2) 3 + (√2)(1) 2 = 3√2
Derivative of Trigonometric Functions
Example:
On a sunny day, a 50 ft flagpole casts a shadow that changes with the angle of elevation of the Sun. Let s be the length
of the shadow and θ the angle of elevation of the Sun. Find the rate at which the length of the shadow is changing with
respect to θ when 𝜃 = 45𝑜 . Express your answer in units of feet/degree.
Solution:
The variables 𝑠 and 𝜃 are related by tan 𝜃 = 50/𝑠 or, equivalently,
𝑠 = 50 cot 𝜃
If θ is measured in radians,
𝑑𝑠
= −50 csc 2 𝜃
𝑑𝜃
which is the rate of change of shadow length with respect to the elevation angle θ in units of feet/radian. When θ = 45
◦ (or equivalently θ = π/4 radians), we obtain
𝑑𝑠 𝑃𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟
ቚ = −50 csc2(𝜋/4) = −100 𝑓𝑒𝑒𝑡/𝑟𝑎𝑑𝑖𝑎𝑛 𝑡𝑎𝑛∅ =
𝑑𝜃 𝜋/4 𝐵𝑎𝑠𝑒
Converting radians (rad) to degrees (deg) yield,
𝐵𝑎𝑠𝑒
𝑐𝑜𝑡∅ =
100𝑓𝑡 𝜋 𝑟𝑎𝑑 5 𝑓𝑡 𝑃𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟
− . =− 𝜋 ≈ −1.75𝑓𝑡/𝑑𝑒𝑔
𝑟𝑎𝑑 180 𝑑𝑒𝑔 9 𝑑𝑒𝑔
Thus, when θ = 45 ◦ , the shadow length is decreasing (because of the minus sign) at an approximate rate of 1.75
ft/deg increase in the angle of elevation.
Derivative of Composition
Suppose you are traveling to school in your car, which gets 20 𝑚𝑖𝑙𝑒𝑠 per gallon of gasoline. The number of miles you can
travel in your car without refueling is a function of the number of gallons of gas you have in the gas tank. In symbols, if 𝑦 is
the number of miles you can travel and 𝑢 is the number of gallons of gas you have initially, then 𝑦 is a function of 𝑢, or 𝑦 =
𝑓(𝑢). As you continue your travels, you note that your local service station is selling gasoline for $4 per gallon. The number
of gallons of gas you have initially is a function of the amount of money you spend for that gas. If 𝑥 is the number of dollars
you spend on gas, then 𝑢 = 𝑔(𝑥). Now 20 𝑚𝑖𝑙𝑒𝑠 per gallon is the rate at which your mileage changes with respect to the
amount of gasoline you use, so
𝑑𝑦
𝑓′ 𝑢 = = 20𝑚𝑖𝑙𝑒𝑠 𝑝𝑒𝑟 𝑔𝑎𝑙𝑙𝑜𝑛
𝑑𝑢
Similarly, since gasoline costs $4 per gallon, each dollar you spend will give you 1/4 of a gallon of gas, and
𝑑𝑢 1
𝑔′ 𝑥 = = 𝑔𝑎𝑙𝑙𝑜𝑛𝑠 𝑝𝑒𝑟 𝑑𝑜𝑙𝑙𝑎𝑟
𝑑𝑥 4
Notice that the number of miles you can travel is also a function of the number of dollars you spend on gasoline. This fact is
expressible as the composition of function
𝑦 = 𝑓(𝑢) = 𝑓(𝑔(𝑥))
You might be interested in how many miles you can travel per dollar, which is 𝑑𝑦/𝑑𝑥. Intuition suggests that rates of
change multiply in this case
Derivative of Composition
The Chain Rule
If g is differentiable at 𝑥 and 𝑓 is differentiable a 𝑔(𝑥), then the composition 𝑓𝑜𝑔 is differentiable at
𝑥. Moreover, if
𝑦 = 𝑓(𝑔(𝑥)) 𝑎𝑛𝑑 𝑢 = 𝑔(𝑥)
then 𝑦 = 𝑓(𝑢) and
𝑑𝑦 𝑑𝑦 𝑑𝑢
= .
𝑑𝑥 𝑑𝑢 𝑑𝑥
Derivative of Composition
Example:
Find 𝑑𝑦/𝑑𝑥 if 𝑦 = cos(𝑥3).
Solution:
Let 𝑢 = 𝑥 3 and express y as 𝑦 = cos 𝑢. Applying Formula yields
𝑑𝑦 𝑑𝑦 𝑑𝑢
= .
𝑑𝑥 𝑑𝑢 𝑑𝑥
𝑑 𝑑 3
= 𝑐𝑜𝑠𝑢 . 𝑥
𝑑𝑢 𝑑𝑥
= (− sin 𝑢) · (3𝑥 2 )
= (− sin(𝑥 3 )) · (3𝑥 2 ) = −3𝑥 2 sin(𝑥 3 )
Derivative of Composition
Derivative of Composition
