HOW TO EVALUATE DERIVATIVES OF PARAMETRIC FUNCTIONS AND SECOND ORDER DERIVATIVES

1. GRADE 12 MATH
DIFFERENTIATION OF
PARAMETRIC FUNCTIONS
SUMAN MATHEWS
MATH CONTENT SPECIALIST
PRESENTS

2. How do you find dy/dx if both x and y
are functions of t ?
How to differentiate one function with
respect to another function?

3. 𝐼𝑓 𝑥 = 𝑓 𝑡 , 𝑦 = 𝑓 𝑡 ,
𝑑𝑦
𝑑𝑥
=
𝑑𝑦
𝑑𝑡
𝑑𝑥
𝑑𝑡
, 𝑝𝑟𝑜𝑣𝑖𝑑𝑒𝑑
𝑑𝑥
𝑑𝑡
≠ 0
Question 1
Find
𝑑𝑦
𝑑𝑥
, 𝑖𝑓 𝑥 = 𝑎 ( 1 − cos 𝜃), 𝑦 = 𝑎 (𝜃 + sin 𝜃)
𝑑𝑦
𝑑𝑥
=
𝑑𝑦
𝑑𝜃
𝑑𝑥
𝑑𝜃
=
𝑎(1 + 𝑐𝑜𝑠𝜃)
𝑎𝑠𝑖𝑛𝜃
=
1 + 𝑐𝑜𝑠𝜃
𝑠𝑖𝑛𝜃
HERE t, is the parameter

4. Question 2
If x = 2 cost – cos 2t, y = 2 sint – sin 2t, find
𝑑𝑦
𝑑𝑥
at t =
𝜋
4
𝑑𝑦
𝑑𝑥
=
2 cos 𝑡 − 2 cos 2𝑡
−2 𝑠𝑖𝑛𝑡 + 2 𝑠𝑖𝑛2𝑡
=
𝑐𝑜𝑠𝑡 − cos 2𝑡
−𝑠𝑖𝑛𝑡 + sin 2𝑡
𝑑𝑦
𝑑𝑥
𝑎𝑡 𝑡 =
𝜋
4
=
cos
𝜋
4
− cos
𝜋
2
−𝑠𝑖𝑛
𝜋
4
+ sin
𝜋
2
=
1
√2
−1
√2
+ 1

5. =
1
2 − 1
=
2 + 1
2 − 1
= 2 + 1
Question 3
𝐷𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑡𝑖𝑎𝑡𝑒 𝑡𝑎𝑛−1
2𝑥
1 − 𝑥2 𝑤𝑖𝑡ℎ 𝑟𝑒𝑠𝑝𝑒𝑐𝑡 𝑡𝑜 𝑡𝑎𝑛−1
𝑥

6. 𝑙𝑒𝑡 𝑦 = 𝑡𝑎𝑛−1
2𝑥
1 − 𝑥2
𝑙𝑒𝑡 𝑧 = 𝑡𝑎𝑛−1
𝑥
y and z are both functions of x
𝑑𝑦
𝑑𝑧
=
𝑑𝑦
𝑑𝑥
𝑑𝑥
𝑑𝑧
-----------------1

7. We now calculate
𝑑𝑦
𝑑𝑥
𝑦 = 𝑡𝑎𝑛−1
2𝑥
1 − 𝑥2
𝑝𝑢𝑡 𝑥 = tan 𝜃
𝑦 = 𝑡𝑎𝑛−1
(
2𝑡𝑎𝑛𝜃
1−𝑡𝑎𝑛2 𝜃
)= 𝑡𝑎𝑛−1
(tan 2 𝜃) = 2𝜃

8. = 2 𝑡𝑎𝑛−1 𝑥
𝑑𝑦
𝑑𝑥
=
y
2
1 + 𝑥2
We now calculate
𝑑𝑧
𝑑𝑥
𝑧 = 𝑡𝑎𝑛−1 𝑥

9. 𝑑𝑧
𝑑𝑥
=
1
1 + 𝑥2
𝑑𝑥
𝑑𝑧
= 1 + 𝑥2
Coming back to 1
𝑑𝑦
𝑑𝑥
= 2
1 + 𝑥2 (1 + 𝑥2
) = 2

10. Question 4
𝑑𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑡𝑖𝑎𝑡𝑒 𝑡𝑎𝑛−1 3𝑥 − 𝑥3)
1−3𝑥2 𝑤. 𝑟. 𝑡. 𝑡𝑎𝑛−1
(
2𝑥
1−𝑥2)
Let y = 𝑡𝑎𝑛−1
(
3𝑥−𝑥3
1−3𝑥2)
Put x = tan 𝜃
𝑦 = 𝑡𝑎𝑛−1
(
3𝑡𝑎𝑛𝜃 − 𝑡𝑎𝑛3
𝜃
1 − 3𝑡𝑎𝑛2 𝜃
)

11. 𝑦 = 𝑡𝑎𝑛−1(tan 3𝜃) = 3𝜃 = 3𝑡𝑎𝑛−1 𝑥
𝑑𝑦
𝑑𝑥
=
3
1 + 𝑥2
-------------------------1
𝑙𝑒𝑡 𝑢 = 𝑡𝑎𝑛−1
(
2𝑥
1 + 𝑥2)
𝑝𝑢𝑡 𝑥 = 𝑡𝑎𝑛𝜃

12. 𝑢 = 𝑡𝑎𝑛−1
2𝑡𝑎𝑛𝜃
1 − 𝑡𝑎𝑛2 𝜃
= 𝑡𝑎𝑛−1(tan 2𝜃) = 2𝜃 = 2𝑡𝑎𝑛−1 𝑥
𝑑𝑢
𝑑𝑥
=
2
1 + 𝑥2
--------------------------2
From 1 and 2 ,
𝑑𝑦
𝑑𝑢
=
𝑑𝑦
𝑑𝑥
𝑑𝑢
𝑑𝑥
=
3
1+𝑥2 ×
1+𝑥2
2
=
3
2
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