1) Find the derivative dy/dx if both x and y are functions of a parameter t. The formula is dy/dx = (dy/dt) / (dx/dt), provided that dx/dt does not equal 0. 2) As an example, if x = a(1 - cosθ) and y = a(θ + sinθ), then dy/dx = (1 + cosθ) / sinθ. 3) For a given problem where x and y are functions of t, you first find the derivatives dx/dt and dy/dt, and then plug into the formula to find dy/dx.

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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