Module 6: TRIGONOMETRIC AND INVERSE TRIGONOMETRIC DIFFERENTIATION
6.1 Derivatives of Trigonometric Functions
If u is a differentiable function of x, then the following are differentiation formulas of the trigonometric
functions
𝑑 𝑑𝑢
(𝑠𝑖𝑛 𝑢) = 𝑐𝑜𝑠 𝑢
𝑑𝑥 𝑑𝑥
𝑑 𝑑𝑢
(𝑐𝑜𝑠 𝑢) = − 𝑠𝑖𝑛 𝑢
𝑑𝑥 𝑑𝑥
𝑑 2 𝑑𝑢
(𝑡𝑎𝑛 𝑢) = 𝑠𝑒𝑐 𝑢
𝑑𝑥 𝑑𝑥
𝑑 𝑑𝑢
(𝑐𝑜𝑡 𝑢) = − 𝑐𝑠𝑐 2 𝑢
𝑑𝑥 𝑑𝑥
𝑑 𝑑𝑢
(𝑠𝑒𝑐 𝑢) = 𝑡𝑎𝑛 𝑢 𝑠𝑒𝑐 𝑢
𝑑𝑥 𝑑𝑥
𝑑 𝑑𝑢
(𝑐𝑠𝑐 𝑢) = − 𝑐𝑜𝑡 𝑢 𝑐𝑠𝑐 𝑢
𝑑𝑥 𝑑𝑥
Example 1 Determine the derivative of each function.
1+𝑐𝑜𝑠 2𝑥 csc x - cot x
(a) 𝑦 = 𝑥 𝑡𝑎𝑛 ( 𝑥 2 - x)
2
(b) 𝑦 = (c) 𝑦 =
𝑠𝑖𝑛 2𝑥 𝑐𝑠𝑐𝑥+𝑐𝑜𝑡 𝑥
Solution:
(a) Differentiating with respect to x, we have
𝑦 ′ = 𝑥(2 tan(𝑥 2 - x))𝑠𝑒𝑐 2 (𝑥 2 - x)(2𝑥 − 1) + 𝑡𝑎𝑛2 ( 𝑥 2 - x)
𝑦 ′ = 𝑡𝑎𝑛( 𝑥 2 - x)[2𝑥(2𝑥 − 1)𝑠𝑒𝑐 2 (𝑥 2 - x) + 𝑡𝑎𝑛( 𝑥 2 - x)]
(b) Differentiating with respect to x, we have
𝑠𝑖𝑛2𝑥(−2𝑠𝑖𝑛2𝑥)−(1+𝑐𝑜𝑠2𝑥)(2𝑐𝑜𝑠2𝑥)
𝑦′ = (𝑠𝑖𝑛2𝑥)2
−2𝑠𝑖𝑛2 2𝑥−2𝑐𝑜𝑠2𝑥−2𝑐𝑜𝑠2 2𝑥
𝑦′ = 𝑠𝑖𝑛2 2𝑥
−2(𝑠𝑖𝑛2 2𝑥+𝑐𝑜𝑠2 2𝑥)−2𝑐𝑜𝑠2𝑥
𝑦′ = but 𝑠𝑖𝑛2 2𝑥 + 𝑐𝑜𝑠 2 2𝑥 = 1
𝑠𝑖𝑛2 2𝑥
−2−2𝑐𝑜𝑠2𝑥 1+𝑐𝑜𝑠2𝑥
So 𝑦 ′ = , however = 𝑐𝑜𝑠 2 𝑥 and 𝑠𝑖𝑛2𝑥 = 2𝑠𝑖𝑛𝑥𝑐𝑜𝑠𝑥
𝑠𝑖𝑛2 2𝑥 2
−2(1+𝑐𝑜𝑠2𝑥) −4𝑐𝑜𝑠2 𝑥
Thus, 𝑦 ′ = (𝑠𝑖𝑛2𝑥)2
= 4𝑠𝑖𝑛2 𝑥𝑐𝑜𝑠2 𝑥
−1
𝑦 ′ = 𝑠𝑖𝑛2 𝑥
𝑦 ′ = −𝑐𝑠𝑐 2 𝑥
(c) Differentiating with respect to x, we have
(𝑐𝑠𝑐𝑥+𝑐𝑜𝑡𝑥)(−𝑐𝑠𝑐𝑥𝑐𝑜𝑡𝑥+𝑐𝑠𝑐 2 𝑥)−(𝑐𝑠𝑐𝑥−𝑐𝑜𝑡𝑥)(−𝑐𝑠𝑐𝑥𝑐𝑜𝑡𝑥−𝑐𝑠𝑐 2 𝑥)
𝑦′ = (𝑐𝑠𝑐𝑥+𝑐𝑜𝑡𝑥)2
𝑐𝑠𝑐𝑥(𝑐𝑠𝑐𝑥+𝑐𝑜𝑡𝑥)(𝑐𝑠𝑐𝑥−𝑐𝑜𝑡𝑥)+𝑐𝑠𝑐𝑥(𝑐𝑠𝑐𝑥−𝑐𝑜𝑡𝑥)(𝑐𝑠𝑐𝑥+𝑐𝑜𝑡𝑥)
𝑦′ = (𝑐𝑠𝑐𝑥+𝑐𝑜𝑡𝑥)2
2𝑐𝑠𝑥(𝑐𝑠𝑐𝑥−𝑐𝑜𝑡𝑥)(𝑐𝑠𝑐𝑥+𝑐𝑜𝑡𝑥)
𝑦′ = (𝑐𝑠𝑐𝑥+𝑐𝑜𝑡𝑥)2
′ 2𝑐𝑠𝑐𝑥(𝑐𝑠𝑐𝑥−𝑐𝑜𝑡𝑥)
𝑦 = 𝑐𝑠𝑐𝑥+𝑐𝑜𝑡𝑥
𝑑𝑦
Example 2 Find 𝑑𝑥
given that y sin x − x 2 cos y − xy = 0
Solution:
From y sin x − x 2 cos y − xy = 0
Differentiating this implicitly with respect to x, we get
𝑦𝑐𝑜𝑠𝑥 + 𝑦 ′ 𝑠𝑖𝑛𝑥 + 𝑥 2 𝑠𝑖𝑛𝑦𝑦 ′ − 2𝑥𝑐𝑜𝑠𝑦 − 𝑥𝑦 ′ − 𝑦 = 0
𝑦 ′ (𝑠𝑖𝑛𝑥 + 𝑥 2 𝑠𝑖𝑛𝑦 − 𝑥) = 𝑦 + 2𝑥𝑐𝑜𝑠𝑦 − 𝑦𝑐𝑜𝑠𝑥
𝑦+2𝑥𝑐𝑜𝑠𝑦−𝑦𝑐𝑜𝑠𝑥
𝑦′ = 𝑠𝑖𝑛𝑥+𝑥 2 𝑠𝑖𝑛𝑦−𝑥
𝑑𝑦 𝑦+2𝑥𝑐𝑜𝑠𝑦−𝑦𝑐𝑜𝑠𝑥
𝑑𝑥
= 𝑠𝑖𝑛𝑥+𝑥 2 𝑠𝑖𝑛𝑦−𝑥
𝑑2 𝑦
6.1 Derivatives of Trigonometric Functions
If u is a differentiable function of x, then the following are differentiation formulas of the trigonometric
functions
𝑑 𝑑𝑢
(𝑠𝑖𝑛 𝑢) = 𝑐𝑜𝑠 𝑢
𝑑𝑥 𝑑𝑥
𝑑 𝑑𝑢
(𝑐𝑜𝑠 𝑢) = − 𝑠𝑖𝑛 𝑢
𝑑𝑥 𝑑𝑥
𝑑 2 𝑑𝑢
(𝑡𝑎𝑛 𝑢) = 𝑠𝑒𝑐 𝑢
𝑑𝑥 𝑑𝑥
𝑑 𝑑𝑢
(𝑐𝑜𝑡 𝑢) = − 𝑐𝑠𝑐 2 𝑢
𝑑𝑥 𝑑𝑥
𝑑 𝑑𝑢
(𝑠𝑒𝑐 𝑢) = 𝑡𝑎𝑛 𝑢 𝑠𝑒𝑐 𝑢
𝑑𝑥 𝑑𝑥
𝑑 𝑑𝑢
(𝑐𝑠𝑐 𝑢) = − 𝑐𝑜𝑡 𝑢 𝑐𝑠𝑐 𝑢
𝑑𝑥 𝑑𝑥
Example 1 Determine the derivative of each function.
1+𝑐𝑜𝑠 2𝑥 csc x - cot x
(a) 𝑦 = 𝑥 𝑡𝑎𝑛 ( 𝑥 2 - x)
2
(b) 𝑦 = (c) 𝑦 =
𝑠𝑖𝑛 2𝑥 𝑐𝑠𝑐𝑥+𝑐𝑜𝑡 𝑥
Solution:
(a) Differentiating with respect to x, we have
𝑦 ′ = 𝑥(2 tan(𝑥 2 - x))𝑠𝑒𝑐 2 (𝑥 2 - x)(2𝑥 − 1) + 𝑡𝑎𝑛2 ( 𝑥 2 - x)
𝑦 ′ = 𝑡𝑎𝑛( 𝑥 2 - x)[2𝑥(2𝑥 − 1)𝑠𝑒𝑐 2 (𝑥 2 - x) + 𝑡𝑎𝑛( 𝑥 2 - x)]
(b) Differentiating with respect to x, we have
𝑠𝑖𝑛2𝑥(−2𝑠𝑖𝑛2𝑥)−(1+𝑐𝑜𝑠2𝑥)(2𝑐𝑜𝑠2𝑥)
𝑦′ = (𝑠𝑖𝑛2𝑥)2
−2𝑠𝑖𝑛2 2𝑥−2𝑐𝑜𝑠2𝑥−2𝑐𝑜𝑠2 2𝑥
𝑦′ = 𝑠𝑖𝑛2 2𝑥
−2(𝑠𝑖𝑛2 2𝑥+𝑐𝑜𝑠2 2𝑥)−2𝑐𝑜𝑠2𝑥
𝑦′ = but 𝑠𝑖𝑛2 2𝑥 + 𝑐𝑜𝑠 2 2𝑥 = 1
𝑠𝑖𝑛2 2𝑥
−2−2𝑐𝑜𝑠2𝑥 1+𝑐𝑜𝑠2𝑥
So 𝑦 ′ = , however = 𝑐𝑜𝑠 2 𝑥 and 𝑠𝑖𝑛2𝑥 = 2𝑠𝑖𝑛𝑥𝑐𝑜𝑠𝑥
𝑠𝑖𝑛2 2𝑥 2
−2(1+𝑐𝑜𝑠2𝑥) −4𝑐𝑜𝑠2 𝑥
Thus, 𝑦 ′ = (𝑠𝑖𝑛2𝑥)2
= 4𝑠𝑖𝑛2 𝑥𝑐𝑜𝑠2 𝑥
−1
𝑦 ′ = 𝑠𝑖𝑛2 𝑥
𝑦 ′ = −𝑐𝑠𝑐 2 𝑥
(c) Differentiating with respect to x, we have
(𝑐𝑠𝑐𝑥+𝑐𝑜𝑡𝑥)(−𝑐𝑠𝑐𝑥𝑐𝑜𝑡𝑥+𝑐𝑠𝑐 2 𝑥)−(𝑐𝑠𝑐𝑥−𝑐𝑜𝑡𝑥)(−𝑐𝑠𝑐𝑥𝑐𝑜𝑡𝑥−𝑐𝑠𝑐 2 𝑥)
𝑦′ = (𝑐𝑠𝑐𝑥+𝑐𝑜𝑡𝑥)2
𝑐𝑠𝑐𝑥(𝑐𝑠𝑐𝑥+𝑐𝑜𝑡𝑥)(𝑐𝑠𝑐𝑥−𝑐𝑜𝑡𝑥)+𝑐𝑠𝑐𝑥(𝑐𝑠𝑐𝑥−𝑐𝑜𝑡𝑥)(𝑐𝑠𝑐𝑥+𝑐𝑜𝑡𝑥)
𝑦′ = (𝑐𝑠𝑐𝑥+𝑐𝑜𝑡𝑥)2
2𝑐𝑠𝑥(𝑐𝑠𝑐𝑥−𝑐𝑜𝑡𝑥)(𝑐𝑠𝑐𝑥+𝑐𝑜𝑡𝑥)
𝑦′ = (𝑐𝑠𝑐𝑥+𝑐𝑜𝑡𝑥)2
′ 2𝑐𝑠𝑐𝑥(𝑐𝑠𝑐𝑥−𝑐𝑜𝑡𝑥)
𝑦 = 𝑐𝑠𝑐𝑥+𝑐𝑜𝑡𝑥
𝑑𝑦
Example 2 Find 𝑑𝑥
given that y sin x − x 2 cos y − xy = 0
Solution:
From y sin x − x 2 cos y − xy = 0
Differentiating this implicitly with respect to x, we get
𝑦𝑐𝑜𝑠𝑥 + 𝑦 ′ 𝑠𝑖𝑛𝑥 + 𝑥 2 𝑠𝑖𝑛𝑦𝑦 ′ − 2𝑥𝑐𝑜𝑠𝑦 − 𝑥𝑦 ′ − 𝑦 = 0
𝑦 ′ (𝑠𝑖𝑛𝑥 + 𝑥 2 𝑠𝑖𝑛𝑦 − 𝑥) = 𝑦 + 2𝑥𝑐𝑜𝑠𝑦 − 𝑦𝑐𝑜𝑠𝑥
𝑦+2𝑥𝑐𝑜𝑠𝑦−𝑦𝑐𝑜𝑠𝑥
𝑦′ = 𝑠𝑖𝑛𝑥+𝑥 2 𝑠𝑖𝑛𝑦−𝑥
𝑑𝑦 𝑦+2𝑥𝑐𝑜𝑠𝑦−𝑦𝑐𝑜𝑠𝑥
𝑑𝑥
= 𝑠𝑖𝑛𝑥+𝑥 2 𝑠𝑖𝑛𝑦−𝑥
𝑑2 𝑦
