3.5 – derivative of trigonometric functions derivative notation review: function notation
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3.5 – Derivative of Trigonometric Functions
Derivative Notation
REVIEW:
𝑢= 𝑓 (𝑥) 𝑣=𝑔(𝑥 )𝑦= 𝑓 (𝑥 )
Function Notation
𝑦 ′= 𝑓 ′ (𝑥 )=𝑑𝑦𝑑𝑥
=𝑑𝑑𝑥
( 𝑓 (𝑥 ) )= 𝑑𝑑𝑥
𝑦=𝐷𝑥 𝑓 (𝑥 )=𝐷𝑥 𝑦
𝑢′= 𝑓 ′ (𝑥 )=𝑑𝑢𝑑𝑥
=𝑑𝑑𝑥
( 𝑓 (𝑥 ) )= 𝑑𝑑𝑥
𝑢=𝐷𝑥 𝑓 (𝑥 )=𝐷𝑥𝑢
𝑠=𝑠 (𝑡)
𝑣 ′=𝑔′ (𝑥 )= 𝑑𝑣𝑑𝑥
=𝑑𝑑𝑥
(𝑔 (𝑥 ) )= 𝑑𝑑𝑥
𝑣=𝐷𝑥 𝑓 (𝑥 )=𝐷𝑥𝑣
𝑠′=𝑠 ′ (𝑡 )=𝑑𝑠𝑑𝑡
=𝑑𝑑𝑡
(𝑠 (𝑡))= 𝑑𝑑𝑡
𝑠=𝐷 𝑡𝑠 (𝑡 )=𝐷𝑡 𝑠
3.5 – Derivative of Trigonometric Functions
Derivative of Cosine
𝑦= 𝑓 (𝑥 )=cos 𝑥
𝑦= 𝑓 (𝑥 )=sin 𝑥Derivative of Sine
𝑦 ′= 𝑓 ′ (𝑥 )=𝑑𝑦𝑑𝑥
=cos𝑥
𝑦 ′= 𝑓 ′ (𝑥 )= 𝑑𝑦𝑑𝑥
=− sin𝑥
Derivative of Tangent
𝑦= 𝑓 (𝑥 )=tan𝑥
𝑦 ′= 𝑓 ′ (𝑥 )= 𝑑𝑦𝑑𝑥
=𝑠𝑒𝑐2𝑥
Derivative of Secant
𝑦= 𝑓 (𝑥 )=sec 𝑥
𝑦= 𝑓 (𝑥 )=csc 𝑥Derivative of Cosecant
𝑦 ′= 𝑓 ′ (𝑥 )=𝑑𝑦𝑑𝑥
=−𝑐𝑜𝑡𝑥 𝑐𝑠𝑐𝑥
𝑦 ′= 𝑓 ′ (𝑥 )= 𝑑𝑦𝑑𝑥
=tan𝑥 sec 𝑥
Derivative of Cotangent
𝑦= 𝑓 (𝑥 )=cot 𝑥
𝑦 ′= 𝑓 ′ (𝑥 )= 𝑑𝑦𝑑𝑥
=−𝑐𝑠𝑐2 𝑥
3.5 – Derivative of Trigonometric Functions
3.5 – Derivative of Trigonometric Functions
3.5 – Derivative of Trigonometric Functions
Practice Problems – Worksheet Derivatives of Trigonometric Functions
3.6 – The Chain RuleReview of the Product Rule:
𝑦=(3𝑥3+2 𝑥2 )2¿ (3𝑥3+2 𝑥2 ) (3 𝑥3+2𝑥2 )
𝑦 ′= (3 𝑥3+2𝑥2 ) (9 𝑥2+4 𝑥 )+(9 𝑥2+4 𝑥 ) (3 𝑥3+2𝑥2 )
𝑦 ′=2 (3𝑥3+2 𝑥2 ) (9 𝑥2+4 𝑥 )
𝑦=(6 𝑥2+𝑥 )3¿ (6 𝑥2+𝑥 ) (6 𝑥2+𝑥 ) (6𝑥2+𝑥 )+
𝑦 ′=3 (6 𝑥2+𝑥 )2 (12𝑥+1 )
𝑦 ′=(6 𝑥2+𝑥 )2 (12 𝑥+1 )+ (6 𝑥2+𝑥 )2 (12𝑥+1 )+(6 𝑥2+𝑥 )2 (12𝑥+1 )
3.6 – The Chain Rule
Additional Problems:
𝑦=(3𝑥3+2 𝑥2 )2 𝑦 ′=2 (3𝑥3+2 𝑥2 ) (9 𝑥2+4 𝑥 )
𝑦=(6 𝑥2+𝑥 )3 𝑦 ′=3 (6 𝑥2+𝑥 )2 (12𝑥+1 )
𝑦=(𝑥3+2 𝑥 )9 (𝑥3+2 𝑥 )89 (3 𝑥2+2 )
𝑦=(5 𝑥2+1 )4 (5 𝑥2+1 )34 (10 𝑥 )
𝑦 ′=¿𝑦 ′=¿
𝑦=(2𝑥5−3𝑥4+𝑥−3 )13 (2 𝑥5−3𝑥4+𝑥−3 )1213 (10 𝑥4−12 𝑥3+1 )𝑦 ′=¿
3.6 – The Chain Rule
Practice Problems – Worksheet The Chain Rule