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