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Calculus 3-7 CHAIN RULE.

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Presentation on theme: "Calculus 3-7 CHAIN RULE."β€” Presentation transcript:

1 Calculus 3-7 CHAIN RULE

2 Chain Rule If 𝑓 and 𝑔 are differentiable, then the composite function π‘“βˆ˜π‘” π‘₯ =𝑓 𝑔 π‘₯ is differentiable and 𝑓 𝑔 π‘₯ β€² = 𝑓 β€² 𝑔 π‘₯ 𝑔 β€² π‘₯

3 𝑓 π‘₯ = π‘₯ 4 +1 𝑓 π‘₯ = 𝑔(π‘₯) 𝑔 π‘₯ = π‘₯ 4 +1 𝑓 β€² π‘₯ = 1 2 𝑔 π‘₯ βˆ’ 1 2 𝑔 β€² π‘₯ 𝑓 β€² π‘₯ = 4 π‘₯ 3 2 π‘₯ 4 +1

4 Leibniz Notation 𝑑𝑦 𝑑π‘₯ = 𝑓 β€² 𝑒 𝑔 β€² π‘₯ = 𝑑𝑓 𝑑𝑒 βˆ— 𝑑𝑒 𝑑π‘₯

5 𝑓 π‘₯ = 8 π‘₯ 𝑓 𝑒 = 𝑒 3 𝑓 β€² 𝑒 =3 𝑒 2 𝑒=𝑔 π‘₯ =8 π‘₯ 4 +5 𝑔 β€² π‘₯ =32 π‘₯ 3 𝑓 β€² π‘₯ = 𝑓 β€² 𝑒 𝑔 β€² π‘₯ 𝑓 β€² π‘₯ = 8 π‘₯ (32 π‘₯ 3 )

6 𝑓 π‘₯ = tan (4βˆ’3π‘₯) sec (3βˆ’4π‘₯) 𝑓 β€² π‘₯ =𝑔 π‘₯ β„Ž β€² π‘₯ +β„Ž π‘₯ 𝑔 β€² π‘₯ Product Rule 𝑔 π‘₯ = tan (4βˆ’3π‘₯) β„Ž π‘₯ = sec (3βˆ’4π‘₯) 𝑔 𝑒 = tan (𝑒) 𝑒=4βˆ’3π‘₯ 𝑔 β€² 𝑒 =𝑒′ sec 2 𝑒 𝑒 β€² =βˆ’3 𝑔 β€² π‘₯ =βˆ’3 sec 2 (4βˆ’3π‘₯)

7 𝑓 π‘₯ = tan (4βˆ’3π‘₯) sec (3βˆ’4π‘₯) β„Ž 𝑣 = sec (𝑣) 𝑣=3βˆ’4π‘₯ β„Ž β€² 𝑣 =𝑣′ sec (𝑣) tan (𝑣) 𝑣 β€² =βˆ’4 β„Ž β€² π‘₯ =βˆ’4 sec 3βˆ’4π‘₯ tan 3βˆ’4π‘₯ 𝑓 β€² π‘₯ = tan 4βˆ’3π‘₯ βˆ’4 sec 3βˆ’4π‘₯ tan 3βˆ’4π‘₯ + sec (3βˆ’4π‘₯) βˆ’3 sec 2 4βˆ’3π‘₯

8 𝑓 π‘₯ = sin ( π‘₯ 2 +4π‘₯) 𝑓 𝑒 = sin 𝑒 𝑓 β€² 𝑒 =𝑒′ cos 𝑒 𝑒= π‘₯ 2 +4π‘₯ 𝑒 β€² =2π‘₯+4 𝑓 β€² π‘₯ = (2x+4)cos ( π‘₯ 2 +4π‘₯)

9 Problems 3.7 #29-61 odd, 73, 75


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