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Calculus 3-7 CHAIN RULE
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Chain Rule If π and π are differentiable, then the composite function πβπ π₯ =π π π₯ is differentiable and π π π₯ β² = π β² π π₯ π β² π₯
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π π₯ = π₯ 4 +1 π π₯ = π(π₯) π π₯ = π₯ 4 +1 π β² π₯ = 1 2 π π₯ β 1 2 π β² π₯ π β² π₯ = 4 π₯ 3 2 π₯ 4 +1
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Leibniz Notation ππ¦ ππ₯ = π β² π’ π β² π₯ = ππ ππ’ β ππ’ ππ₯
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π π₯ = 8 π₯ π π’ = π’ 3 π β² π’ =3 π’ 2 π’=π π₯ =8 π₯ 4 +5 π β² π₯ =32 π₯ 3 π β² π₯ = π β² π’ π β² π₯ π β² π₯ = 8 π₯ (32 π₯ 3 )
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π π₯ = tan (4β3π₯) sec (3β4π₯) π β² π₯ =π π₯ β β² π₯ +β π₯ π β² π₯ Product Rule π π₯ = tan (4β3π₯) β π₯ = sec (3β4π₯) π π’ = tan (π’) π’=4β3π₯ π β² π’ =π’β² sec 2 π’ π’ β² =β3 π β² π₯ =β3 sec 2 (4β3π₯)
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π π₯ = tan (4β3π₯) sec (3β4π₯) β π£ = sec (π£) π£=3β4π₯ β β² π£ =π£β² sec (π£) tan (π£) π£ β² =β4 β β² π₯ =β4 sec 3β4π₯ tan 3β4π₯ π β² π₯ = tan 4β3π₯ β4 sec 3β4π₯ tan 3β4π₯ + sec (3β4π₯) β3 sec 2 4β3π₯
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π π₯ = sin ( π₯ 2 +4π₯) π π’ = sin π’ π β² π’ =π’β² cos π’ π’= π₯ 2 +4π₯ π’ β² =2π₯+4 π β² π₯ = (2x+4)cos ( π₯ 2 +4π₯)
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Problems 3.7 #29-61 odd, 73, 75
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