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Quotient Rule Polynomials
Silent Teacher Intelligent Practice Narration Your Turn π ππ₯ 2π₯ π₯β1 π ππ₯ π₯ 2 βπ₯+5 π₯ 3 β2π₯β1 Practice
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Find ππ¦ ππ₯ of 9 π₯ 2 β1 3π₯ Find ππ¦ ππ₯ of 2 π₯ 2 βπ₯ 2π₯
Worked Example Your Turn Find ππ¦ ππ₯ of 9 π₯ 2 β1 3π₯ Find ππ¦ ππ₯ of 2 π₯ 2 βπ₯ 2π₯ Differentiate 2 π₯ 2 β2π₯+1 3π₯β1 Differentiate 3 π₯ 2 +5π₯β1 4π₯+1 The teacher can show both ways of doing these too, via simplifying or via the quotient rule and then simplifying. @DrChris_Baker
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1. π ππ₯ 4π₯β1 π₯ 8. π ππ₯ 4π₯ π₯β1 2 2. π ππ₯ π₯ 4π₯β1 9. π ππ₯ 4π₯β1 π₯β1 2
1. π ππ₯ 4π₯β1 π₯ 2. π ππ₯ π₯ 4π₯β1 3. π ππ₯ π₯ 2 β4π₯β1 π₯ 2 4. π ππ₯ π₯ 2 π₯ 2 β4π₯β1 5. π ππ₯ 2π₯+1 4π₯β1 6. π ππ₯ 2π₯+2 4π₯β1 7. π ππ₯ 2π₯+1 4π₯+1 8. π ππ₯ 4π₯ π₯β1 2 9. π ππ₯ 4π₯β1 π₯β1 2 10. π ππ₯ 4π₯β1 π₯β1 3 11. π ππ₯ β2 π₯ 2 +4π₯β1 π₯β1 2 12. π ππ₯ 5 π₯ 3 β2 π₯ 2 +4π₯β1 π₯β1 2 π ππ₯ 4π₯β1 π(π₯) 14. π ππ₯ π π₯ 4π₯β1 @DrChris_Baker
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13. π ππ₯ 4π₯β1 π(π₯) = 1β4π₯ π β² π₯ +4π(π₯) π π₯ 2
8. π ππ₯ 4π₯ π₯β1 2 =β 4(π₯+1) π₯β1 3 9. π ππ₯ 4π₯β1 π₯β1 2 =β 2(2π₯+1) π₯β1 3 10. π ππ₯ 4π₯β1 π₯β1 3 =β 8π₯+1 π₯β1 4 11. π ππ₯ β2 π₯ 2 +4π₯β1 π₯β1 2 =β 2 π₯β1 3 12. π ππ₯ 5 π₯ 3 β2 π₯ 2 +4π₯β1 π₯β1 2 = 5 π₯ 3 β15 π₯ 2 β2 π₯β1 3 π ππ₯ 4π₯β1 π(π₯) = 1β4π₯ π β² π₯ +4π(π₯) π π₯ 2 14. π ππ₯ π π₯ 4π₯β1 = 4π₯β1 π β² π₯ βπ(π₯) 4π₯β1 2 1. π ππ₯ 4π₯β1 π₯ = 1 π₯ 2 2. π ππ₯ π₯ 4π₯β1 =β 1 4π₯β1 2 3. π ππ₯ π₯ 2 β4π₯β1 π₯ 2 = 4π₯+2 π₯ 3 4. π ππ₯ π₯ 2 π₯ 2 β4π₯β1 =β 2π₯(2π₯+1) π₯ 2 β4π₯β1 2 5. π ππ₯ 2π₯+1 4π₯β1 =β 6 4π₯β1 2 6. π ππ₯ 2π₯+2 4π₯β1 =β π₯β1 2 7. π ππ₯ 2π₯+1 4π₯+1 =β 2 4π₯+1 2 @DrChris_Baker
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