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calculus
- 1. S Geometric Interpretation ππ = ππ β ππ π ππ β π ππ π· ππ, π ππ πΈ ππ, π ππ πππππ = π = Ξπ¦ Ξπ₯ = π ππ β π ππ ππ β ππ πππ ππβπ π ππ β π ππ ππ β ππ πππππ π¨π ππ‘π πππ§π ππ§π π₯π’π§π: β π¦ β π₯ = πππ ππβπ π π₯ + Ξπ₯ β π ππ Ξπ₯ π = π π
- 2. 2.2 The Derivative of a Function The derivative of a function f given by y = f (x) with respect to x at any x in its domain is the number: β π¦ β π₯ = lim Ξπ₯β0 π π₯ + Ξπ₯ β π π₯ Ξπ₯ Other notations for β π¦ β π₯ are βΆ fβ²(x), yβ² To avoid confusion, let us denote our Ξx to h. β π¦ β π₯ = lim ββ0 π π₯ + β β π π₯ β We will be using this formula by definition of the derivative.
- 3. Derivative by Definition EXAMPLE 1: π¦ = π₯2 + 5π₯ + 6 Formula to be used: β π¦ β π₯ = lim ββ0 π π₯ + β β π π₯ β Solution: β π¦ β π₯ = lim ββ0 π₯ + β 2 + 5 π₯ + β + 6 β (π₯2+5π₯ + 6) β = lim ββ0 π₯2 + 2π₯β + β2 + 5π₯ + 5β + 6 β π₯2 β 5π₯ β 6 β = lim ββ0 2π₯β + β2 + 5β β = lim ββ0 β(2π₯ + β + 5) β = lim ββ0 2π₯ + β + 5 = 2π₯ + 0 + 5 β π¦ β π₯ = 2π₯ + 5
- 4. Derivative by Definition EXAMPLE 2: π¦ = 2π₯ β 2 Formula to be used: β π¦ β π₯ = lim ββ0 π π₯ + β β π π₯ β Solution: β π¦ β π₯ = lim ββ0 2(π₯ + β) β 2 β 2π₯ β 2 β = lim ββ0 2(π₯ + β) β 2 β 2π₯ β 2 β π₯ 2(π₯ + β) β 2 + 2π₯ β 2 2(π₯ + β) β 2 + 2π₯ β 2 = lim ββ0 2(π₯ + β) β 2 β (2π₯ β 2) β[ 2(π₯ + β) β 2 + 2π₯ β 2] = lim ββ0 2π₯ + 2β β 2 β 2π₯ + 2 β[ 2π₯ + 2β β 2 + 2π₯ β 2] = lim ββ0 2β β[ 2π₯ + 2β β 2 + 2π₯ β 2] = lim ββ0 2 2π₯ + 2β β 2 + 2π₯ β 2 = 2 2π₯ + 2(0) β 2 + 2π₯ β 2 = 2 2π₯ + 0 β 2 + 2π₯ β 2 = 2 2π₯ β 2 + 2π₯ β 2 β π¦ β π₯ = 2 2 2π₯ β 2
- 6. DIFFERENTATION RULES There are standard formulas called differentiation formulas or differentiation rules which will enable us to find the derivative of even complicated functions as rapidly as we can write.
- 7. ο§ Rule 1: The Constant Rule If f(x) = c where c is constant, then fβ(x) = 0. The derivative of a constant is just equal to zero. β β π₯ (c) = 0 EXAMPLE 1 : y = 5 EXAMPLE 2: EXAMPLE 3: y = 5Ο EXAMPLE 4: y = 1000 = β β π₯ (1000) yβ = 0 y = π = β β π₯ ( π) yβ = 0 = β β π₯ (5) yβ = 0 = β β π₯ (5Ο) yβ = 0 DiFFERENTiATiON RULES
- 8. ο§ Rule 2: The Power Rule If f(x) = π₯π, then fβ (x) = nπ₯πβ1 β β π₯ (π₯π ) = nπ₯πβ1 EXAMPLE 1: = β β π₯ (π₯4) EXAMPLE 3: y = π EXAMPLE 2: y = π₯6 EXAMPLE 4: y = 1 π₯2 y = π₯4 yβ = 4 π₯4β1 yβ = 4 π₯3 yβ = (6) π₯6β1 yβ = 6 π₯5 = β β π₯ ( π₯6) = (π₯) 1 2 = π₯β2 yβ = β2π₯βπ yβ = π π (π₯) 1 2 β1 = π π (π₯)β 1 2 yβ = 1 2 π₯ 1 2 = 1 2 π₯ yβ = β2π₯β2β1 or = β2 π₯3 DiFFERENTiATiON RULES
- 9. ο§ Rule 3: The Identity Function Rule DiFFERENTiATiON RULES β β π₯ (x) = 1 If f(x) = x where x is not raised to any power, then fβ(x) = 1. The derivative of x is always 1. EXAMPLE 1 : y = x = β β π₯ (x) = 1
- 10. ο§ Rule 4: The Constant Multiple Rule EXAMPLE 1 : y = 5π₯8 EXAMPLE 2 : EXAMPLE 3 : y = 4π₯3 y = 1 4 π₯8 yβ = 8 5π₯8β1 yβ = 3 4π₯3β1 β β π₯ (c f (x) = c fβ (x)) . . yβ = 40π₯7 yβ = 12π₯2 yβ = 8 1 4 π₯8β1 yβ = 2π₯7 If f(x) = c f (x), where c is constant, then fβ (x) = c fβ (x) . . DiFFERENTiATiON RULES
- 11. ο§ Rule 5: The Sum/Difference Rule EXAMPLE 1 : y = 5x + 3 EXAMPLE 2 : EXAMPLE 3 : y = 4π₯3 β ππ y = πππ + πππ yβ = π π + π yβ = 3 4π₯3β1 β π(π) β β π₯ (f(x)Β±g(x))= fβ(x) Β±gβ²(x) yβ = π yβ = 12π₯2 β π yβ = π ππ₯πβ1 + π yβ = 6π DiFFERENTiATiON RULES
- 12. ο§ Rule 6 : The Product Rule EXAMPLE 1 : y = 4π₯2 + 9 5π₯3 β 6 πβ² π₯ = 8π₯ f π₯ = 4π₯2 + 9 g π₯ = 5π₯3 β 6 gβ π₯ = 15π₯2 π¦β² = 4π₯2 + 9 15π₯2 + 8π₯ 5π₯3 β 6 π¦β² = 60π₯4 + 135π₯2 + 40π₯4 β 48π₯ πβ² = πππππ + πππππ β πππ EXAMPLE 2 :y = ππ₯2 β 4 π₯π β 3π₯ f π₯ = ππ₯2 β 4 g π₯ = π₯π β 3π₯ πβ² π₯ = 6π₯ πβ² π₯ = 2π₯ β 3 π¦β² = ππ₯2 β 4 2π₯ β 3 + 6π₯ π₯π β 3π₯ π¦β² = 6π₯3 β 9π₯2 β 8π₯ + 12 + 6π₯3 β 18π₯3 πβ² = βπππ β πππ β ππ + ππ DiFFERENTiATiON RULES
- 13. ο§ Rule 7: The Quotient Rule EXAMPLE 1: π¦ = 10π₯3 β 3π₯ π₯ πβ² π₯ = 30π₯2 β 3 f π₯ = 10π₯π β ππ g π₯ = π πβ² π₯ = π π¦β² = π₯ 30π₯2 β 3 β 10π₯3 β 3π₯ β 1 π₯2 π¦β² = 30π₯3 β 3π₯ β 10π₯3 + 3π₯ π₯2 π¦β² = 20π₯3 π₯2 = 20π₯ DiFFERENTiATiON RULES
- 14. ο§ Rule 8: The Chain Rule EXAMPLE 1: y = 3π₯ + 1 π EXAMPLE 2: yβ = 7 3π₯ + 1 πβπ 3 1 + 0 yβ = 7 3π₯ + 1 πβπ 3 yβ = 21 3π₯ + 1 π y = 5π₯2 β 3π₯π π yβ = 5 5π₯3 β 3π₯π πβπ 3 5π₯3β1 β 2 3π₯2β1 yβ = 5 5π₯3 β 3π₯π πβπ 3 5π₯3β1 β 2 3π₯2β1 yβ = 5 5π₯3 β 3π₯π π 15π₯2 β 6π₯ yβ = 5 5π₯3 β 3π₯π π 15π₯2 β 6π₯ DiFFERENTiATiON RULES
- 15. Differentiation Rules β π¦ β π₯ = lim Ξπ₯β0 π π₯ + Ξπ₯ β π π₯ Ξπ₯ ο§ Rule 1: The Constant Rule β β π₯ (c) = 0 ο§ Rule 2: The Power Rule β β π₯ (π₯π) = nπ₯πβ1 ο§ Rule 3: The Identity Function Rule β β π₯ (x) = 1 ο§ Rule 4: The Constant Multiple Rule β β π₯ (c f (x) = c fβ (x) . . ο§ Rule 5: The Sum/Difference Rule β β π₯ (f(x)Β±g(x))= fβ(x) Β±gβ²(x) ο§ Rule 6 : The Product Rule ο§ Rule 7: The Quotient Rule