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Differentiation (Part 1).pptx

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SakibAhmed402053

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Department of GED Daffodil International University
Function 2 If π‘₯ and 𝑦 are two variables related to one another in such a way that each value of π‘₯ determines exactly one value of 𝑦, then we say that 𝑦 is a function of π‘₯ and it is simply written as 𝑦 = 𝑓 π‘₯ , where π‘₯ is an independent variable and 𝑦 is a dependent variable. The value of 𝑦 or 𝑓(π‘₯) is called functional value. 𝑦 = π‘₯2 is a function For π‘₯ = Β±1, Β±2, Β±3, … … … We get, 𝑦 = 1, 4, 9, … … … 𝑦2 = π‘₯ π‘œπ‘Ÿ, (𝑦 = Β± π‘₯) is not a function For π‘₯ = 1, 4, 9, … … … We get, 𝑦 = Β±1, Β±2, Β±3, … … … 1 1 -1 x y 4 2 -2 9 3 -3 1 1 -1 y x 4 2 -2 9 3 -3
Concept of Derivative 3 Differentiation β†’ Instantaneous rate of change of a function with respect to one of its variables β‰ˆ to find the slope of the tangent line to the function at a point. 𝑆𝑝𝑒𝑒𝑑 = π‘β„Žπ‘Žπ‘›π‘”π‘’ π‘œπ‘“ π‘π‘œπ‘ π‘–π‘‘π‘–π‘œπ‘› π‘β„Žπ‘Žπ‘›π‘”π‘’ π‘œπ‘“ π‘‘π‘–π‘šπ‘’ = Ξ”π‘₯ Δ𝑑 Δ𝑑 Ξ”π‘₯ 𝑑 π‘₯ 5 20 25 15 0 10 0 2 3 4 7 8 5 6 1 Time Position 9 𝑑 π‘₯ 5 20 25 15 0 10 0 2 3 4 7 8 5 6 1 Time Position 9
Geometrical Representation of Derivative 4 Gradient/Slope = change in 𝑦 change in π‘₯ = 𝑓 π‘₯ + β„Ž βˆ’ 𝑓(π‘₯) π‘₯ + β„Ž βˆ’ π‘₯ = 𝑓 π‘₯ + β„Ž βˆ’ 𝑓(π‘₯) h π‘₯ π‘₯ + β„Ž π‘₯ 𝑦 π‘₯ + β„Ž, 𝑓 π‘₯ + β„Ž π‘₯, 𝑓 π‘₯ 𝑓 π‘₯ + β„Ž 𝑓 π‘₯ β„Ž Secant Line
Geometrical Representation of Derivative 5 π‘₯ π‘₯ + β„Ž π‘₯ 𝑦 π‘₯ + β„Ž, 𝑓 π‘₯ + β„Ž π‘₯, 𝑓 π‘₯ 𝑓 π‘₯ + β„Ž 𝑓 π‘₯ β„Ž Gradient/Slope = change in 𝑦 change in π‘₯ = 𝑓 π‘₯ + β„Ž βˆ’ 𝑓(π‘₯) π‘₯ + β„Ž βˆ’ π‘₯ = 𝑓 π‘₯ + β„Ž βˆ’ 𝑓(π‘₯) h Secant Line
Geometrical Representation of Derivative 6 π‘₯ π‘₯ + β„Ž π‘₯ 𝑦 π‘₯ + β„Ž, 𝑓 π‘₯ + β„Ž π‘₯, 𝑓 π‘₯ 𝑓 π‘₯ + β„Ž 𝑓 π‘₯ β„Ž Gradient/Slope = change in 𝑦 change in π‘₯ = 𝑓 π‘₯ + β„Ž βˆ’ 𝑓(π‘₯) π‘₯ + β„Ž βˆ’ π‘₯ = 𝑓 π‘₯ + β„Ž βˆ’ 𝑓(π‘₯) β„Ž Secant Line
Geometrical Representation of Derivative 7 First Principle: A function 𝑓(π‘₯) is differentiable at a point π‘₯ if lim β„Žβ†’0 𝑓 π‘₯ + β„Ž βˆ’ 𝑓(π‘₯) β„Ž exists (is finite). ∴ 𝑑𝑦 𝑑π‘₯ = 𝑓′ π‘₯ = lim β„Žβ†’0 𝑓 π‘₯ + β„Ž βˆ’ 𝑓(π‘₯) β„Ž π‘₯ π‘₯ 𝑦 π‘₯, 𝑓 π‘₯ 𝑓 π‘₯ β„Ž β†’ 0 Gradient/Slope = change in 𝑦 change in π‘₯ = 𝑓 π‘₯ + β„Ž βˆ’ 𝑓(π‘₯) π‘₯ + β„Ž βˆ’ π‘₯ = 𝑓 π‘₯ + β„Ž βˆ’ 𝑓(π‘₯) β„Ž Secant Line Tangent Line
Example 1: Find the derivative of 𝒇 𝒙 = 𝐜 using first principle 8 Let 𝑦 = 𝑓 π‘₯ = 𝑐 𝑓 π‘₯ = 𝑐 𝑓 π‘₯ + β„Ž = 𝑐 = lim β„Žβ†’0 𝑐 βˆ’ 𝑐 β„Ž From First Principle, 𝑑𝑦 𝑑π‘₯ = 𝑓′(π‘₯) = lim β„Žβ†’0 𝑓 π‘₯ + β„Ž βˆ’ 𝑓(π‘₯) β„Ž = lim β„Žβ†’0 0 β„Ž = lim β„Žβ†’0 0 = 0 ∴ π’…π’š 𝒅𝒙 = 𝒅 𝒅𝒙 𝒄 = 𝟎
Example 2: Find the derivative of 𝒇 𝒙 = 𝒙 using first principle 9 Let 𝑦 = 𝑓 π‘₯ = π‘₯ 𝑓 π‘₯ = π‘₯ 𝑓 π‘₯ + β„Ž = π‘₯ + β„Ž = lim β„Žβ†’0 π‘₯ + β„Ž βˆ’ π‘₯ β„Ž From First Principle, 𝑑𝑦 𝑑π‘₯ = 𝑓′(π‘₯) = lim β„Žβ†’0 𝑓 π‘₯ + β„Ž βˆ’ 𝑓(π‘₯) β„Ž = lim β„Žβ†’0 β„Ž β„Ž = lim β„Žβ†’0 1 = 1 ∴ π’…π’š 𝒅𝒙 = 𝒅 𝒅𝒙 𝒙 = 𝟏
Example 3: Find the derivative of 𝒇 𝒙 = 𝒙𝒏 using first principle 10 Let 𝑦 = 𝑓 π‘₯ = π‘₯𝑛 𝑓 π‘₯ = π‘₯𝑛 𝑓 π‘₯ + β„Ž = π‘₯ + β„Ž 𝑛 = lim β„Žβ†’0 π‘₯ + β„Ž 𝑛 βˆ’ π‘₯𝑛 β„Ž From First Principle, 𝑑𝑦 𝑑π‘₯ = 𝑓′(π‘₯) = lim β„Žβ†’0 𝑓 π‘₯ + β„Ž βˆ’ 𝑓(π‘₯) β„Ž = lim β„Žβ†’0 π‘₯ 1 + β„Ž π‘₯ 𝑛 βˆ’ π‘₯𝑛 β„Ž = lim β„Žβ†’0 π‘₯𝑛 1 + β„Ž π‘₯ 𝑛 βˆ’ π‘₯𝑛 β„Ž ∴ 𝑑𝑦 𝑑π‘₯ = 𝑑 𝑑π‘₯ π‘₯𝑛 = 𝑛π‘₯π‘›βˆ’1 = π‘₯𝑛 lim β„Žβ†’0 1 + β„Ž π‘₯ 𝑛 βˆ’ 1 β„Ž 𝑑𝑦 𝑑π‘₯ = π‘₯𝑛 lim β„Žβ†’0 1 + 𝑛 β„Ž π‘₯ + 𝑛 𝑛 βˆ’ 1 2! β„Ž π‘₯ 2 + 𝑛 𝑛 βˆ’ 1 (𝑛 βˆ’ 2) 3! β„Ž π‘₯ 3 + β‹― βˆ’ 1 β„Ž = π‘₯𝑛 lim β„Žβ†’0 𝑛 β„Ž π‘₯ + 𝑛 𝑛 βˆ’ 1 2! β„Ž π‘₯ 2 + 𝑛 𝑛 βˆ’ 1 (𝑛 βˆ’ 2) 3! β„Ž π‘₯ 3 + β‹― β„Ž = π‘₯𝑛 lim β„Žβ†’0 β„Ž 𝑛 1 π‘₯ + 𝑛 𝑛 βˆ’ 1 2! β„Ž π‘₯2 + 𝑛 𝑛 βˆ’ 1 (𝑛 βˆ’ 2) 3! β„Ž2 π‘₯3 + β‹― β„Ž = π‘₯𝑛 lim β„Žβ†’0 𝑛 1 π‘₯ + 𝑛 𝑛 βˆ’ 1 2! β„Ž π‘₯2 + 𝑛 𝑛 βˆ’ 1 (𝑛 βˆ’ 2) 3! β„Ž2 π‘₯3 + β‹― = π‘₯𝑛 𝑛 1 π‘₯ = π‘₯𝑛 𝑛π‘₯βˆ’1 = 𝑛π‘₯π‘›βˆ’1
Example 4: Find the derivative of 𝒇 𝒙 = 𝒆𝒙 using first principle 11 Let 𝑦 = 𝑓 π‘₯ = 𝑒π‘₯ 𝑓 π‘₯ = 𝑒π‘₯ 𝑓 π‘₯ + β„Ž = 𝑒 π‘₯+β„Ž = lim β„Žβ†’0 𝑒 π‘₯+β„Ž βˆ’ 𝑒π‘₯ β„Ž From First Principle, 𝑑𝑦 𝑑π‘₯ = 𝑓′(π‘₯) = lim β„Žβ†’0 𝑓 π‘₯ + β„Ž βˆ’ 𝑓(π‘₯) β„Ž = lim β„Žβ†’0 𝑒π‘₯. π‘’β„Ž βˆ’ 𝑒π‘₯ β„Ž = lim β„Žβ†’0 𝑒π‘₯ π‘’β„Ž βˆ’ 1 β„Ž = 𝑒π‘₯ lim β„Žβ†’0 1 + β„Ž + β„Ž2 2! + β„Ž3 3! + β„Ž4 4! + β‹― βˆ’ 1 β„Ž 𝑑𝑦 𝑑π‘₯ = 𝑒π‘₯ lim β„Žβ†’0 β„Ž + β„Ž2 2! + β„Ž3 3! + β„Ž4 4! + β‹― β„Ž ∡ 𝑒π‘₯ = 1 + π‘₯ + π‘₯2 2! + π‘₯3 3! + π‘₯4 4! + β‹― ∴ π’…π’š 𝒅𝒙 = 𝒅 𝒅𝒙 𝒆𝒙 = 𝒆𝒙 = 𝑒π‘₯ lim β„Žβ†’0 β„Ž 1 + β„Ž 2! + β„Ž2 3! + β„Ž3 4! + β‹― β„Ž = 𝑒π‘₯ lim β„Žβ†’0 1 + β„Ž 2! + β„Ž2 3! + β„Ž3 4! + β‹― = 𝑒π‘₯. 1
Example 5: Find the derivative of 𝒇 𝒙 = ln π‘₯ using first principle 12 Let 𝑦 = 𝑓 π‘₯ = ln π‘₯ 𝑓 π‘₯ = ln π‘₯ 𝑓 π‘₯ + β„Ž = ln π‘₯ + β„Ž = lim β„Žβ†’0 ln π‘₯ + β„Ž π‘₯ β„Ž From First Principle, 𝑑𝑦 𝑑π‘₯ = 𝑓′(π‘₯) = lim β„Žβ†’0 𝑓 π‘₯ + β„Ž βˆ’ 𝑓(π‘₯) β„Ž = lim β„Žβ†’0 ln 1 + β„Ž π‘₯ β„Ž 𝑑𝑦 𝑑π‘₯ = lim β„Žβ†’0 β„Ž 1 π‘₯ βˆ’ 1 2 β„Ž π‘₯2 + 1 3 β„Ž2 π‘₯3 βˆ’ 1 4 β„Ž3 π‘₯4 + β‹― β„Ž ∡ ln 1 + π‘₯ = π‘₯ βˆ’ π‘₯2 2 + π‘₯3 3 βˆ’ π‘₯4 4 + β‹― ∴ π’…π’š 𝒅𝒙 = 𝒅 𝒅𝒙 ln π‘₯ = 1 π‘₯ = lim β„Žβ†’0 1 π‘₯ βˆ’ 1 2 β„Ž π‘₯2 + 1 3 β„Ž2 π‘₯3 βˆ’ 1 4 β„Ž3 π‘₯4 + β‹― = 1 π‘₯ = lim β„Žβ†’0 β„Ž π‘₯ βˆ’ 1 2 β„Ž2 π‘₯2 + 1 3 β„Ž3 π‘₯3 βˆ’ 1 4 β„Ž4 π‘₯4 + β‹― β„Ž
Example 6: Find the derivative of 𝒇 𝒙 = cos 𝒙 using first principle 13 Let 𝑦 = 𝑓 π‘₯ = cos π‘₯ 𝑓 π‘₯ = cos π‘₯ 𝑓 π‘₯ + β„Ž = cos π‘₯ + β„Ž = lim β„Žβ†’0 cos π‘₯ + β„Ž βˆ’ cos π‘₯ β„Ž From First Principle, 𝑑𝑦 𝑑π‘₯ = 𝑓′(π‘₯) = lim β„Žβ†’0 𝑓 π‘₯ + β„Ž βˆ’ 𝑓(π‘₯) β„Ž = lim β„Žβ†’0 βˆ’2 sin 2π‘₯ + β„Ž 2 sin β„Ž 2 β„Ž = βˆ’ lim β„Žβ†’0 sin 2π‘₯ + β„Ž 2 Γ— lim β„Ž 2 β†’0 sin β„Ž 2 β„Ž 2 = βˆ’ sin 2π‘₯ + 0 2 Γ— 1 ∴ π’…π’š 𝒅𝒙 = 𝒅 𝒅𝒙 cos 𝒙 = βˆ’ 𝐬𝐒𝐧 𝒙 cos π‘Ž βˆ’ cos 𝑏 = βˆ’2 sin π‘Ž + 𝑏 2 sin π‘Ž βˆ’ 𝑏 2 lim β„Žβ†’0 sin β„Ž β„Ž = lim β„Žβ†’0 β„Ž sin β„Ž = 1 ∡ β„Ž β†’ 0 ⟹ β„Ž 2 β†’ 0???? β„Ž β†’ 0 ⟹ 1 2 Γ— β„Ž β†’ 1 2 Γ— 0 ⟹ β„Ž 2 β†’ 0
Example 7: Find the derivative of 𝒇 𝒙 = 𝐬𝐒𝐧 𝒂𝒙 using first principle 14 Let 𝑦 = 𝑓 π‘₯ = sin π‘Žπ‘₯ 𝑓 π‘₯ = sin π‘Žπ‘₯ 𝑓 π‘₯ + β„Ž = sin π‘Ž π‘₯ + β„Ž = sin π‘Žπ‘₯ + π‘Žβ„Ž = lim β„Žβ†’0 sin π‘Žπ‘₯ + π‘Žβ„Ž βˆ’ sin π‘Žπ‘₯ β„Ž From First Principle, 𝑑𝑦 𝑑π‘₯ = 𝑓′(π‘₯) = lim β„Žβ†’0 𝑓 π‘₯ + β„Ž βˆ’ 𝑓(π‘₯) β„Ž = lim β„Žβ†’0 2 cos 2π‘Žπ‘₯ + π‘Žβ„Ž 2 sin π‘Žβ„Ž 2 β„Ž = lim β„Žβ†’0 cos 2π‘Žπ‘₯ + π‘Žβ„Ž 2 Γ— lim β„Žβ†’0 sin π‘Žβ„Ž 2 β„Ž 2 = cos 2π‘Žπ‘₯ + π‘Ž. 0 2 Γ— lim β„Žβ†’0 π‘Ž sin π‘Žβ„Ž 2 π‘Žβ„Ž 2 𝑑𝑦 𝑑π‘₯ = (cos π‘Žπ‘₯) Γ— π‘Ž lim π‘Žβ„Ž 2 β†’0 sin π‘Žβ„Ž 2 π‘Žβ„Ž 2 sin π‘Ž βˆ’ sin 𝑏 = 2 π‘π‘œπ‘  π‘Ž + 𝑏 2 𝑠𝑖𝑛 π‘Ž βˆ’ 𝑏 2 ∴ π’…π’š 𝒅𝒙 = 𝒅 𝒅𝒙 𝐬𝐒𝐧 𝒂𝒙 = 𝒂 𝐜𝐨𝐬 𝒂𝒙 = (cos π‘Žπ‘₯) Γ— π‘Ž 1 ∡ β„Ž β†’ 0 ⟹ π‘Žβ„Ž 2 β†’ 0 lim β„Žβ†’0 sin β„Ž β„Ž = lim β„Žβ†’0 β„Ž sin β„Ž = 1
Example 8: Find the derivative of 𝒇 𝒙 = 𝒕𝒂𝒏 𝒙 using first principle 15 Let 𝑦 = 𝑓 π‘₯ = tan π‘₯ 𝑓 π‘₯ = tan π‘₯ 𝑓 π‘₯ + β„Ž = tan π‘₯ + β„Ž = lim β„Žβ†’0 tan π‘₯ + β„Ž βˆ’ tan π‘₯ β„Ž From First Principle, 𝑑𝑦 𝑑π‘₯ = 𝑓′(π‘₯) = lim β„Žβ†’0 𝑓 π‘₯ + β„Ž βˆ’ 𝑓(π‘₯) β„Ž = lim β„Žβ†’0 sin π‘₯ + β„Ž cos π‘₯ + β„Ž βˆ’ sin π‘₯ cos π‘₯ β„Ž = lim β„Žβ†’0 sin π‘₯ + β„Ž cos π‘₯ βˆ’ sin π‘₯ cos π‘₯ + β„Ž cos π‘₯ + β„Ž cos π‘₯ β„Ž = lim β„Žβ†’0 sin π‘₯ + β„Ž cos π‘₯ βˆ’ cos π‘₯ + β„Ž sin π‘₯ β„Ž cos π‘₯ + β„Ž cos π‘₯ 𝑑𝑦 𝑑π‘₯ = lim β„Žβ†’0 sin β„Ž β„Ž cos π‘₯ + β„Ž cos π‘₯ sin π‘Ž cos 𝑏 βˆ’ cos π‘Ž sin 𝑏 = sin π‘Ž βˆ’ 𝑏 ∴ π’…π’š 𝒅𝒙 = 𝒅 𝒅𝒙 𝐭𝐚𝐧 𝒙 = 𝐬𝐞𝐜𝟐 𝒙 = lim β„Žβ†’0 sin β„Ž β„Ž Γ— lim β„Žβ†’0 1 cos π‘₯ + β„Ž cos π‘₯ = 1 Γ— 1 cos π‘₯ + 0 cos π‘₯ = 1 cos2 π‘₯ = sec2 π‘₯ = lim β„Žβ†’0 sin π‘₯ + β„Ž βˆ’ π‘₯ β„Ž cos π‘₯ + β„Ž cos π‘₯
Example 9: Find the derivative of 𝒇 𝒙 = π­πšπ§βˆ’πŸ 𝒙 using first principle 16 Let 𝑦 = 𝑓 π‘₯ = tanβˆ’1 π‘₯ 𝑓 π‘₯ = tanβˆ’1 π‘₯ 𝑓 π‘₯ + β„Ž = tanβˆ’1 π‘₯ + β„Ž 𝑑𝑦 𝑑π‘₯ = lim β„Žβ†’0 tanβˆ’1 π‘₯ + β„Ž βˆ’ tanβˆ’1 π‘₯ β„Ž β‹― (𝑖) From First Principle, 𝑑𝑦 𝑑π‘₯ = 𝑓′(π‘₯) = lim β„Žβ†’0 𝑓 π‘₯ + β„Ž βˆ’ 𝑓(π‘₯) β„Ž Supposetanβˆ’1 π‘₯ = 𝑦 ⟹ π‘₯ = tan 𝑦 andtanβˆ’1 π‘₯ + β„Ž = 𝑦 + π‘˜ ⟹ π‘₯ + β„Ž = tan 𝑦 + π‘˜ ⟹ β„Ž = tan 𝑦 + π‘˜ βˆ’ π‘₯ ⟹ β„Ž = tan 𝑦 + π‘˜ βˆ’ tan 𝑦 π‘₯ + β„Ž = tan 𝑦 + π‘˜ Whenβ„Ž β†’ 0, π‘₯ = tan 𝑦 + π‘˜ ⟹ tanβˆ’1 π‘₯ = 𝑦 + π‘˜ ⟹ tanβˆ’1 π‘₯ = tanβˆ’1 π‘₯ + π‘˜ ⟹ π‘˜ β†’ 0 Thus, β„Ž β†’ 0 ⟹ π‘˜ β†’ 0 In concept of Limit 𝑑𝑦 𝑑π‘₯ = lim β„Žβ†’0 𝑦 + π‘˜ βˆ’ 𝑦 β„Ž 𝑑𝑦 𝑑π‘₯ = lim π‘˜β†’0 π‘˜ tan 𝑦 + π‘˜ βˆ’ tan 𝑦 [∡ β„Ž β†’ 0 ⟹ π‘˜ β†’ 0 ? ? ? ]
Example 9: Find the derivative of 𝒇 𝒙 = π­πšπ§βˆ’πŸ 𝒙 using first principle (cont.…) 17 sin π‘Ž cos 𝑏 βˆ’ cos π‘Ž sin 𝑏 = sin π‘Ž βˆ’ 𝑏 𝑑𝑦 𝑑π‘₯ = lim π‘˜β†’0 π‘˜ tan 𝑦 + π‘˜ βˆ’ tan 𝑦 [∡ β„Ž β†’ 0 ⟹ π‘˜ β†’ 0] = lim π‘˜β†’0 π‘˜ sin 𝑦 + π‘˜ cos 𝑦 + π‘˜ βˆ’ sin 𝑦 cos 𝑦 = lim π‘˜β†’0 π‘˜ sin 𝑦 + π‘˜ cos 𝑦 βˆ’ sin 𝑦 cos 𝑦 + π‘˜ cos 𝑦 + π‘˜ cos 𝑦 = lim π‘˜β†’0 π‘˜ cos 𝑦 + π‘˜ cos 𝑦 sin 𝑦 + π‘˜ cos 𝑦 βˆ’ cos 𝑦 + π‘˜ sin 𝑦 = lim π‘˜β†’0 π‘˜ cos 𝑦 + π‘˜ cos 𝑦 sin 𝑦 + π‘˜ βˆ’ 𝑦 𝑑𝑦 𝑑π‘₯ = lim π‘˜β†’0 π‘˜ sin π‘˜ Γ— lim π‘˜β†’0 cos 𝑦 + π‘˜ cos 𝑦 = 1 Γ— cos 𝑦 + 0 cos 𝑦 = cos2 𝑦 = 1 sec2 𝑦 = 1 1 + tan2 𝑦 = 1 1 + π‘₯2 = lim π‘˜β†’0 π‘˜ cos 𝑦 + π‘˜ cos 𝑦 sin π‘˜ ∴ π’…π’š 𝒅𝒙 = 𝒅 𝒅𝒙 π­πšπ§βˆ’πŸ 𝒙 = 𝟏 𝟏 + π’™πŸ ∡ π‘‘π‘Žπ‘› 𝑦 = π‘₯ lim π‘˜β†’0 sin π‘˜ π‘˜ = lim π‘˜β†’0 π‘˜ sin π‘˜ = 1
Example 10: Find π’…π’š 𝒅𝒙 𝐟𝐨𝐫𝐲 = 𝒙 using first principle 18 Let 𝑦 = 𝑓 π‘₯ = π‘₯ 𝑓 π‘₯ = π‘₯ 𝑓 π‘₯ + β„Ž = π‘₯ + β„Ž = lim β„Žβ†’0 π‘₯ + β„Ž βˆ’ π‘₯ π‘₯ + β„Ž + π‘₯ β„Ž π‘₯ + β„Ž + π‘₯ From First Principle, 𝑑𝑦 𝑑π‘₯ = 𝑓′(π‘₯) = lim β„Žβ†’0 𝑓 π‘₯ + β„Ž βˆ’ 𝑓(π‘₯) β„Ž = lim β„Žβ†’0 π‘₯ + β„Ž 2 βˆ’ π‘₯ 2 β„Ž π‘₯ + β„Ž + π‘₯ = lim β„Žβ†’0 π‘₯ + β„Ž βˆ’ π‘₯ β„Ž π‘₯ + β„Ž + π‘₯ = 1 π‘₯ + 0 + π‘₯ ∴ π’…π’š 𝒅𝒙 = 𝒅 𝒅𝒙 𝒙 = 𝟏 𝟐 𝒙 = lim β„Žβ†’0 β„Ž β„Ž π‘₯ + β„Ž + π‘₯ 𝑑𝑦 𝑑π‘₯ = lim β„Žβ†’0 1 π‘₯ + β„Ž + π‘₯ = 1 π‘₯ + π‘₯ = lim β„Žβ†’0 π‘₯ + β„Ž βˆ’ π‘₯ β„Ž
Exercise 19 Find the derivative of the following functions using first principle: 1. f x = ax 2. f x = cos ax 3. f x = sinβˆ’1 x 4. f x = cosβˆ’1 x 5. f x = sin x2

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Differentiation (Part 1).pptx

  • 1. Department of GED Daffodil International University
  • 2. Function 2 If π‘₯ and 𝑦 are two variables related to one another in such a way that each value of π‘₯ determines exactly one value of 𝑦, then we say that 𝑦 is a function of π‘₯ and it is simply written as 𝑦 = 𝑓 π‘₯ , where π‘₯ is an independent variable and 𝑦 is a dependent variable. The value of 𝑦 or 𝑓(π‘₯) is called functional value. 𝑦 = π‘₯2 is a function For π‘₯ = Β±1, Β±2, Β±3, … … … We get, 𝑦 = 1, 4, 9, … … … 𝑦2 = π‘₯ π‘œπ‘Ÿ, (𝑦 = Β± π‘₯) is not a function For π‘₯ = 1, 4, 9, … … … We get, 𝑦 = Β±1, Β±2, Β±3, … … … 1 1 -1 x y 4 2 -2 9 3 -3 1 1 -1 y x 4 2 -2 9 3 -3
  • 3. Concept of Derivative 3 Differentiation β†’ Instantaneous rate of change of a function with respect to one of its variables β‰ˆ to find the slope of the tangent line to the function at a point. 𝑆𝑝𝑒𝑒𝑑 = π‘β„Žπ‘Žπ‘›π‘”π‘’ π‘œπ‘“ π‘π‘œπ‘ π‘–π‘‘π‘–π‘œπ‘› π‘β„Žπ‘Žπ‘›π‘”π‘’ π‘œπ‘“ π‘‘π‘–π‘šπ‘’ = Ξ”π‘₯ Δ𝑑 Δ𝑑 Ξ”π‘₯ 𝑑 π‘₯ 5 20 25 15 0 10 0 2 3 4 7 8 5 6 1 Time Position 9 𝑑 π‘₯ 5 20 25 15 0 10 0 2 3 4 7 8 5 6 1 Time Position 9
  • 4. Geometrical Representation of Derivative 4 Gradient/Slope = change in 𝑦 change in π‘₯ = 𝑓 π‘₯ + β„Ž βˆ’ 𝑓(π‘₯) π‘₯ + β„Ž βˆ’ π‘₯ = 𝑓 π‘₯ + β„Ž βˆ’ 𝑓(π‘₯) h π‘₯ π‘₯ + β„Ž π‘₯ 𝑦 π‘₯ + β„Ž, 𝑓 π‘₯ + β„Ž π‘₯, 𝑓 π‘₯ 𝑓 π‘₯ + β„Ž 𝑓 π‘₯ β„Ž Secant Line
  • 5. Geometrical Representation of Derivative 5 π‘₯ π‘₯ + β„Ž π‘₯ 𝑦 π‘₯ + β„Ž, 𝑓 π‘₯ + β„Ž π‘₯, 𝑓 π‘₯ 𝑓 π‘₯ + β„Ž 𝑓 π‘₯ β„Ž Gradient/Slope = change in 𝑦 change in π‘₯ = 𝑓 π‘₯ + β„Ž βˆ’ 𝑓(π‘₯) π‘₯ + β„Ž βˆ’ π‘₯ = 𝑓 π‘₯ + β„Ž βˆ’ 𝑓(π‘₯) h Secant Line
  • 6. Geometrical Representation of Derivative 6 π‘₯ π‘₯ + β„Ž π‘₯ 𝑦 π‘₯ + β„Ž, 𝑓 π‘₯ + β„Ž π‘₯, 𝑓 π‘₯ 𝑓 π‘₯ + β„Ž 𝑓 π‘₯ β„Ž Gradient/Slope = change in 𝑦 change in π‘₯ = 𝑓 π‘₯ + β„Ž βˆ’ 𝑓(π‘₯) π‘₯ + β„Ž βˆ’ π‘₯ = 𝑓 π‘₯ + β„Ž βˆ’ 𝑓(π‘₯) β„Ž Secant Line
  • 7. Geometrical Representation of Derivative 7 First Principle: A function 𝑓(π‘₯) is differentiable at a point π‘₯ if lim β„Žβ†’0 𝑓 π‘₯ + β„Ž βˆ’ 𝑓(π‘₯) β„Ž exists (is finite). ∴ 𝑑𝑦 𝑑π‘₯ = 𝑓′ π‘₯ = lim β„Žβ†’0 𝑓 π‘₯ + β„Ž βˆ’ 𝑓(π‘₯) β„Ž π‘₯ π‘₯ 𝑦 π‘₯, 𝑓 π‘₯ 𝑓 π‘₯ β„Ž β†’ 0 Gradient/Slope = change in 𝑦 change in π‘₯ = 𝑓 π‘₯ + β„Ž βˆ’ 𝑓(π‘₯) π‘₯ + β„Ž βˆ’ π‘₯ = 𝑓 π‘₯ + β„Ž βˆ’ 𝑓(π‘₯) β„Ž Secant Line Tangent Line
  • 8. Example 1: Find the derivative of 𝒇 𝒙 = 𝐜 using first principle 8 Let 𝑦 = 𝑓 π‘₯ = 𝑐 𝑓 π‘₯ = 𝑐 𝑓 π‘₯ + β„Ž = 𝑐 = lim β„Žβ†’0 𝑐 βˆ’ 𝑐 β„Ž From First Principle, 𝑑𝑦 𝑑π‘₯ = 𝑓′(π‘₯) = lim β„Žβ†’0 𝑓 π‘₯ + β„Ž βˆ’ 𝑓(π‘₯) β„Ž = lim β„Žβ†’0 0 β„Ž = lim β„Žβ†’0 0 = 0 ∴ π’…π’š 𝒅𝒙 = 𝒅 𝒅𝒙 𝒄 = 𝟎
  • 9. Example 2: Find the derivative of 𝒇 𝒙 = 𝒙 using first principle 9 Let 𝑦 = 𝑓 π‘₯ = π‘₯ 𝑓 π‘₯ = π‘₯ 𝑓 π‘₯ + β„Ž = π‘₯ + β„Ž = lim β„Žβ†’0 π‘₯ + β„Ž βˆ’ π‘₯ β„Ž From First Principle, 𝑑𝑦 𝑑π‘₯ = 𝑓′(π‘₯) = lim β„Žβ†’0 𝑓 π‘₯ + β„Ž βˆ’ 𝑓(π‘₯) β„Ž = lim β„Žβ†’0 β„Ž β„Ž = lim β„Žβ†’0 1 = 1 ∴ π’…π’š 𝒅𝒙 = 𝒅 𝒅𝒙 𝒙 = 𝟏
  • 10. Example 3: Find the derivative of 𝒇 𝒙 = 𝒙𝒏 using first principle 10 Let 𝑦 = 𝑓 π‘₯ = π‘₯𝑛 𝑓 π‘₯ = π‘₯𝑛 𝑓 π‘₯ + β„Ž = π‘₯ + β„Ž 𝑛 = lim β„Žβ†’0 π‘₯ + β„Ž 𝑛 βˆ’ π‘₯𝑛 β„Ž From First Principle, 𝑑𝑦 𝑑π‘₯ = 𝑓′(π‘₯) = lim β„Žβ†’0 𝑓 π‘₯ + β„Ž βˆ’ 𝑓(π‘₯) β„Ž = lim β„Žβ†’0 π‘₯ 1 + β„Ž π‘₯ 𝑛 βˆ’ π‘₯𝑛 β„Ž = lim β„Žβ†’0 π‘₯𝑛 1 + β„Ž π‘₯ 𝑛 βˆ’ π‘₯𝑛 β„Ž ∴ 𝑑𝑦 𝑑π‘₯ = 𝑑 𝑑π‘₯ π‘₯𝑛 = 𝑛π‘₯π‘›βˆ’1 = π‘₯𝑛 lim β„Žβ†’0 1 + β„Ž π‘₯ 𝑛 βˆ’ 1 β„Ž 𝑑𝑦 𝑑π‘₯ = π‘₯𝑛 lim β„Žβ†’0 1 + 𝑛 β„Ž π‘₯ + 𝑛 𝑛 βˆ’ 1 2! β„Ž π‘₯ 2 + 𝑛 𝑛 βˆ’ 1 (𝑛 βˆ’ 2) 3! β„Ž π‘₯ 3 + β‹― βˆ’ 1 β„Ž = π‘₯𝑛 lim β„Žβ†’0 𝑛 β„Ž π‘₯ + 𝑛 𝑛 βˆ’ 1 2! β„Ž π‘₯ 2 + 𝑛 𝑛 βˆ’ 1 (𝑛 βˆ’ 2) 3! β„Ž π‘₯ 3 + β‹― β„Ž = π‘₯𝑛 lim β„Žβ†’0 β„Ž 𝑛 1 π‘₯ + 𝑛 𝑛 βˆ’ 1 2! β„Ž π‘₯2 + 𝑛 𝑛 βˆ’ 1 (𝑛 βˆ’ 2) 3! β„Ž2 π‘₯3 + β‹― β„Ž = π‘₯𝑛 lim β„Žβ†’0 𝑛 1 π‘₯ + 𝑛 𝑛 βˆ’ 1 2! β„Ž π‘₯2 + 𝑛 𝑛 βˆ’ 1 (𝑛 βˆ’ 2) 3! β„Ž2 π‘₯3 + β‹― = π‘₯𝑛 𝑛 1 π‘₯ = π‘₯𝑛 𝑛π‘₯βˆ’1 = 𝑛π‘₯π‘›βˆ’1
  • 11. Example 4: Find the derivative of 𝒇 𝒙 = 𝒆𝒙 using first principle 11 Let 𝑦 = 𝑓 π‘₯ = 𝑒π‘₯ 𝑓 π‘₯ = 𝑒π‘₯ 𝑓 π‘₯ + β„Ž = 𝑒 π‘₯+β„Ž = lim β„Žβ†’0 𝑒 π‘₯+β„Ž βˆ’ 𝑒π‘₯ β„Ž From First Principle, 𝑑𝑦 𝑑π‘₯ = 𝑓′(π‘₯) = lim β„Žβ†’0 𝑓 π‘₯ + β„Ž βˆ’ 𝑓(π‘₯) β„Ž = lim β„Žβ†’0 𝑒π‘₯. π‘’β„Ž βˆ’ 𝑒π‘₯ β„Ž = lim β„Žβ†’0 𝑒π‘₯ π‘’β„Ž βˆ’ 1 β„Ž = 𝑒π‘₯ lim β„Žβ†’0 1 + β„Ž + β„Ž2 2! + β„Ž3 3! + β„Ž4 4! + β‹― βˆ’ 1 β„Ž 𝑑𝑦 𝑑π‘₯ = 𝑒π‘₯ lim β„Žβ†’0 β„Ž + β„Ž2 2! + β„Ž3 3! + β„Ž4 4! + β‹― β„Ž ∡ 𝑒π‘₯ = 1 + π‘₯ + π‘₯2 2! + π‘₯3 3! + π‘₯4 4! + β‹― ∴ π’…π’š 𝒅𝒙 = 𝒅 𝒅𝒙 𝒆𝒙 = 𝒆𝒙 = 𝑒π‘₯ lim β„Žβ†’0 β„Ž 1 + β„Ž 2! + β„Ž2 3! + β„Ž3 4! + β‹― β„Ž = 𝑒π‘₯ lim β„Žβ†’0 1 + β„Ž 2! + β„Ž2 3! + β„Ž3 4! + β‹― = 𝑒π‘₯. 1
  • 12. Example 5: Find the derivative of 𝒇 𝒙 = ln π‘₯ using first principle 12 Let 𝑦 = 𝑓 π‘₯ = ln π‘₯ 𝑓 π‘₯ = ln π‘₯ 𝑓 π‘₯ + β„Ž = ln π‘₯ + β„Ž = lim β„Žβ†’0 ln π‘₯ + β„Ž π‘₯ β„Ž From First Principle, 𝑑𝑦 𝑑π‘₯ = 𝑓′(π‘₯) = lim β„Žβ†’0 𝑓 π‘₯ + β„Ž βˆ’ 𝑓(π‘₯) β„Ž = lim β„Žβ†’0 ln 1 + β„Ž π‘₯ β„Ž 𝑑𝑦 𝑑π‘₯ = lim β„Žβ†’0 β„Ž 1 π‘₯ βˆ’ 1 2 β„Ž π‘₯2 + 1 3 β„Ž2 π‘₯3 βˆ’ 1 4 β„Ž3 π‘₯4 + β‹― β„Ž ∡ ln 1 + π‘₯ = π‘₯ βˆ’ π‘₯2 2 + π‘₯3 3 βˆ’ π‘₯4 4 + β‹― ∴ π’…π’š 𝒅𝒙 = 𝒅 𝒅𝒙 ln π‘₯ = 1 π‘₯ = lim β„Žβ†’0 1 π‘₯ βˆ’ 1 2 β„Ž π‘₯2 + 1 3 β„Ž2 π‘₯3 βˆ’ 1 4 β„Ž3 π‘₯4 + β‹― = 1 π‘₯ = lim β„Žβ†’0 β„Ž π‘₯ βˆ’ 1 2 β„Ž2 π‘₯2 + 1 3 β„Ž3 π‘₯3 βˆ’ 1 4 β„Ž4 π‘₯4 + β‹― β„Ž
  • 13. Example 6: Find the derivative of 𝒇 𝒙 = cos 𝒙 using first principle 13 Let 𝑦 = 𝑓 π‘₯ = cos π‘₯ 𝑓 π‘₯ = cos π‘₯ 𝑓 π‘₯ + β„Ž = cos π‘₯ + β„Ž = lim β„Žβ†’0 cos π‘₯ + β„Ž βˆ’ cos π‘₯ β„Ž From First Principle, 𝑑𝑦 𝑑π‘₯ = 𝑓′(π‘₯) = lim β„Žβ†’0 𝑓 π‘₯ + β„Ž βˆ’ 𝑓(π‘₯) β„Ž = lim β„Žβ†’0 βˆ’2 sin 2π‘₯ + β„Ž 2 sin β„Ž 2 β„Ž = βˆ’ lim β„Žβ†’0 sin 2π‘₯ + β„Ž 2 Γ— lim β„Ž 2 β†’0 sin β„Ž 2 β„Ž 2 = βˆ’ sin 2π‘₯ + 0 2 Γ— 1 ∴ π’…π’š 𝒅𝒙 = 𝒅 𝒅𝒙 cos 𝒙 = βˆ’ 𝐬𝐒𝐧 𝒙 cos π‘Ž βˆ’ cos 𝑏 = βˆ’2 sin π‘Ž + 𝑏 2 sin π‘Ž βˆ’ 𝑏 2 lim β„Žβ†’0 sin β„Ž β„Ž = lim β„Žβ†’0 β„Ž sin β„Ž = 1 ∡ β„Ž β†’ 0 ⟹ β„Ž 2 β†’ 0???? β„Ž β†’ 0 ⟹ 1 2 Γ— β„Ž β†’ 1 2 Γ— 0 ⟹ β„Ž 2 β†’ 0
  • 14. Example 7: Find the derivative of 𝒇 𝒙 = 𝐬𝐒𝐧 𝒂𝒙 using first principle 14 Let 𝑦 = 𝑓 π‘₯ = sin π‘Žπ‘₯ 𝑓 π‘₯ = sin π‘Žπ‘₯ 𝑓 π‘₯ + β„Ž = sin π‘Ž π‘₯ + β„Ž = sin π‘Žπ‘₯ + π‘Žβ„Ž = lim β„Žβ†’0 sin π‘Žπ‘₯ + π‘Žβ„Ž βˆ’ sin π‘Žπ‘₯ β„Ž From First Principle, 𝑑𝑦 𝑑π‘₯ = 𝑓′(π‘₯) = lim β„Žβ†’0 𝑓 π‘₯ + β„Ž βˆ’ 𝑓(π‘₯) β„Ž = lim β„Žβ†’0 2 cos 2π‘Žπ‘₯ + π‘Žβ„Ž 2 sin π‘Žβ„Ž 2 β„Ž = lim β„Žβ†’0 cos 2π‘Žπ‘₯ + π‘Žβ„Ž 2 Γ— lim β„Žβ†’0 sin π‘Žβ„Ž 2 β„Ž 2 = cos 2π‘Žπ‘₯ + π‘Ž. 0 2 Γ— lim β„Žβ†’0 π‘Ž sin π‘Žβ„Ž 2 π‘Žβ„Ž 2 𝑑𝑦 𝑑π‘₯ = (cos π‘Žπ‘₯) Γ— π‘Ž lim π‘Žβ„Ž 2 β†’0 sin π‘Žβ„Ž 2 π‘Žβ„Ž 2 sin π‘Ž βˆ’ sin 𝑏 = 2 π‘π‘œπ‘  π‘Ž + 𝑏 2 𝑠𝑖𝑛 π‘Ž βˆ’ 𝑏 2 ∴ π’…π’š 𝒅𝒙 = 𝒅 𝒅𝒙 𝐬𝐒𝐧 𝒂𝒙 = 𝒂 𝐜𝐨𝐬 𝒂𝒙 = (cos π‘Žπ‘₯) Γ— π‘Ž 1 ∡ β„Ž β†’ 0 ⟹ π‘Žβ„Ž 2 β†’ 0 lim β„Žβ†’0 sin β„Ž β„Ž = lim β„Žβ†’0 β„Ž sin β„Ž = 1
  • 15. Example 8: Find the derivative of 𝒇 𝒙 = 𝒕𝒂𝒏 𝒙 using first principle 15 Let 𝑦 = 𝑓 π‘₯ = tan π‘₯ 𝑓 π‘₯ = tan π‘₯ 𝑓 π‘₯ + β„Ž = tan π‘₯ + β„Ž = lim β„Žβ†’0 tan π‘₯ + β„Ž βˆ’ tan π‘₯ β„Ž From First Principle, 𝑑𝑦 𝑑π‘₯ = 𝑓′(π‘₯) = lim β„Žβ†’0 𝑓 π‘₯ + β„Ž βˆ’ 𝑓(π‘₯) β„Ž = lim β„Žβ†’0 sin π‘₯ + β„Ž cos π‘₯ + β„Ž βˆ’ sin π‘₯ cos π‘₯ β„Ž = lim β„Žβ†’0 sin π‘₯ + β„Ž cos π‘₯ βˆ’ sin π‘₯ cos π‘₯ + β„Ž cos π‘₯ + β„Ž cos π‘₯ β„Ž = lim β„Žβ†’0 sin π‘₯ + β„Ž cos π‘₯ βˆ’ cos π‘₯ + β„Ž sin π‘₯ β„Ž cos π‘₯ + β„Ž cos π‘₯ 𝑑𝑦 𝑑π‘₯ = lim β„Žβ†’0 sin β„Ž β„Ž cos π‘₯ + β„Ž cos π‘₯ sin π‘Ž cos 𝑏 βˆ’ cos π‘Ž sin 𝑏 = sin π‘Ž βˆ’ 𝑏 ∴ π’…π’š 𝒅𝒙 = 𝒅 𝒅𝒙 𝐭𝐚𝐧 𝒙 = 𝐬𝐞𝐜𝟐 𝒙 = lim β„Žβ†’0 sin β„Ž β„Ž Γ— lim β„Žβ†’0 1 cos π‘₯ + β„Ž cos π‘₯ = 1 Γ— 1 cos π‘₯ + 0 cos π‘₯ = 1 cos2 π‘₯ = sec2 π‘₯ = lim β„Žβ†’0 sin π‘₯ + β„Ž βˆ’ π‘₯ β„Ž cos π‘₯ + β„Ž cos π‘₯
  • 16. Example 9: Find the derivative of 𝒇 𝒙 = π­πšπ§βˆ’πŸ 𝒙 using first principle 16 Let 𝑦 = 𝑓 π‘₯ = tanβˆ’1 π‘₯ 𝑓 π‘₯ = tanβˆ’1 π‘₯ 𝑓 π‘₯ + β„Ž = tanβˆ’1 π‘₯ + β„Ž 𝑑𝑦 𝑑π‘₯ = lim β„Žβ†’0 tanβˆ’1 π‘₯ + β„Ž βˆ’ tanβˆ’1 π‘₯ β„Ž β‹― (𝑖) From First Principle, 𝑑𝑦 𝑑π‘₯ = 𝑓′(π‘₯) = lim β„Žβ†’0 𝑓 π‘₯ + β„Ž βˆ’ 𝑓(π‘₯) β„Ž Supposetanβˆ’1 π‘₯ = 𝑦 ⟹ π‘₯ = tan 𝑦 andtanβˆ’1 π‘₯ + β„Ž = 𝑦 + π‘˜ ⟹ π‘₯ + β„Ž = tan 𝑦 + π‘˜ ⟹ β„Ž = tan 𝑦 + π‘˜ βˆ’ π‘₯ ⟹ β„Ž = tan 𝑦 + π‘˜ βˆ’ tan 𝑦 π‘₯ + β„Ž = tan 𝑦 + π‘˜ Whenβ„Ž β†’ 0, π‘₯ = tan 𝑦 + π‘˜ ⟹ tanβˆ’1 π‘₯ = 𝑦 + π‘˜ ⟹ tanβˆ’1 π‘₯ = tanβˆ’1 π‘₯ + π‘˜ ⟹ π‘˜ β†’ 0 Thus, β„Ž β†’ 0 ⟹ π‘˜ β†’ 0 In concept of Limit 𝑑𝑦 𝑑π‘₯ = lim β„Žβ†’0 𝑦 + π‘˜ βˆ’ 𝑦 β„Ž 𝑑𝑦 𝑑π‘₯ = lim π‘˜β†’0 π‘˜ tan 𝑦 + π‘˜ βˆ’ tan 𝑦 [∡ β„Ž β†’ 0 ⟹ π‘˜ β†’ 0 ? ? ? ]
  • 17. Example 9: Find the derivative of 𝒇 𝒙 = π­πšπ§βˆ’πŸ 𝒙 using first principle (cont.…) 17 sin π‘Ž cos 𝑏 βˆ’ cos π‘Ž sin 𝑏 = sin π‘Ž βˆ’ 𝑏 𝑑𝑦 𝑑π‘₯ = lim π‘˜β†’0 π‘˜ tan 𝑦 + π‘˜ βˆ’ tan 𝑦 [∡ β„Ž β†’ 0 ⟹ π‘˜ β†’ 0] = lim π‘˜β†’0 π‘˜ sin 𝑦 + π‘˜ cos 𝑦 + π‘˜ βˆ’ sin 𝑦 cos 𝑦 = lim π‘˜β†’0 π‘˜ sin 𝑦 + π‘˜ cos 𝑦 βˆ’ sin 𝑦 cos 𝑦 + π‘˜ cos 𝑦 + π‘˜ cos 𝑦 = lim π‘˜β†’0 π‘˜ cos 𝑦 + π‘˜ cos 𝑦 sin 𝑦 + π‘˜ cos 𝑦 βˆ’ cos 𝑦 + π‘˜ sin 𝑦 = lim π‘˜β†’0 π‘˜ cos 𝑦 + π‘˜ cos 𝑦 sin 𝑦 + π‘˜ βˆ’ 𝑦 𝑑𝑦 𝑑π‘₯ = lim π‘˜β†’0 π‘˜ sin π‘˜ Γ— lim π‘˜β†’0 cos 𝑦 + π‘˜ cos 𝑦 = 1 Γ— cos 𝑦 + 0 cos 𝑦 = cos2 𝑦 = 1 sec2 𝑦 = 1 1 + tan2 𝑦 = 1 1 + π‘₯2 = lim π‘˜β†’0 π‘˜ cos 𝑦 + π‘˜ cos 𝑦 sin π‘˜ ∴ π’…π’š 𝒅𝒙 = 𝒅 𝒅𝒙 π­πšπ§βˆ’πŸ 𝒙 = 𝟏 𝟏 + π’™πŸ ∡ π‘‘π‘Žπ‘› 𝑦 = π‘₯ lim π‘˜β†’0 sin π‘˜ π‘˜ = lim π‘˜β†’0 π‘˜ sin π‘˜ = 1
  • 18. Example 10: Find π’…π’š 𝒅𝒙 𝐟𝐨𝐫𝐲 = 𝒙 using first principle 18 Let 𝑦 = 𝑓 π‘₯ = π‘₯ 𝑓 π‘₯ = π‘₯ 𝑓 π‘₯ + β„Ž = π‘₯ + β„Ž = lim β„Žβ†’0 π‘₯ + β„Ž βˆ’ π‘₯ π‘₯ + β„Ž + π‘₯ β„Ž π‘₯ + β„Ž + π‘₯ From First Principle, 𝑑𝑦 𝑑π‘₯ = 𝑓′(π‘₯) = lim β„Žβ†’0 𝑓 π‘₯ + β„Ž βˆ’ 𝑓(π‘₯) β„Ž = lim β„Žβ†’0 π‘₯ + β„Ž 2 βˆ’ π‘₯ 2 β„Ž π‘₯ + β„Ž + π‘₯ = lim β„Žβ†’0 π‘₯ + β„Ž βˆ’ π‘₯ β„Ž π‘₯ + β„Ž + π‘₯ = 1 π‘₯ + 0 + π‘₯ ∴ π’…π’š 𝒅𝒙 = 𝒅 𝒅𝒙 𝒙 = 𝟏 𝟐 𝒙 = lim β„Žβ†’0 β„Ž β„Ž π‘₯ + β„Ž + π‘₯ 𝑑𝑦 𝑑π‘₯ = lim β„Žβ†’0 1 π‘₯ + β„Ž + π‘₯ = 1 π‘₯ + π‘₯ = lim β„Žβ†’0 π‘₯ + β„Ž βˆ’ π‘₯ β„Ž
  • 19. Exercise 19 Find the derivative of the following functions using first principle: 1. f x = ax 2. f x = cos ax 3. f x = sinβˆ’1 x 4. f x = cosβˆ’1 x 5. f x = sin x2