Differentiation by first principles
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The document discusses the concept of differentiability using limits and four examples: 1) The function f(x) = x - 5 is not differentiable at x = 5 since the left and right hand derivatives are not equal. 2) The function f(x) = x^x, 0 ≤ x < 2 and f(x) = (x-1)x, 2 ≤ x < 3 is not differentiable at x = 2 since the left and right hand derivatives are not equal. 3) The function f(x) = x^2, x ≤ 1 and f(x) = 1/x, x > 1 is not differentiable at x = 1 since the left and
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This document provides a table of 100 functions with their derivatives worked out. It begins with introductory rules for deriving various types of functions like constants, polynomials, exponentials, logarithmic, trigonometric, and combined functions. Then it lists 100 specific functions and their derivatives to serve as worked examples applying the basic rules.
100derivadasresueltasyosoytuprofe
This document provides a table of 100 functions with their derivatives worked out. It begins with introductory rules for deriving various types of functions like constants, polynomials, exponentials, logarithmic, trigonometric, and combined functions. Then it lists 100 specific functions and their derivatives, ranging from simple polynomials and exponentials to more complex expressions combining multiple functions. The purpose is to offer examples of applying the fundamental derivative rules to evaluate the derivatives of diverse mathematical expressions.
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Differentiation by first principles
- 2. A function f is said to be derivable at x = c, if lim ℎ→0+ 𝑓 𝑐 + ℎ − 𝑓(𝑐) ℎ = lim ℎ→0− 𝑓 𝑐 + ℎ − 𝑓(𝑐) ℎ Right hand derivative Left hand derivative Or lim ℎ→0 𝑓 𝑐 + ℎ − 𝑓(𝑐) ℎ exists and is denoted by 𝑓′ (𝑐)
- 3. Question 1 Discuss the differentiability of f(x) = 𝑥 − 5 at x = 5 Right hand derivative = lim ℎ→0+ 𝑓 5+ℎ −𝑓(5) ℎ = = lim ℎ→0+ ℎ ℎ = lim ℎ→0 ℎ ℎ = 1 = lim ℎ→0+ 5 + ℎ − 5 − 5 − 5 ℎ
- 4. Left hand derivative = lim ℎ→0− 𝑓 5+ℎ −𝑓(5) ℎ = lim ℎ→0− 5+ℎ−5 − 5−5 ℎ Since right hand derivative ≠ 𝑙𝑒𝑓𝑡 ℎ𝑎𝑛𝑑 𝑑𝑒𝑟𝑖𝑣𝑎𝑡𝑖𝑣𝑒, 𝑓 𝑖𝑠 𝑛𝑜𝑡 𝑑𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑡𝑖𝑎𝑏𝑙𝑒 at x = 5 = lim ℎ→0− ℎ ℎ = lim ℎ→0 −ℎ ℎ = −1
- 5. Question 2 Examine the function f defined by 𝑓 𝑥 = 𝑥 𝑥 , 0 ≤ 𝑥 < 2 = (x-1)x. 2≤ 𝑥 < 3 for differentiability at x = 2 Left hand derivative = lim ℎ→0− 𝑓 2+ℎ −𝑓(2) ℎ = lim ℎ→0 2+ℎ 2+ℎ −2 ℎ = lim ℎ→0 2 + ℎ − 2 ℎ = lim ℎ→0 ℎ ℎ = 1
- 6. Right hand derivative lim ℎ→0+ 𝑓 2 + ℎ − 𝑓(2) ℎ = lim ℎ→0 2 + ℎ − 1 2 + ℎ − 2 ℎ =lim ℎ→0 1+ℎ 2+ℎ −2 ℎ = lim ℎ→0 2+3ℎ+ℎ2−2 ℎ = lim ℎ→0 ℎ(ℎ + 3) ℎ = lim ℎ→0 ℎ + 3 = 3
- 7. Left hand derivative ≠ right hand derivative The function is not differentiable at x = 2 Question 3 𝑓 𝑥 = 𝑥2 , 𝑥 ≤ 1 = 1 𝑥 , x >1 Discuss the differentiability of f at x = 1
- 8. lim ℎ→0+ 𝑓 1 + ℎ − 𝑓(1) ℎ = lim ℎ→0 1 1 + ℎ − 1 ℎ Right hand derivative = lim ℎ→0 1 − (1 + ℎ) ℎ(1 + ℎ) = lim ℎ→0 −ℎ ℎ(1 + ℎ) = −1
- 9. Left hand derivative lim ℎ→0− 𝑓 1 + ℎ − 𝑓(1) ℎ = lim ℎ→0 (1 + ℎ)2−1 ℎ = lim ℎ→0 1 + 2ℎ + ℎ2 − 1 ℎ = lim ℎ→0 ℎ + 2 =2 Right hand derivative ≠ 𝑙𝑒𝑓𝑡 ℎ𝑎𝑛𝑑 𝑑𝑒𝑟𝑖𝑣𝑎𝑡𝑖𝑣𝑒 f is not differentiable at x = 1
- 10. Question 4 Find the right hand and left hand derivatives at x = 1. Hence find the condition for the existence of the derivative at x = 1 𝑓 𝑥 = 𝑎𝑥2 + 1, 𝑥 ≥ 1 = x +a, x <1 lim ℎ→0+ 𝑓 1 + ℎ − 𝑓(1) ℎ = lim ℎ→0 𝑎(1 + ℎ)2 + 1 − (𝑎 + 1) ℎ = lim ℎ→0 𝑎 1 + 2ℎ + ℎ2 + 1 − 𝑎 − 1 ℎ Right hand derivative
- 11. = lim ℎ→0 𝑎 + 2𝑎ℎ + 𝑎ℎ2 + 1 − 𝑎 − 1 ℎ = lim ℎ→0 2𝑎 + 𝑎ℎ =2a Left hand derivative lim ℎ→0− 𝑓 1 + ℎ − 𝑓(1) ℎ = lim ℎ→0 1 + ℎ + 𝑎 − (𝑎 + 1) ℎ = lim ℎ→0 1 + ℎ + 𝑎 − 𝑎 − 1 ℎ = 1
- 12. 2𝑎 = 1 Since f is derivable at x = 1 𝑎 = 1 2 **********************************