Derivatives Applications PDF

Derivatives Applications Mohamed Sief [email protected] Nile Valley University November 11, 2024 Mohamed Sief (NVU) Mathematics November 11, 2024 1 / 43 Lecture Outline 1 Higher Derivat...

Derivatives Applications Mohamed Sief [email protected] Nile Valley University November 11, 2024 Mohamed Sief (NVU) Mathematics November 11, 2024 1 / 43 Lecture Outline 1 Higher Derivatives 2 Increasing / Decreasing (I/D) Test for Monotonicity 3 Maximum and Minimum Values 4 Second Derivative Test 5 L’Hôpital’s Rule 6 Exercise 7 Conclusion Mohamed Sief (NVU) Mathematics November 11, 2024 2 / 43 Outline 1 Higher Derivatives 2 Increasing / Decreasing (I/D) Test for Monotonicity 3 Maximum and Minimum Values 4 Second Derivative Test 5 L’Hôpital’s Rule 6 Exercise 7 Conclusion Mohamed Sief (NVU) Mathematics November 11, 2024 3 / 43 Higher Derivatives Definition If y = f (x) is a differentiable function. Then First Derivative dy = y ′ = f ′ (x) is the first derivative of f, dx Second Derivative d2 y = y ′′ = f ′′ (x) is the second derivative of f, dx2 nth Derivative dn y = y (n) = f (n) (x) dxn is the nth derivative of f, where n is positive integer number. Mohamed Sief (NVU) Mathematics November 11, 2024 4 / 43 Example 1: Higher Derivatives Problem Find the 4th derivative of the following functions: (i) f (x) = 3x (ii) f (x) = ex √ (iii) f (x) = x (iv) f (x) = x3 + 2x Mohamed Sief (NVU) Mathematics November 11, 2024 5 / 43 Example 1 (continued) Solution for (i) f (x) = 3x f ′ (x)= 3x ln 3 f ′′ (x)= 3x (ln 3)2 f ′′′ (x)= 3x (ln 3)3 f (4) (x)= 3x (ln 3)4 Key Observations Each derivative multiplies by another ln 3 Pattern continues for higher derivatives Mohamed Sief (NVU) Mathematics November 11, 2024 6 / 43 Example 1 (continued) Solution for (ii) f (x) = ex f ′ (x)= ex f ′′ (x)= ex f ′′′ (x)= ex f (4) (x)= ex Key Observations All derivatives of ex are equal to ex Mohamed Sief (NVU) Mathematics November 11, 2024 7 / 43 Example 1 (continued) √ 1 Solution for (iii) f (x) = x = x2 1 1 f ′ (x)= x− 2 2 1 3 f ′′ (x)= − x− 2 4 ′′′ 3 −5 f (x)= x 2 8 15 7 f (4) (x)= − x− 2 16 Key Observations Exponent decreases by 1 each time Signs alternate between positive and negative Mohamed Sief (NVU) Mathematics November 11, 2024 8 / 43 Example 1 (continued) Solution for (iv) f (x) = x3 + 2x f ′ (x)= 3x2 + 2 f ′′ (x)= 6x f ′′′ (x)= 6 f (4) (x)= 0 Key Observations Power terms reduce degree by 1 each time Linear term disappears after second derivative All derivatives after 4th will be zero Mohamed Sief (NVU) Mathematics November 11, 2024 9 / 43 Outline 1 Higher Derivatives 2 Increasing / Decreasing (I/D) Test for Monotonicity 3 Maximum and Minimum Values 4 Second Derivative Test 5 L’Hôpital’s Rule 6 Exercise 7 Conclusion Mohamed Sief (NVU) Mathematics November 11, 2024 10 / 43 Increasing / Decreasing (I/D) Test for Monotonicity Increasing / Decreasing (I/D)Test Let f be differentiable on an interval I: If f ′ (x) > 0 for all x ∈ I, then f is increasing on I If f ′ (x) < 0 for all x ∈ I, then f is decreasing on I Mohamed Sief (NVU) Mathematics November 11, 2024 11 / 43 Example 3: I/D Test for Monotonic Functions Example Determine whether each of the following functions is increasing or decreasing. 1 f (x) = ln x 2 g(x) = x2 Mohamed Sief (NVU) Mathematics November 11, 2024 12 / 43 Solution: Natural Logarithm Solution for f (x) = ln x Find derivative: f ′ (x) = 1 x Domain of ln x is (0, ∞) For x > 0, we have f ′ (x) = 1 x >0 Therefore, ln x is strictly increasing on (0, ∞) Graph Analysis y y = ln x x Mohamed Sief (NVU) Mathematics November 11, 2024 13 / 43 Solution: Quadratic Function Solution for g(x) = x2 Find derivative: g ′ (x) = 2x Analyze where g ′ (x) is positive/negative: When x < 0: g ′ (x) < 0 (decreasing) When x = 0: g ′ (0) = 0 (critical point) When x > 0: g ′ (x) > 0 (increasing) Graph Analysis y y = x2 x Mohamed Sief (NVU) Mathematics November 11, 2024 14 / 43 Solution Summary Key Points ln x is strictly increasing on its entire domain x2 is decreasing for x < 0 and increasing for x > 0 I/D test is a powerful tool for determining monotonicity Mohamed Sief (NVU) Mathematics November 11, 2024 15 / 43 Outline 1 Higher Derivatives 2 Increasing / Decreasing (I/D) Test for Monotonicity 3 Maximum and Minimum Values 4 Second Derivative Test 5 L’Hôpital’s Rule 6 Exercise 7 Conclusion Mohamed Sief (NVU) Mathematics November 11, 2024 16 / 43 Local Extreme Values Definition: Local and Global Extrema Let f be a function defined and c is an interior point of the domain. The function f is said to have Local Maximum at c provided that: f (c) ≥ f (x) for all x near c Local Minimum at c provided that: f (c) ≤ f (x) for all x near c Mohamed Sief (NVU) Mathematics November 11, 2024 17 / 43 Local Extreme Values Definition: Local and Global Extrema Figure: Local Extreme Values Mohamed Sief (NVU) Mathematics November 11, 2024 18 / 43 Fermat’s Theorem and Critical Points Fermat’s Theorem Suppose that c is an interior point of the domain of f. If f has a local extremum at c, then: f ′ (c) = 0 Or f ′ (c) does not exist. Critical Points A point c in the domain of f is a critical point if either: f ′ (c) = 0 (horizontal tangent) f ′ (c) does not exist (corner or vertical tangent) Mohamed Sief (NVU) Mathematics November 11, 2024 19 / 43 The First Derivative Test The First Derivative Test Suppose that c is a critical point of a continuous function f. Case 1: Local Maximum If f ′ changes from positive to negative at c, then f has a local maximum at c. Case 2: Local Minimum If f ′ changes from negative to positive at c, then f has a local minimum at c. Case 3: Neither If f ′ does not change sign at c, then f has no local extremum at c. Mohamed Sief (NVU) Mathematics November 11, 2024 20 / 43 Visual Summary Case 1: Local Maximum Case 2: Local Minimum Mohamed Sief (NVU) Mathematics November 11, 2024 21 / 43 Example 1: First Derivative Test Problem Statement For the function f (x) = x3 − 3x2 − 9x + 5, find: Where f is increasing and where it is decreasing All local extrema of f (x) Strategy 1 Find f ′ (x) 2 Solve f ′ (x) = 0 to find critical points 3 Use the First Derivative Test to analyze intervals 4 Evaluate function at critical points Mohamed Sief (NVU) Mathematics November 11, 2024 22 / 43 Finding Critical Points Step 1: Finding the Derivative f ′ (x) = 3x2 − 6x − 9 = 3(x2 − 2x − 3) = 3(x − 3)(x + 1) Step 2: Finding Critical Points Set f ′ (x) = 0: 3(x − 3)(x + 1) = 0 x = −1 or x = 3 Mohamed Sief (NVU) Mathematics November 11, 2024 23 / 43 Interval Analysis First Derivative Test:f ′ (x) = 3(x − 3)(x + 1) Interval Test Point Sign of f ′ Behavior (−∞, −1) −2 + Increasing (−1, 3) 0 − Decreasing (3, ∞) 4 + Increasing Conclusions f is increasing on (−∞, −1) f is decreasing on (−1, 3) f is increasing on (3, ∞) x = −1 is a local minimum point x = 3 is a local maximum point Mohamed Sief (NVU) Mathematics November 11, 2024 24 / 43 Example 2: First Derivative Test Problem Statement For the function f (x) = x1/3 (x − 4) = x4/3 − 4x1/3 , find: Critical points Intervals where f is increasing and decreasing Local and absolute extreme values Strategy 1 Find f ′ (x) using chain rule 2 Solve f ′ (x) = 0 and check where f ′ is undefined 3 Use First Derivative Test for analysis 4 Evaluate function at critical points Mohamed Sief (NVU) Mathematics November 11, 2024 25 / 43 Finding Critical Points Step 1: Finding the Derivative 4 1 f ′ (x) = x1/3 − 4 · x−2/3 3 3 4 −2/3 = x (x − 1) 3 4 = 2/3 (x − 1) 3x Step 2: Finding Critical Points f ′ (x) = 0 when: x = 1 (from numerator) f ′ (x) is undefined at x = 0 (from denominator) Therefore, critical points are at x = 0 and x = 1 Mohamed Sief (NVU) Mathematics November 11, 2024 26 / 43 Interval Analysis First Derivative Test f ′ (x) = 4 3x2/3 (x − 1) Interval Test Point Sign of f ′ Behavior (−∞, 0) −1 − Decreasing (0, 1) 0.5 − Decreasing (1, ∞) 2 + Increasing Extreme Values At x = 0: f (0) = 0 At x = 1: f (1) = −3 x = 1 is a local minimum point The function has no local maximum Mohamed Sief (NVU) Mathematics November 11, 2024 27 / 43 Conclusions Final Results Critical points: x = 0 and x = 1 f is decreasing on (−∞, 0) and (0, 1) f is increasing on (1, ∞) Local minimum: f (1) = −3 No local maximum Mohamed Sief (NVU) Mathematics November 11, 2024 28 / 43 Outline 1 Higher Derivatives 2 Increasing / Decreasing (I/D) Test for Monotonicity 3 Maximum and Minimum Values 4 Second Derivative Test 5 L’Hôpital’s Rule 6 Exercise 7 Conclusion Mohamed Sief (NVU) Mathematics November 11, 2024 29 / 43 Second Derivative Test Theorem: Second Derivative Test Suppose that f ′ (c) = 0 and f ′′ (c) exist. Then: If Then f ′′ (c) >0 Local Minimum f ′′ (c) < 0 Local Maximum Important Notes Only works at critical points Requires twice differentiability Faster than first derivative test Mohamed Sief (NVU) Mathematics November 11, 2024 30 / 43 Example: Second Derivative Test Problem Statement Use the second derivative test to find local extrema of f (x) = x3 − 3x2 + 1 Step-by-Step Solution First derivative: f ′ (x) = 3x2 − 6x = 3x(x − 2) Critical points: x = 0 and x = 2 Second derivative: f ′′ (x) = 6x − 6 = 6(x − 1) Mohamed Sief (NVU) Mathematics November 11, 2024 31 / 43 Analysis Using Second Derivative Test Analysis at x = 0 At x = 0: f ′′ (0) = −6 < 0 =⇒ Local Maximum f (0) = 1 is a local maximum value Analysis at x = 2 At x = 2: f ′′ (2) = 6 > 0 =⇒ Local Minimum f (2) = −3 is a local minimum value Mohamed Sief (NVU) Mathematics November 11, 2024 32 / 43 Outline 1 Higher Derivatives 2 Increasing / Decreasing (I/D) Test for Monotonicity 3 Maximum and Minimum Values 4 Second Derivative Test 5 L’Hôpital’s Rule 6 Exercise 7 Conclusion Mohamed Sief (NVU) Mathematics November 11, 2024 33 / 43 L’Hôpital’s Rule Theorem If: lim f (x) = 0 and lim g(x) = 0, or both limits are ∞ x→a x→a Then: 0 ∞ Type 1: Type 2: ∞ 0 Both numerator and Both approach ∞ denominator approach 0 x2 Example: lim x x→∞ e sin x Example: lim x→0 x Solution: lim 2xx =0 cos x x→∞ e Solution: lim =1 x→0 1 Mohamed Sief (NVU) Mathematics November 11, 2024 34 / 43 L’Hôpital’s Rule Theorem If: lim f (x) = 0 and lim g(x) = 0, or both limits are ∞ x→a x→a Then: 0 ∞ Type 1: Type 2: ∞ 0 Both numerator and Both approach ∞ denominator approach 0 x2 Example: lim x x→∞ e sin x Example: lim x→0 x Solution: lim 2xx =0 cos x x→∞ e Solution: lim =1 x→0 1 Mohamed Sief (NVU) Mathematics November 11, 2024 34 / 43 L’Hôpital’s Rule Theorem If: lim f (x) = 0 and lim g(x) = 0, or both limits are ∞ x→a x→a Then: f (x) f ′ (x) lim = lim ′ x→a g(x) x→a g (x) 0 ∞ Type 1: Type 2: ∞ 0 Both numerator and Both approach ∞ denominator approach 0 x2 Example: lim x x→∞ e sin x Example: lim x→0 x Solution: lim 2xx =0 cos x x→∞ e Solution: lim =1 x→0 1 Mohamed Sief (NVU) Mathematics November 11, 2024 34 / 43 L’Hôpital’s Rule Key Notes Must verify indeterminate form before applying Can be applied multiple times if necessary Only works for quotients in indeterminate form Mohamed Sief (NVU) Mathematics November 11, 2024 35 / 43 Example 1: Basic Application of L’Hôpital’s Rule Problem Statement Evaluate the limit: sin x lim x→0 x Solution Using L’Hôpital’s Rule d sin x dx (sin x) lim = lim d x→0 x dx (x) x→0 cos x = lim x→0 1 =1 Mohamed Sief (NVU) Mathematics November 11, 2024 36 / 43 Example 2: Multiple Applications Problem Statement Evaluate the limit: 1 − cos x lim x→0 x2 First Application d 1 − cos x dx (1 − cos x) lim = lim x→0 x2 x→0 d 2 dx (x ) sin x 0 = lim (Still form) x→0 2x 0 Mohamed Sief (NVU) Mathematics November 11, 2024 37 / 43 Example 2: Multiple Applications Second Application sin x cos x lim = lim x→0 2x x→0 2 1 = 2 Key Observation Sometimes we need to apply L’Hôpital’s Rule multiple times until we get a determinate form. Mohamed Sief (NVU) Mathematics November 11, 2024 38 / 43 Outline 1 Higher Derivatives 2 Increasing / Decreasing (I/D) Test for Monotonicity 3 Maximum and Minimum Values 4 Second Derivative Test 5 L’Hôpital’s Rule 6 Exercise 7 Conclusion Mohamed Sief (NVU) Mathematics November 11, 2024 39 / 43 Extreme Values: Problems and Key Answers Problems and Solutions Find the extreme values of the following functions: Problems: Key Answers: (i) f (x) = 13 − 15x + 9x2 − x3 (i) Local min: f (1) = 6 (ii) f (x) = x4 Local max: f (5) = 38 (iii) f (x) = x2 − 2 ln x (ii) Absolute min: f (0) = 0 No local max (iv) f (x) = x3 + 2 (iii) Local min: f (1) = 1 No local max (iv) No local extrema Mohamed Sief (NVU) Mathematics November 11, 2024 40 / 43 Outline 1 Higher Derivatives 2 Increasing / Decreasing (I/D) Test for Monotonicity 3 Maximum and Minimum Values 4 Second Derivative Test 5 L’Hôpital’s Rule 6 Exercise 7 Conclusion Mohamed Sief (NVU) Mathematics November 11, 2024 41 / 43 Lecture Summary - Key Concepts Key Topics Covered 1 Derivatives and Their Applications Higher-order derivatives Critical points and extrema 2 Function Behavior Analysis Increasing/Decreasing intervals Local maxima and minima 3 Advanced Topics First and Second Derivative Tests L’Hôpital’s Rule Mohamed Sief (NVU) Mathematics November 11, 2024 42 / 43 Mohamed Sief (NVU) Mathematics November 11, 2024 43 / 43