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Semelhante a CHAPTER 2 (2.4 - 2.8) (1).ppsx
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- 3. APPLICATION OF DIFFERENTIATION • Calculus (differentiation and integration) was developed to improve the understanding of application of differentiation. • Differentiation and integration can help us solve many types of real-world problems. We use the derivative to determine the maximum and minimum values of particular functions. • (e.g. cost, strength, amount of material used in a building, profit, loss, etc.).
- 4. TURNING POINTS/STATIONARY POINTS/CRITICAL POINTS • A point where a function changes from an increasing to a decreasing function or visa-versa is known as a turning point. • To determine whether a function is increasing or decreasing, we use differentiation. • If the derivative is positive, a function is increasing meanwhile if the derivative is negative, a function is decreasing. Increasing function positive gradient Decreasing function negative gradient point P & Q Stationary point/Critical point/Turning point point P minimum point point Q maximum point
- 5. TURNING POINTS/STATIONARY POINTS/CRITICAL POINTS • Stationary points are points on a curve where the gradient is zero. • This is where the curve reaches a minimum or maximum. • A maximum is a high point and a minimum is a low point. • Stationary point is also known as turning point or critical point. • There are three types of stationary points: maximum, minimum and points of inflection (/inflexion).
- 6. MAXIMUM, MINIMUM or INFLECTION POINT • The gradient is zero at all the stationary point. • To determine whether the point is a maximum point, a minimum point or a point of inflection, second derivative will tell the nature of the curve.
- 7. APPLICATION OF DIFFERENTIATION TURNING POINTS/STATIONARY POINTS/CRITICAL POINTS • gradient (kecerunan) 𝑑𝑦 𝑑𝑥 = 0 • so bila 𝑑𝑦 𝑑𝑥 = 0, maka TUJUAN kita ialah mencari nilai titik kecerunan tersebut iaitu titik 𝑥 dan 𝑦 MAXIMUM, MINIMUM or INFLECTION POINT • Differentiate kali ke-2 𝑑2𝑦 𝑑𝑥2 berTUJUAN untuk mencari nature of the point • 𝑑2𝑦 𝑑𝑥2 > 0 minimum point • 𝑑2𝑦 𝑑𝑥2 < 0 maximum point • 𝑑2𝑦 𝑑𝑥2 = 0 inflection point
- 8. TURNING POINTS/STATIONARY POINTS/CRITICAL POINTS Find the stationary points on the graph of 𝑦 = 𝑥2 + 4𝑥 + 3 and state their nature. 𝑦 = 𝑥2 + 4𝑥 + 3 𝑑𝑦 𝑑𝑥 = 0 𝑑𝑦 𝑑𝑥 = 2𝑥 + 4 ∴ 2𝑥 + 4 = 0 𝑥 = −2 𝑤ℎ𝑒𝑛 𝑥 = −2 𝑦 = −2 2 + 4 −2 + 3 𝑦 = −1 𝑡𝑢𝑟𝑛𝑖𝑛𝑔 𝑝𝑜𝑖𝑛𝑡 𝑖𝑠 −2, −1 𝑑2𝑦 𝑑𝑥2 = 2 > 0 𝑃𝑜𝑖𝑛𝑡 −2, −1 𝑖𝑠 𝑎 𝑚𝑖𝑛𝑖𝑚𝑢𝑚 𝑝𝑜𝑖𝑛𝑡 write the function 𝑑𝑦 𝑑𝑥 = 0 Differentiate kali pertama Find value of 𝑥 Find value of 𝑦 State the turning point Differentiate kali ke-2 State nature of the point
- 9. Find the stationary points on the graph of 𝑦 = 𝑥3 − 2𝑥2 − 4𝑥 + 5 and state their nature. 𝑦 = 𝑥3 − 2𝑥2 − 4𝑥 + 5 𝑑𝑦 𝑑𝑥 = 0 𝑑𝑦 𝑑𝑥 = 3𝑥2 − 4𝑥 − 4 ∴ 3𝑥2 − 4𝑥 − 4 = 0 𝑥 = 2, − 2 3 𝑤ℎ𝑒𝑛 𝑥 = − 2 3 𝑦 = − 2 3 3 − 2 − 2 3 2 − 4 − 2 3 + 5 ∴ 𝑦 = 175 27 𝑡𝑢𝑟𝑛𝑖𝑛𝑔 𝑝𝑜𝑖𝑛𝑡 𝑎𝑟𝑒 2, −3 & − 2 3 , 175 27 𝑑2 𝑦 𝑑𝑥2 = 6𝑥 − 4 𝑃𝑜𝑖𝑛𝑡 2, −3 𝑖𝑠 𝑎 𝑚𝑖𝑛𝑖𝑚𝑢𝑚 𝑝𝑜𝑖𝑛𝑡 & 𝑃𝑜𝑖𝑛𝑡 − 2 3 , 175 27 𝑖𝑠 𝑎 𝑚𝑎𝑥𝑖𝑚𝑢𝑚 𝑝𝑜𝑖𝑛𝑡 write the function 𝑑𝑦 𝑑𝑥 = 0 Differentiate kali pertama Find value of 𝑥 State the turning point Differentiate kali ke-2 State nature of the point 𝑤ℎ𝑒𝑛 𝑥 = 2 𝑦 = 2 3 − 2 2 2 − 4 2 + 5 ∴ 𝑦 = −3 Find value of 𝑦 𝑤ℎ𝑒𝑛 𝑥 = − 2 3 𝑤ℎ𝑒𝑛 𝑥 = 2 𝑑2𝑦 𝑑𝑥2 = 6 2 − 4 = 8 > 0 𝑑2 𝑦 𝑑𝑥2 = 6 − 2 3 − 4 = −8 < 0 Find the value of 𝑑2𝑦 𝑑𝑥2
- 10. 𝑦 = 2𝑥2 − 8𝑥 + 8 𝑑𝑦 𝑑𝑥 = 0 𝑑𝑦 𝑑𝑥 = 4𝑥 − 8 ∴ 4𝑥 − 8 = 0 𝑥 = 2 𝑤ℎ𝑒𝑛 𝑥 = 2 𝑦 = 2 2 − 8 2 + 8 𝑦 = 0 𝑡𝑢𝑟𝑛𝑖𝑛𝑔 𝑝𝑜𝑖𝑛𝑡 𝑖𝑠 2, 0 𝑑2𝑦 𝑑𝑥2 = 4 > 0 𝑃𝑜𝑖𝑛𝑡 −2, 3 𝑖𝑠 𝑎 𝑚𝑖𝑛𝑖𝑚𝑢𝑚 𝑝𝑜𝑖𝑛𝑡 write the function 𝑑𝑦 𝑑𝑥 = 0 Differentiate kali pertama Find value of 𝑥 Find value of 𝑦 State the turning point Differentiate kali ke-2 State nature of the point Find the stationary points on the graph of 𝑦 = 2𝑥2 − 8𝑥 + 8 and state their nature. Sketch the graph How to plot the graph
- 13. RATES OF CHANGE Human everyday problems related to rates of change a) An electrical engineer is interested in knowing the rate of change in an electric circuit. b) Construction engineers want to know the rate of change of concrete expansion on a bridge. c) A mechanical engineer should know the rate of shrinkage and expansion of the valve in a pump used for drilling oil. Introduction to rate of change • Differentiation is used to calculate rates of change. • The rate of change of a function 𝑓 𝑥 with respect to 𝑥 can be found by finding the derived function 𝑓′ 𝑥 . • For example in mechanics, the rate of change of displacement (with respect to time) is the velocity. The rate of change of velocity (with respect to time) is the acceleration.
- 14. RATES OF CHANGE 1. Read the problem carefully. Understand the relationships between the variable quantities. ⇒ What data is given? What is to be found/calculate? 2. Make a sketch (diagram/shape) if needed. 3. Define/Substitute any symbols you want to use that are not defined in the statement of the problem. Then, express given and required quantities and rates in terms of these symbols. [example: sides of a cube as 𝑥 ] 4. From a careful reading of the problem or consideration of the sketch, identify one or more equations linking the variable quantities. Differentiate the equation. 5. Substitute any given values for the quantities and their rates, and then solve the resulting equation for the unknown quantities and rates. 6. Conclude the answer. • Rates which contain y = f(x) • Rates which contain radius
- 15. RATES OF CHANGE If the expression of y given by 𝑦 = 𝑥3 + 𝑥2 − 8. Find the rate of change of 𝑦 if 𝑥 increase at 0. 5 𝑢𝑛𝑖𝑡 𝑠𝑒𝑐𝑜𝑛𝑑 when 𝑥 = 5. 𝑦 = 𝑥3 + 𝑥2 − 8 𝑑𝑦 𝑑𝑡 = 𝑑𝑦 𝑑𝑥 × 𝑑𝑥 𝑑𝑡 𝑑𝑦 𝑑𝑥 = 3𝑥2 + 2𝑥 𝑑𝑦 𝑑𝑡 = 3𝑥2 + 2𝑥 × 0.5 = 42.5 𝑢𝑛𝑖𝑡 𝑠𝑒𝑐 write the function related chain rule differentiate the given function substitute value 𝑑𝑥 𝑑𝑡 = 0.5 𝑥 = 5 given info from question = 3 5 2 + 2 5 × 0.5 ∴ Rate of y increase at 42.5 𝑢𝑛𝑖𝑡 𝑠𝑒𝑐𝑜𝑛𝑑 𝑑𝑦 𝑑𝑡 = ?
- 16. RATES OF CHANGE The area 𝐴 of a circle is increasing at a constant rate of 1.5𝑐𝑚2 𝑠−1 . Calculate the rate at which the radius, 𝑟 of the circle is increasing when the perimeter of the circle is 6 𝑐𝑚. 𝐴 = 𝜋𝑟2 𝑑𝐴 𝑑𝑡 = 𝑑𝐴 𝑑𝑟 × 𝑑𝑟 𝑑𝑡 𝑑𝐴 𝑑𝑟 = 2𝜋𝑟 1.5 = 2𝜋𝑟 × 𝑑𝑟 𝑑𝑡 = 0.25 𝑐𝑚2 𝑠𝑒𝑐 write the related formula related chain rule differentiate the related formula substitute value 𝑑𝐴 𝑑𝑡 = 1.5 𝑃 = 6 given info from question 𝑑𝑟 𝑑𝑡 = 1.5 2𝜋𝑟 𝑙 = 2𝜋𝑟 solve related formula to find value ∴ 2𝜋𝑟 = 6 = 1.5 6 ∴ Rate of the area increase at 0.25 𝑐𝑚2 𝑠𝑒𝑐 𝑑𝑟 𝑑𝑡 = ?
- 17. RATES OF CHANGE The length of a rectangle is always four times its width. If the width is decreasing at a rate of 0.1𝑐𝑚𝑠−1, find the rate of change of the area when its perimeter is 12 𝑐𝑚. 𝐴 = 𝑥 4𝑥 = 4𝑥2 𝑑𝐴 𝑑𝑡 = 𝑑𝐴 𝑑𝑥 × 𝑑𝑥 𝑑𝑡 𝑑𝐴 𝑑𝑥 = 8𝑥 𝑑𝐴 𝑑𝑡 = 8𝑥 × −0.1 = −0.96 𝑐𝑚2 𝑠𝑒𝑐 write the related formula related chain rule differentiate the related formula substitute value 𝑑𝑥 𝑑𝑡 = −0.1 𝑃 = 12 given info from question 𝑑𝑟 𝑑𝑡 = 8 6 5 × −0.1 𝑙 = 𝑥 + 𝑥 + 4𝑥 + 4𝑥 solve related formula to find value ∴ 10𝑥 = 12 ∴ Rate of the change of the area decreases at −0.96 𝑐𝑚2 4𝑥 𝑥 𝑥 = 6 5 𝑑𝐴 𝑑𝑡 = ? 𝑙 = 10𝑥
- 19. Parametric Equation Differentiation • A parametric equation is where the 𝑥 and 𝑦 coordinates are both written in terms of another letter. • The third variable is called a parameter and is usually given the letter 𝑡 or 𝜃. • The differentiation of functions given in parametric form is carried out using the Chain Rule. 𝑑𝑦 𝑑𝑥 = 𝑑𝑦 𝑑𝑡 × 𝑑𝑡 𝑑𝑥 where 𝑑𝑡 𝑑𝑥 = 1 𝑑𝑥 𝑑𝑡
- 20. Parametric Equation Differentiation • Steps to solve problem on parametric equation: i. Differentiate each of the parametric equations for the parameter. ii. Substitute the resulting expression for the parameter into the Chain Rule. iii. Simplify the equation.
- 21. Parametric Equation Differentiation Find 𝑑𝑦 𝑑𝑥 when 𝑥 = 𝑡2 + 𝑡 and 𝑦 = 2𝑡 − 1 𝑑𝑥 𝑑𝑡 = 2𝑡 + 1 𝑑𝑦 𝑑𝑥 = 𝑑𝑦 𝑑𝑡 × 𝑑𝑡 𝑑𝑥 𝑑𝑦 𝑑𝑥 = 2 × 1 2𝑡 + 1 related chain rule differentiate the given function substitute value 𝑥 = 𝑡2 + 𝑡 𝑦 = 2𝑡 − 1 Write both the function = 2 2𝑡 + 1 𝑑𝑦 𝑑𝑡 = 2 𝑑𝑡 𝑑𝑥 = 1 2𝑡 + 1
- 22. Parametric Equation Differentiation Given 𝑥 = 2𝑠𝑖𝑛𝜃 and 𝑦 = 4𝑐𝑜𝑠𝜃, then find 𝑑𝑦 𝑑𝑥 𝑑𝑥 𝑑𝜃 = 2ܿߠݏ 𝑑𝑦 𝑑𝑥 = 𝑑𝑦 𝑑𝜃 × 𝑑𝜃 𝑑𝑥 𝑑𝑦 𝑑𝑥 = −4𝑠𝑖𝑛𝜃 × 1 2ܿߠݏ related chain rule differentiate the given function substitute value 𝑥 = 2𝑠𝑖𝑛𝜃 𝑦 = 4ܿߠݏ Write both the function = −4𝑠𝑖𝑛𝜃 2ܿߠݏ 𝑑𝑦 𝑑𝜃 = −4𝑠𝑖𝑛𝜃 𝑑𝜃 𝑑𝑥 = 1 2ܿߠݏ = −2𝑡𝑎𝑛𝜃
- 23. Parametric Equation Differentiation 𝑑𝑥 𝑑𝑡 = −8𝑡−5 𝑑𝑦 𝑑𝑥 = 𝑑𝑦 𝑑𝑡 × 𝑑𝑡 𝑑𝑥 𝑑𝑦 𝑑𝑥 = − 3 𝑡2 − 3𝑡2 × − 𝑡5 8 related chain rule differentiate the given function substitute value 𝑥 = 2 𝑡4 𝑦 = 3 𝑡 − 𝑡3 Write both the function = −3 − 3𝑡4 𝑡2 × − 𝑡5 8 𝑑𝑦 𝑑𝑡 = −3𝑡−2 − 3𝑡2 𝑑𝑡 𝑑𝑥 = − 𝑡5 8 The parametric equations of a curve are 𝑥 = 2 𝑡4 and 𝑦 = 3 𝑡 − 𝑡3, find 𝑑𝑦 𝑑𝑥 terms of 𝑡. = − 3𝑡3 8 1 + 𝑡4 𝑥 = 2𝑡−4 = − 8 𝑡5 𝑦 = 3𝑡−1 − 𝑡3 = − 3 𝑡2 − 3𝑡2
- 25. Implicit Differentiation • Implicit differentiation is the process of finding the derivative of a dependent variable in an implicit function by differentiating each term separately, by expressing the derivative of the dependent variable as a symbol, and by solving the resulting expression for the symbol. • The technique of implicit differentiation allows you to find the derivative of 𝑦 with respect to 𝑥 without having to solve the given equation for 𝑦. • The chain rule must be used whenever the function 𝑦 is being differentiated because of our assumption that 𝑦 may be expressed as a function of 𝑥
- 26. Implicit Differentiation • Steps to solve problem on implicit differentiation: i. Differentiate 𝑥 and 𝑦 with respect to 𝑥 [differentiate 𝑦 followed by 𝑑𝑦 𝑑𝑥 ]. ii. Put all 𝑑𝑦 𝑑𝑥 expression on the left side of the equal sign, while the other expression on the right side of the equal sign. iii. Factorize 𝑑𝑦 𝑑𝑥 expression on the left side. iv. Find 𝑑𝑦 𝑑𝑥 [move all the expression on the left to the right side].
- 27. Implicit Differentiation • Basic implicit differentiation: Function Differentiation 𝑥 𝑦 𝑥𝑦 𝑑𝑥 𝑑𝑥 = 1 𝑑𝑦 𝑑𝑥 𝑦 𝑑𝑥 𝑑𝑥 + 𝑥 𝑑𝑦 𝑑𝑥 = 𝑦 + 𝑥 𝑑𝑦 𝑑𝑥
- 28. Implicit Differentiation Factorize 𝑑𝑦 𝑑𝑥 on the left All 𝑑𝑦 𝑑𝑥 on the left 𝑑𝑦 𝑑𝑥 as subject matter 3𝑥2 + 5𝑦 = 8 − 𝑦4 + 3𝑥 Differentiate 𝑦 followed by 𝑑𝑦 𝑑𝑥 6𝑥 + 5 𝑑𝑦 𝑑𝑥 = 0 − 4𝑦3 𝑑𝑦 𝑑𝑥 + 3 Find 𝑑𝑦 𝑑𝑥 for all the following: 5 𝑑𝑦 𝑑𝑥 + 4𝑦3 𝑑𝑦 𝑑𝑥 = 3 − 6𝑥 𝑑𝑦 𝑑𝑥 5 + 4𝑦3 = 3 − 6𝑥 𝑑𝑦 𝑑𝑥 = 3 − 6𝑥 5 + 4𝑦3
- 29. Implicit Differentiation Factorize 𝑑𝑦 𝑑𝑥 on the left All 𝑑𝑦 𝑑𝑥 on the left 𝑑𝑦 𝑑𝑥 as subject matter 𝑥4 + 10 = 𝑥2 𝑦3 − 5𝑦2 Differentiate 𝑦 followed by 𝑑𝑦 𝑑𝑥 4𝑥3 + 0 = 2𝑥𝑦3 + 3𝑥2 𝑦2 𝑑𝑦 𝑑𝑥 − 10𝑦 𝑑𝑦 𝑑𝑥 Find 𝑑𝑦 𝑑𝑥 for all the following: 3𝑥2𝑦2 𝑑𝑦 𝑑𝑥 − 10𝑦 𝑑𝑦 𝑑𝑥 = 4𝑥3 − 2𝑥𝑦3 𝑑𝑦 𝑑𝑥 3𝑥2 𝑦2 − 10𝑦 = 4𝑥3 − 2𝑥𝑦3 𝑑𝑦 𝑑𝑥 = 4𝑥3 − 2𝑥𝑦3 3𝑥2𝑦2 − 10𝑦 Differentiate using Product Rule 𝑑𝑦 𝑑𝑥 = 𝑣 𝑑𝑢 𝑑𝑥 + 𝑢 𝑑𝑣 𝑑𝑥 𝑑𝑦 𝑑𝑥 = 2𝑥 2𝑥2 − 𝑦3 𝑦(3𝑥2𝑦 − 10)
- 30. Implicit Differentiation Factorize 𝑑𝑦 𝑑𝑥 on the left All 𝑑𝑦 𝑑𝑥 on the left 𝑑𝑦 𝑑𝑥 as subject matter 𝑥4 − 3𝑒2𝑥+2𝑦 = 4𝑦 + 5 Differentiate 𝑦 followed by 𝑑𝑦 𝑑𝑥 4𝑥3 − 3𝑒2𝑥+2𝑦 2 + 2 𝑑𝑦 𝑑𝑥 = 4 𝑑𝑦 𝑑𝑥 + 0 Find 𝑑𝑦 𝑑𝑥 for all the following: 4𝑥3 − 6𝑒2𝑥+2𝑦 − 6𝑒2𝑥+2𝑦 𝑑𝑦 𝑑𝑥 = 4 𝑑𝑦 𝑑𝑥 𝑑𝑦 𝑑𝑥 4 + 6𝑒2𝑥+2𝑦 = 4𝑥3 − 6𝑒2𝑥+2𝑦 𝑑𝑦 𝑑𝑥 = 4𝑥3 − 6𝑒2𝑥+2𝑦 4 + 6𝑒2𝑥+2𝑦 𝑑𝑦 𝑑𝑥 = 2 2𝑥3 − 3𝑒2𝑥+2𝑦 2(2 + 3𝑒2𝑥+2𝑦) 4 𝑑𝑦 𝑑𝑥 + 6𝑒2𝑥+2𝑦 𝑑𝑦 𝑑𝑥 = 4𝑥3 − 6𝑒2𝑥+2𝑦 = 2𝑥3 − 3𝑒2𝑥+2𝑦 2 + 3𝑒2𝑥+2𝑦
- 31. 4𝑥3 = 8𝑦 − 8𝑥 𝑑𝑦 𝑑𝑥 4𝑦 2 + 5 𝑑𝑦 𝑑𝑥 Implicit Differentiation Factorize 𝑑𝑦 𝑑𝑥 on the left All 𝑑𝑦 𝑑𝑥 on the left 𝑑𝑦 𝑑𝑥 as subject matter 𝑥4 − 10 = 2𝑥 4𝑦 + 5𝑦 Differentiate 𝑦 followed by 𝑑𝑦 𝑑𝑥 Find 𝑑𝑦 𝑑𝑥 for all the following: 4𝑥3 = 8𝑦 − 8𝑥 𝑑𝑦 𝑑𝑥 + 80𝑦2 𝑑𝑦 𝑑𝑥 16𝑦2 𝑑𝑦 𝑑𝑥 8𝑥 − 80𝑦2 = 8𝑦 − 64𝑥3 𝑦2 𝑑𝑦 𝑑𝑥 = 8𝑦 − 64𝑥3 𝑦2 8𝑥 − 80𝑦2 8𝑥 𝑑𝑦 𝑑𝑥 + 80𝑦2 𝑑𝑦 𝑑𝑥 = 8𝑦 − 64𝑥3 𝑦2 Differentiate using Quotient Rule 𝑑𝑦 𝑑𝑥 = 𝑣 𝑑𝑢 𝑑𝑥 − 𝑢 𝑑𝑣 𝑑𝑥 𝑣2 64𝑥3 𝑦2 = 8𝑦 − 8𝑥 𝑑𝑦 𝑑𝑥 + 80𝑦2 𝑑𝑦 𝑑𝑥
- 32. 𝑠𝑖𝑛 𝑥2 + 𝑦2 = 𝑐𝑜𝑠 𝑥2 + 𝑦2 ∙ 2𝑥 + 2𝑦 𝑑𝑦 𝑑𝑥 = 2𝑥 𝑐𝑜𝑠 𝑥2 + 𝑦2 + 2𝑦 𝑐𝑜𝑠 𝑥2 + 𝑦2 𝑑𝑦 𝑑𝑥 𝑠𝑖𝑛 𝑥2 𝑦2 = 𝑐𝑜𝑠 𝑥2𝑦2 ∙ 2𝑥𝑦2 + 2𝑥2𝑦 𝑑𝑦 𝑑𝑥 = 2𝑥𝑦2 𝑐𝑜𝑠 𝑥2 + 𝑦2 + 2𝑥2 𝑦𝑐𝑜𝑠 𝑥2 𝑦2 𝑑𝑦 𝑑𝑥 𝑙𝑛 𝑥2 + 𝑦2 = 1 𝑥2 + 𝑦2 ∙ 2𝑥 + 2𝑦 𝑑𝑦 𝑑𝑥 𝑙𝑛 𝑥2 𝑦2 = 1 𝑥2𝑦2 ∙ 2𝑥𝑦2 + 2𝑥2𝑦 𝑑𝑦 𝑑𝑥 𝑒 𝑥2+𝑦2 = 𝑒 𝑥2+𝑦2 ∙ 2𝑥 + 2𝑦 𝑑𝑦 𝑑𝑥 = 2𝑥𝑒 𝑥2+𝑦2 +2𝑦𝑒 𝑥2+𝑦2 𝑑𝑦 𝑑𝑥 𝑒 𝑥2𝑦2 = 𝑒 𝑥2𝑦2 ∙ 2𝑥𝑦2 + 2𝑥2𝑦 𝑑𝑦 𝑑𝑥 = 2𝑥𝑦2 𝑒 𝑥2𝑦2 + 2𝑥2 𝑦𝑒 𝑥2𝑦2 𝑑𝑦 𝑑𝑥 = 2𝑥 𝑥2 + 𝑦2 + 2𝑦 𝑥2 + 𝑦2 𝑑𝑦 𝑑𝑥 = 2𝑥𝑦2 𝑥2𝑦2 + 2𝑥2 𝑦 𝑥2𝑦2 𝑑𝑦 𝑑𝑥 = 2 𝑥 + 2 𝑦 𝑑𝑦 𝑑𝑥
- 34. Partial Differentiation • If you know how to take a derivative, then you can take partial derivatives. • Partial derivatives are defined as derivatives of a function where the variable of differentiation is indicated. • All other variables are held constant. • The chain rule of partial derivatives evaluates the derivative of a function of functions (composite function) without having to substitute, simplify, and then differentiate. • You do sometimes see 𝜕𝑧 𝜕𝑥 . which means that 𝑧 is a function of several variables, including 𝑥, and you are taking the partial derivative of 𝑧 with respect to 𝑥.
- 35. Partial Differentiation: First Order • The first partial derivatives of the equation 𝑧 = 𝑓 𝑥, 𝑦 be a function with two variables with respect to 𝑥 ∶ 𝑓𝑥 = 𝜕𝑓 𝜕𝑥 with respect to 𝑦 : 𝑓𝑦 = 𝜕𝑓 𝜕𝑦
- 36. Partial Differentiation: First Order Given 𝑧 = 𝑥3 + 5𝑦2 − 8 𝜕𝑧 𝜕𝑥 = 3𝑥2 + 0 + 0 𝜕𝑧 𝜕𝑦 = 0 + 10𝑦 + 0 Differentiate 𝑥, 𝑦 jadi PEMALAR Differentiate 𝑦, 𝑥 jadi PEMALAR Jika 𝒙 & 𝒚 adalah SINGLE (sendirian) !!
- 37. Partial Differentiation: First Order Given 𝑧 = 𝑥3 + 5𝑦2 − 8𝑥3𝑦2 𝜕𝑧 𝜕𝑥 = 3𝑥2 + 0 − 24𝑥2 𝑦2 𝜕𝑧 𝜕𝑦 = 0 + 10𝑦 − 16𝑦𝑥3 Differentiate 𝑥, 𝑦 salin semula Differentiate 𝑦, 𝑥 salin semula Jika 𝒙 & 𝒚 adalah COUPLE (bersama2) !! = 10𝑦 − 16𝑥3 𝑦
- 38. Differentiate 𝑥, 𝑦 jadi PEMALAR Differentiate 𝑦, 𝑥 jadi PEMALAR Jika 𝒙 & 𝒚 adalah SINGLE (sendirian) !! Differentiate 𝑥, 𝑦 salin semula Differentiate 𝑦, 𝑥 salin semula Jika 𝒙 & 𝒚 adalah COUPLE (bersama2) !!
- 39. a) 𝑧 = 𝑥4 + 5 − 3𝑦2 Find 𝜕𝑧 𝜕𝑥 and 𝜕𝑧 𝜕𝑦 for all the following: Partial Differentiation 𝜕𝑧 𝜕𝑥 = 4𝑥3 + 0 + 0 𝜕𝑧 𝜕𝑦 = 0 + 0 − 6𝑦 b) 𝑧 = 4𝑥 + 2𝑥4 𝑦2 − 𝑦5 𝜕𝑧 𝜕𝑥 = 4 + 8𝑥3 𝑦2 − 0 𝜕𝑧 𝜕𝑦 = 0 + 4𝑦𝑥4 − 5𝑦4 𝜕𝑧 𝜕𝑥 = 4𝑥3 𝜕𝑧 𝜕𝑦 = −6𝑦 terhadap 𝑥 terhadap 𝑦 Differentiate 𝑥, 𝑦 jadi PEMALAR Differentiate 𝑦, 𝑥 jadi PEMALAR terhadap 𝑥 terhadap 𝑦 𝜕𝑧 𝜕𝑥 = 4 + 8𝑥3𝑦2 𝜕𝑧 𝜕𝑦 = 4𝑥4 𝑦 − 5𝑦4 Differentiate 𝑥, 𝑦 jadi PEMALAR Differentiate 𝑦, 𝑥 jadi PEMALAR Differentiate 𝑥, 𝑦 salin semula Differentiate 𝑦, 𝑥 salin semula
- 40. Find 𝜕𝑧 𝜕𝑥 and 𝜕𝑧 𝜕𝑦 for all the following: Partial Differentiation c) 𝑧 = 𝑥2 + 2 1 − 3𝑦 𝜕𝑧 𝜕𝑥 = 2𝑥 − 6𝑥𝑦 𝜕𝑧 𝜕𝑦 = −3𝑥2 − 6 = 𝑥2 − 3𝑥2 𝑦 + 2 − 6𝑦 d) 𝑧 = 3𝑥 + 𝑥 𝑐𝑜𝑠2𝑦 − 5𝑦 𝜕𝑧 𝜕𝑥 = 3 + 𝑐𝑜𝑠2𝑦 𝜕𝑧 𝜕𝑦 = −2𝑥 𝑠𝑖𝑛2𝑦 − 5 e) 𝑧 = 5𝑒𝑥 𝑦 + 3 − 𝑒2𝑥+3𝑦 𝜕𝑧 𝜕𝑥 = 5𝑒𝑥𝑦 − 2𝑒2𝑥+3𝑦 𝜕𝑧 𝜕𝑦 = 5𝑒𝑥 − 3𝑒2𝑥+3𝑦 f) 𝑧 = 𝑙𝑛 𝑥2 + 𝑦2 + 4𝑥 − 3𝑦 𝜕𝑧 𝜕𝑥 = 1 𝑥2 + 𝑦2 ∙ 2𝑥 + 4 𝜕𝑧 𝜕𝑦 = 1 𝑥2 + 𝑦2 ∙ 2𝑦 − 3 = 2𝑥 𝑥2 + 𝑦2 + 4 = 2𝑦 𝑥2 + 𝑦2 − 3 expand terhadap 𝑥 terhadap 𝑦 terhadap 𝑥 terhadap 𝑦 terhadap 𝑥 terhadap 𝑦 terhadap 𝑥 terhadap 𝑦
- 41. 𝑧 = 𝑠𝑖𝑛 𝑥2 + 𝑦2 𝜕𝑧 𝜕𝑥 = 𝑐𝑜𝑠 𝑥2 + 𝑦2 ∙ 2𝑥 𝜕𝑧 𝜕𝑦 = 𝑐𝑜𝑠 𝑥2 + 𝑦2 ∙ 2𝑦 = 2𝑥 𝑐𝑜𝑠 𝑥2 + 𝑦2 = 2𝑦 𝑐𝑜𝑠 𝑥2 + 𝑦2 𝑧 = 𝑠𝑖𝑛 𝑥2 𝑦2 𝜕𝑧 𝜕𝑥 = 𝑐𝑜𝑠 𝑥2𝑦2 ∙ 2𝑥𝑦2 𝜕𝑧 𝜕𝑦 = 𝑐𝑜𝑠 𝑥2 𝑦2 ∙ 2𝑥2 𝑦 = 2𝑥𝑦2 𝑐𝑜𝑠 𝑥2 + 𝑦2 = 2𝑥2 𝑦𝑐𝑜𝑠 𝑥2 + 𝑦2 𝑧 = 𝑙𝑛 𝑥2 + 𝑦2 𝜕𝑧 𝜕𝑥 = 1 𝑥2 + 𝑦2 ∙ 2𝑥 𝜕𝑧 𝜕𝑦 = 1 𝑥2 + 𝑦2 ∙ 2𝑦 = 2𝑥 𝑥2 + 𝑦2 = 2𝑦 𝑥2 + 𝑦2 𝑧 = 𝑙𝑛 𝑥2𝑦2 𝜕𝑧 𝜕𝑥 = 1 𝑥2𝑦2 ∙ 2𝑥𝑦2 𝜕𝑧 𝜕𝑦 = 1 𝑥2𝑦2 ∙ 2𝑥2 𝑦 = 2 𝑥 = 2 𝑦 𝑧 = 𝑒 𝑥2+𝑦2 𝜕𝑧 𝜕𝑥 = 𝑒 𝑥2+𝑦2 ∙ 2𝑥 𝜕𝑧 𝜕𝑦 = 𝑒 𝑥2+𝑦2 ∙ 2𝑦 = 2𝑥𝑒 𝑥2+𝑦2 = 2𝑦𝑒 𝑥2+𝑦2 𝑧 = 𝑒 𝑥2𝑦2 𝜕𝑧 𝜕𝑥 = 𝑒 𝑥2𝑦2 ∙ 2𝑥𝑦2 𝜕𝑧 𝜕𝑦 = 𝑒 𝑥2𝑦2 ∙ 2𝑥2 𝑦 = 2𝑥𝑦2𝑒 𝑥2𝑦2 = 2𝑥2𝑦𝑒 𝑥2𝑦2
- 43. Partial Differentiation: Second Order • Given a function 𝑓 𝑥, 𝑦 , the function is said to be differentiable if 𝑓𝑥 and 𝑓𝑦 exist. • If the function is differentiable its first-order derivatives can be differentiated again and we can define the second-order partial derivatives of z = 𝑓 𝑥, 𝑦 as follows:
- 44. Differentiate 𝑥, 𝑦 jadi PEMALAR Differentiate 𝑦, 𝑥 jadi PEMALAR Jika 𝒙 & 𝒚 adalah SINGLE (sendirian) !! Differentiate 𝑥, 𝑦 salin semula Differentiate 𝑦, 𝑥 salin semula Jika 𝒙 & 𝒚 adalah COUPLE (bersama2) !!
- 45. Partial Differentiation: Second Order Given 𝑧 = 𝑥3 + 5𝑦2 − 8𝑥3𝑦2 𝜕𝑧 𝜕𝑥 = 3𝑥2 + 0 − 24𝑥2𝑦2 𝜕𝑧 𝜕𝑦 = 0 + 10𝑦 − 16𝑦𝑥3 𝜕2𝑧 𝜕𝑥2 = 6𝑥 − 48𝑥𝑦2 𝜕2 𝑧 𝜕𝑦𝜕𝑥 = 0 − 48𝑦𝑥2 𝜕2𝑧 𝜕𝑦2 = 10 − 16𝑥3 𝜕2 𝑧 𝜕𝑥𝜕𝑦 = 0 − 48𝑥2𝑦 First Order Partial Differentiation Second Order Partial Differentiation terhadap 𝑥 terhadap 𝑦 terhadap 𝑦 terhadap 𝑥 terhadap 𝑥 terhadap 𝑦 Jawapan mesti SAMA !
- 46. Partial Differentiation : Second Order Find the second order of partial differentiation for all the following: a) 𝑧 = 5𝑥 + 𝑥5 𝑦6 + 4𝑦 + 25 𝜕𝑧 𝜕𝑥 = 5 + 5𝑥4 𝑦6 𝜕𝑧 𝜕𝑦 = 6𝑦5𝑥5 − 4 𝜕2𝑧 𝜕𝑥2 = 20𝑥3 𝑦6 𝜕2 𝑧 𝜕𝑦𝜕𝑥 = 30𝑦5 𝑥4 𝜕2 𝑧 𝜕𝑦2 = 30𝑦4𝑥5 𝜕2𝑧 𝜕𝑥𝜕𝑦 = 30𝑥4𝑦5 = 30𝑥4 𝑦5 = 30𝑥5𝑦4 = 6𝑥5𝑦5 − 4 terhadap 𝑥 terhadap 𝑦 terhadap 𝑦 terhadap 𝑥 terhadap 𝑥 terhadap 𝑦
- 47. Partial Differentiation : Second Order Find the second order of partial differentiation for all the following: b) 𝑓 𝑥, 𝑦 = 2𝑥𝑒4𝑦 + 𝑐𝑜𝑠 3𝑥2 + 𝑦3 𝑓𝑥 = 2𝑒4𝑦 + 𝑐𝑜𝑠 3𝑥2 + 𝑦3 ∙ 6𝑥 𝑓𝑥𝑥 = 𝑐𝑜𝑠 3𝑥2 + 𝑦3 ∙ 6 + 6𝑥 ∙ −𝑠𝑖𝑛 3𝑥2 + 𝑦3 𝑓𝑥𝑦 = 2𝑒4𝑦 ∙ 4 + 6𝑥 𝑐𝑜𝑠 3𝑥2 + 𝑦3 ∙ 3𝑦2 = 2𝑒4𝑦 + 6𝑥 𝑐𝑜𝑠 3𝑥2 + 𝑦3 = 6𝑐𝑜𝑠 3𝑥2 + 𝑦3 − 6𝑥 𝑠𝑖𝑛 3𝑥2 + 𝑦3 = 8𝑒4𝑦 + 18𝑥𝑦2𝑐𝑜𝑠 3𝑥2 + 𝑦3 product rule
- 48. Partial Differentiation : Second Order Find the second order of partial differentiation for all the following: b) 𝑓 𝑥, 𝑦 = 2𝑥𝑒4𝑦 + 𝑐𝑜𝑠 3𝑥2 + 𝑦3 𝑓𝑦 = 2𝑥𝑒4𝑦 ∙ 4 + 𝑐𝑜𝑠 3𝑥2 + 𝑦3 ∙ 3𝑦2 𝑓𝑦𝑦 = 8𝑥𝑒4𝑦 ∙ 4 + 𝑐𝑜𝑠 3𝑥2 + 𝑦3 ∙ 6𝑦 + 3𝑦2 ∙ −𝑠𝑖𝑛 3𝑥2 + 𝑦3 ∙ 3𝑦2 𝑓𝑦𝑥 = 2𝑒4𝑦 ∙ 4 + 6𝑥 𝑐𝑜𝑠 3𝑥2 + 𝑦3 ∙ 3𝑦2 = 8𝑒4𝑦 + 18𝑥𝑦2 𝑐𝑜𝑠 3𝑥2 + 𝑦3 = 8𝑥𝑒4𝑦 + 3𝑦2 𝑐𝑜𝑠 3𝑥2 + 𝑦3 = 32𝑥𝑒4𝑦 + 6𝑦 𝑐𝑜𝑠 3𝑥2 + 𝑦3 − 9𝑦4𝑠𝑖𝑛 3𝑥2 + 𝑦3 product rule
- 49. Partial Differentiation : Second Order Find the second order of partial differentiation for all the following: c) 𝑧 = 𝑥2 𝑙𝑛 3 − 𝑦4 𝜕𝑧 𝜕𝑥 = 2𝑥 𝑙𝑛 3 − 𝑦4 𝜕𝑧 𝜕𝑦 = 𝑥2 𝑙𝑛 3 − 𝑦4 ∙ −4𝑦3 𝜕2 𝑧 𝜕𝑥2 = 2 𝑙𝑛 3 − 𝑦4 𝜕2𝑧 𝜕𝑦𝜕𝑥 = 2𝑥 𝑙𝑛 3 − 𝑦4 ∙ −4𝑦3 𝜕2𝑧 𝜕𝑦2 = −4𝑥2 𝑦3 𝑙𝑛 3 − 𝑦4 ∙ −4𝑦3 𝜕2 𝑧 𝜕𝑥𝜕𝑦 = 2𝑥 𝑙𝑛 3 − 𝑦4 ∙ −4𝑦3 = −8𝑥𝑦3 𝑙𝑛 3 − 𝑦4 = −4𝑥2 𝑦3 𝑙𝑛 3 − 𝑦4 = 16𝑥2 𝑦6 𝑙𝑛 3 − 𝑦4 = −8𝑥𝑦3 𝑙𝑛 3 − 𝑦4
- 51. Technique of Total Differentiation • A total derivative of a multivariable function of several variables, each of which is a function of another argument, is the derivative of the function with respect to said argument. • It is equal to the sum of the partial derivatives with respect to each variable times the derivative of that variable with respect to the independent variable.
- 52. Technique of Total Differentiation • A function 𝑤 = 𝑓 𝑥, 𝑦, 𝑧 and 𝑡 with 𝑥, 𝑦, 𝑧 being functions of 𝑡, • Then the total derivatives of 𝑤 with respect to 𝑡 is given by • Total derivatives are often used in related rates problems; for example, finding the rate of change of volume when two parameters are changing with time. 𝑑𝑤 𝑑𝑡 = 𝜕𝑤 𝜕𝑥 ∙ 𝑑𝑥 𝑑𝑡 + 𝜕𝑤 𝜕𝑦 ∙ 𝑑𝑦 𝑑𝑡 + 𝜕𝑤 𝜕𝑧 ∙ 𝑑𝑧 𝑑𝑡
- 53. Technique of Total Differentiation Solve all the following: a) 𝑧 = 𝑥2 𝑦3 − 2𝑥 + 3 5𝑦 𝜕𝑧 𝜕𝑥 = 2𝑥𝑦3 − 2 𝜕𝑧 𝜕𝑦 = 3𝑥2 𝑦2 − 3 5𝑦2 𝑑𝑧 = 𝜕𝑧 𝜕𝑥 ∙ 𝑑𝑥 + 𝜕𝑧 𝜕𝑦 ∙ 𝑑𝑦 partial differentiation Chain rule total derivative = 2𝑥𝑦3 − 2 ∙ 𝑑𝑥 + 3𝑥2 𝑦2 − 3 5𝑦2 ∙ 𝑑𝑦 substitute value
- 54. Technique of Total Differentiation Solve all the following: b) 𝑠 = 𝑥𝑦 + 𝑠𝑖𝑛𝑦2 − 𝑒3𝑥 𝜕𝑠 𝜕𝑥 = 𝑦 − 3𝑒3𝑥 𝜕𝑠 𝜕𝑦 = 𝑥 + 2𝑦 𝑐𝑜𝑠𝑦2 𝑑𝑠 = 𝜕𝑠 𝜕𝑥 ∙ 𝑑𝑥 + 𝜕𝑠 𝜕𝑦 ∙ 𝑑𝑦 partial differentiation Chain rule total derivative = 𝑦 − 3𝑒3𝑥 ∙ 𝑑𝑥 + 𝑥 + 2𝑦 𝑐𝑜𝑠𝑦2 ∙ 𝑑𝑦 substitute value
- 55. Technique of Total Differentiation Solve all the following: c) The radius and height of a cylinder are both 2 𝑐𝑚. The radius is decreased at 1 𝑐𝑚 𝑠𝑒𝑐 and the height is increasing at 2 𝑐𝑚 𝑠𝑒𝑐 . What is the change in volume with respect to time at this instant? 𝜕𝑉 𝜕𝑟 = 2𝜋𝑟ℎ 𝜕𝑉 𝜕ℎ = 𝜋𝑟2 𝑑𝑉 𝑑𝑡 = 𝜕𝑉 𝜕𝑟 ∙ 𝑑𝑟 𝑑𝑡 + 𝜕𝑉 𝜕ℎ ∙ 𝑑ℎ 𝑑𝑡 partial differentiation Chain rule total derivative 𝑑𝑉 𝑑𝑡 = 2𝜋𝑟ℎ ∙ 𝑑𝑟 𝑑𝑡 + (𝜋𝑟2) ∙ 𝑑ℎ 𝑑𝑡 substitute partial differentiation The volume of cylinder is 𝑉 = 𝜋𝑟2ℎ The total derivative of this with respect to time is = 2𝜋 2 2 ∙ −1 + 𝜋 2 2 ∙ 2 = −8𝜋 + 8𝜋 = 0 ∴ there is no changes in volume of the cylinder 𝑑𝑟 𝑑𝑡 = −1 given info from question 𝑑ℎ 𝑑𝑡 = 2 𝑟 = ℎ = 2 substitute value
- 56. Technique of Total Differentiation Solve all the following: d) If 𝑧 = 1 − 2𝑥2 + 3𝑥𝑦 + 𝑦3 , find the total derivative of 𝑧 when 𝑥, 𝑦 changes from 1, 2 to 0.8, 2.3 . 𝜕𝑧 𝜕𝑥 = −4𝑥 + 3𝑦 𝜕𝑧 𝜕𝑦 = −3𝑥 + 3𝑦2 𝑑𝑧 = 𝜕𝑧 𝜕𝑥 ∙ 𝑑𝑥 + 𝜕𝑧 𝜕𝑦 ∙ 𝑑𝑦 partial differentiation Chain rule total derivative 𝑑𝑧 = −4𝑥 + 3𝑦 ∙ 𝑑𝑥 + (−3𝑥 + 3𝑦2 ) ∙ 𝑑𝑦 substitute partial differentiation = −4 1 + 3 2 ∙ −0.2 + +(−3 1 + 3 2 2 ) ∙ 0.3 = 2.9 𝑢𝑛𝑖𝑡 ∴ there is 2.9 𝑢𝑛𝑖𝑡 changes in 𝑧 𝑑𝑥 = −0.2 given info from question 𝑑𝑦 = 0.3 substitute value 1, 2 0.8, 2.3 change of 𝑥 change of 𝑦 −0.2 0.3