Differentiation and Integration - Chain Rule

Chain Rule The Chain Rule is a formula for computing the derivative of the composition of two functions. d/dx (y) = d/du (y) * d/dx (u) Functions 2 -Differentiation Continued & Integration Created by Dr Paraic Treacy Chain Rule Examples of functions that are differentiated using the chain rule: 1. y = (3x ^ 2 + 17) ^ 4 2. y = e ^ (4x ^ 2 + 6x) 3. y = ln(2x ^ 3 + 7x) 4. y = sqrt(3x ^ 3 + 1) 5. y = 3sin(2x ^ 2 + 7) 6. y = - 4cos(x ^ 3 - 4x) Chain Rule - Ex. 1 Differentiate y = (2x - 3) ^ 5 with respect to x. Let u = 2x - 3 d/dx (u) = 2 Now: y = u ^ 5 d/du (y) = 5u ^ 4 Chain Rule - Ex. 1 d/dx (y) = (5u ^ 4)(2) d/dx (y) = d/du (y) * d/dx (u) d/dx (y) = 10u ^ 4 = 10 * (2x - 3) ^ 4 Answer: d/dx (y) = 10 * (2x - 3) ^ 4 Chain Rule - Ex. 2 Differentiate y = ln(cos x ^ 2) with respect to x. Let u = cos x ^ 2 dx = -2xsinx² Now: y = ln(u) d/du (y) = 1/u Special Cases (See Log Tables page 25) f(x) f' * (x) = d/dx [f(x)] x ^ n n * x ^ (n - 1) In x 1/x COSX -sinx sin x COS X epsilon ^ x e ^ (ax) e ^ x Chain Rule - Ex. 2 dy dx =()(-2xsinx²) f' * (x) = (- 2x * sin x ^ 2)/(cos x ^ 2) d/dx (y) = d/du (y) * d/dx (u) = (- 2x * sin x ^ 2)/u = (- 2x * sin x ^ 2)/(cos x ^ 2) Steps for applying the Chain Rule 1. Identify u and rewrite y in terms of u. 2. Differentiate u to get and differentiate y to get d/dx (u) d/du (y) 3. Sub for d/du (y) and d/dx (u) in the formula d/dx (y) = d/du (y) * d/dx (u) 4. Simplify the equation. Replace u with what it was originally. Chain Rule - A Quicker Way - Ex. 1 Differentiate y = (2x - 3) ^ 5 with respect to x. Reduce power by 1 Multiply by the original power Multiply by the diff of what's inside the brackets d/dx (y) = 5 * (2x - 3) ^ 4 * (2) d/dx (y) = 10 * (2x - 3) ^ 4 Chain Rule - A Quicker Way - Ex. 2 Differentiate y = sqrt(4x - 1) with respect to x. Rewrite: y = (4x - 1) ^ (1/7) Multiply in the rig Reduce power by I d/dx (y) = 1/2 * (4x - 1) ^ (z_{1}/2) * (4) Muttipity by the dit of what's hande the br d/dx (y) = 2 * (4x - 1) ^ (- 1/2) = 2/(sqrt(4x - 1)) Chain Rule - Ex. 3 The revenue received from the sale of electric fans is seasonal, with maximum revenue received in the summer. The revenue received from the sale of fans can be approximated by: R(x) = 100cos((pi*x)/6) + 120 where x is time in months measured from July 1st Find R' * (x) and evaluate R' * (x) for (i) August (ii) May 1 ^ st ; and interpret your answers. 1 ^ pi Chain Rule - Ex. 3 R(x) = 100cos((pi*x)/6) + 120 Apply the chain rule: Let u = (pi*x)/6 Now: R(u) = 100cos(u) + 120 d/du (R) = - 100sin(u) d/dx (u) = pi/6 u = (pi*x)/6 We know: d/dx (R) = - 100sin(u) * pi/6 d/dx (R) = d/du (R) * d/dx (u) Chain Rule - Ex. 3 d/dx (R) = - 100sin(u) * pi/6 d/dx (R) = - (50pi)/3 * sin(u) d/dx (R) = - (50pi)/3 * sin((pi*x)/6) Chain Rule - Ex. 3 Short Approach to Chain Rule: R' * (x) = (pi/6) * 100(- sin((pi*x)/6)) R' * (x) = - (50pi)/3 * sin((pi*x)/6) R(x) = 100cos((pi*x)/6) + 120 Note: R ^ r * (x) means the same as d/dx (R) Chain Rule - Ex. 3 R' * (x) = - (50pi)/3 * sin((pi*x)/6) (i) R' * (x) for August 1 ^ a x = 1 as August 1 is one month after july 1 ^ a R' * (1) = - (50pi)/3 * sin((pi(1))/6) R' * (1) = - 26.18 Rate of change is negative -this means that the revenue received from the sale of fans is decreasing in August. Chain Rule - Ex. 3 R' * (x) = - (50pi)/3 * sin((pi*x)/6) R' * (10) = 45.345 (ii) R' * (x) for May 1 Rate of change is positive - this means that the revenue received from the sale of fans is increasing in May. x = 10 as May 1" is ten months after July 1 R' * (10) = - (50pi)/3 * sin((pi(10))/6) Optimisation - Ex. 3 Square corners are cut from a piece of 12 cm by 12 cm tinplate which is then bent to form an open dish. What size squares should be removed to maximise the capacity of the dish? m (12 - 2x) * cns Optimisation - Ex. 3 Capacity = length x width x depth C = (12 - 2x)(12 - 2x)(x) C = 144x - 48x ^ 2 + 4x ^ 3 (0-2) 1. Differentiate Capacity to find d/dx (C) 2. Let d/dx (c) = 0 3. Find value for x Optimisation - Ex. 3 We take x = 2 because would not be possible (refer to the diagram) x = 6 Max capacity at x = 2 C = 144x - 48x ^ 2 + 4x ^ 3 Maximum Capacity = 128c * m ^ 3 C = 144(2) - 48 * (2) ^ 2 + 4 * (2) ^ 3 (12-2) Calculus Problem in the Real World Two variables change simultaneously: the radius of the circle and its area.. The first key thing we must do is establish some kind of relationship between these two variables and we can draw on our knowledge of geometry to do this: A = pi * r ^ 2 We are asked to identify how fast the area of the patch is increasing ie. d/dx (A) We are told the radius increases at a rate of 30 m/hr. d/dt (r) = 30 Optimisation - Ex. 3 [Divide across by 12 to Let d/dx (C) = 0 12x ^ 2 - 96x + 144 = 0 d/dx (C) = 144 - 96x + 12x ^ 2 C = 144x - 48x ^ 2 + 4x ^ 3 144 - 96x + 12x ^ 2 = 0 simplify] (x - 2)(x - 6) = 0 x = 2 or x = 6 x ^ 2 - 8x + 12 = 0 Calculus Problem in the Real World An oil rig springs a leak in calm seas and the oil spreads in a circular patch around the rig. If the radius of the oil patch increases at a rate of 30 metres/hour, how fast is the area of the patch increasing when the patch has a radius of 100 metres. Also consider what will happen as the radius increases? Calculus Problem in the Real World Now we will try to write in terms of (dA)/(dt) = dr dt *? d/dt (A) = d/dt (r) * d/dr (A) d/dt (r) d/dt (A) We know A = pi * r ^ 2 If we differentiate this: Now we can solve d/dr (A) = 2pi*r d/dt (A) = (30)(2pi*r) d/dt (A) = d/dt (r) * d/dr (A) Calculus Problem in the Real World d/dt (A) = 60pi*r d/dt (A) = (30)(2pi*r) Conclusion: the area of the patch is increasing at a rate of 60nr We are asked for the rate of increase in area when the radius is 100 metres = 18 ,850 m^ 2 /hr d/dt (A) = 60pi*r = 60pi(100) When the radius of the oil patch is 100 m, the oil patch is increasing at a rate of 18, 850m ^ 2 / h * r What is Integration? Integration is the reverse process of differentiation. integrate x ^ n dx = (x ^ (n + 1))/(n + 1) + c power Add one to the power Divide by new "c" represents the constant that may be present (to be explained later). Applications of Integration Integration is used to calculate the following: Moments of Inertia. Volume of a solid revolution. Electric Charges. Force by a liquid pressure. Area under a curve. Area between two curves. Work by a variable force. displacement, Velocity, Acceleration. Integration and Differentiation Differentiate Integration and Differentiation If we start with the function f(x) = 3x ^ 2 + 4x - 10 Differentiate f(x) d/dx (f) = 6x + 4 Integrate to reverse this process: integrate 1 df = integrate 6x + 4 dx f(x) = 3x ^ 2 + 4x + c f(x) = (6x ^ 2)/2 + 4x + c Integration and Differentiation We started with f(x) = 3x ^ 2 + 4x - 10 Integration reversed the process of differentiation and we ended up with f(x) = 3x ^ 2 + 4x + c The constant (-10) was eliminated during differentiation thus the reason for including "c" when integrating.