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Page 1 of 12 Calculus 1 – Differential Calculus Professor: Dr. Clemente M. Aguinaldo Jr., PNU North Luzon Disclaimer: This learning material is the property of Dr. Clemente M. Aguinaldo Jr. of PNU North Luzon and is intended for use in his Calculus 1 class. Unauthorized reproduction, distribution, or use—including taking photos of any part of this material—is strictly prohibited. Name: Lesson 4: Differentiation Formulas In plane geometry, a line that touches a circle at exactly one point is called a tangent line. But what if a line T touches the graph of a function at a single point? Is T a tangent line? The answer is yes. In this lesson, we will define T as the tangent line to a curve at a point P if it passes through P and has slope m, where m is the limit of the slopes of the secant lines as other points in the curve approach P. This slope will be defined later as the derivative of a function. Let us discuss some terms first before doing differentiations. AVERAGE RATE OF CHANGE Let’s consider a function y = f(x) and two inputs x1 and x2. The change in input, or the change in x, is x2 – x1. The change in output, or the change in y, is y2 – y1, where y1 = f(x1) and y2 = f(x2). Now, we can get the ratio of the change in output and the change in input and call it the average rate of change, defined below. Definition 4.1: (Average Rate of Change) The average rate of change of y with respect to x, as x changes from x1 to x2, is the ratio of the change in output to the change in input, written as 𝑦2−𝑦1 𝑥2−𝑥1 , where x2 x1. If we look at the graph of a function on the right, we see that 𝑦2−𝑦1 𝑥2−𝑥1 = 𝑓(𝑥2)−𝑓(𝑥1) 𝑥2−𝑥1 , which is both the average rate of change and the slope of the line from P(x1, y1) to Q(x2, y2). The line passing through P and Q is called a secant line. The slope of the secant line QP is interpreted as the average rate of change of f from x1 to x2. Example: For y = f(x) = x2, find the average rate of change as: 1) x changes from 1 to 3, 2) x changes from 2 to 4. Solution: 1) 𝑦2−𝑦1 𝑥2−𝑥1 = 𝑓(𝑥2)−𝑓(𝑥1) 𝑥2−𝑥1 = = = ____. This can also be interpreted as: the slope of the secant line PQ, passing through P(1, 1) and Q(3, 9), is ____. 2) 𝑦2−𝑦1 𝑥2−𝑥1 = 𝑓(𝑥2)−𝑓(𝑥1) 𝑥2−𝑥1 = = = ____. We now develop a notation for average rates of change that do not require subscripts. Instead of x1, we will write simply x; in place of x2, we will write x + h. It may help to think of the h as the horizontal distance between the inputs x1 and x2. That is, to get from x1, or x, to x2, we move a distance h. Thus, x2 = x + h. Then the average rate of change, also called a difference quotient, is given by 𝑦2−𝑦1 𝑥2−𝑥1 = 𝑓(𝑥2)−𝑓(𝑥1) 𝑥2−𝑥1 = 𝑓(𝑥+ℎ)−𝑓(𝑥) (𝑥+ℎ)−𝑥 = 𝑓(𝑥+ℎ)−𝑓(𝑥) ℎ .
Page 2 of 12 Calculus 1 – Differential Calculus Professor: Dr. Clemente M. Aguinaldo Jr., PNU North Luzon Disclaimer: This learning material is the property of Dr. Clemente M. Aguinaldo Jr. of PNU North Luzon and is intended for use in his Calculus 1 class. Unauthorized reproduction, distribution, or use—including taking photos of any part of this material—is strictly prohibited. Definition 4.2: (Difference Quotient) The average rate of change of f with respect to x is also called the difference quotient. It is given by 𝑓(𝑥+ℎ)−𝑓(𝑥) ℎ , where h 0. The difference quotient is equal to the slope of the secant line from (x, f(x)) to (x + h, f(x + h)). Keep in mind that, in general, f(x + h) f(x) + f(h). In other texts, x is used instead of h. Example: For f(x) = x2, find the difference quotient when: 1) x = 4 and h = 1. 2) x = 4 and h = 0.1. 3) x = 4 and h = 0.01. Solution: 1) We substitute x = 4 and h = 1 into the formula: 𝑓(𝑥+ℎ)−𝑓(𝑥) ℎ = 𝑓(4+1)−𝑓(4) 1 = 𝑓(5)−𝑓(4) 1 = 𝑓(5)−𝑓(4) 1 = = = ______. The difference quotient is ___. It is also the slope of the line from (4, 16) to (5, 25). 2) We substitute x = 4 and h = 0.1 into the formula: 𝑓(𝑥+ℎ)−𝑓(𝑥) ℎ = 𝑓(4+0.1)−𝑓(4) 0.1 = 𝑓(4.1)−𝑓(4) 0.1 . Since f(4.1) = (4.1)2 = 16.81 and f(4) = 42 = 16, we have 𝑓(4.1)−𝑓(4) 0.1 = = = _____. 3) We substitute x = 4 and h = 0.01 into the formula: 𝑓(𝑥+ℎ)−𝑓(𝑥) ℎ = 𝑓(4+0.01)−𝑓(4) 0.01 = 𝑓(4.01)−𝑓(4) 0.01 = (4.01)2−(4)2 0.01 = = = ____. Here, the slope of the line from (4.01, 16.0801) to (4, 16) to = ______. Have you noticed that the difference quotient here is getting closer and closer to a certain value as h gets closer and closer to zero? This means that as the distance between x and h becomes smaller and smaller, h also approaches 0. When this happens, the secant lines PQi’s that intersect the graph of a function at two points become a tangent line T that touches P. This is illustrated in the figure at the right. The limit of the difference equation as h approaches zero is the slope m of the tangent line at P(x, f(x)). This limit is also called the instantaneous rate of change of f(x) at x or the derivative of y = f(x) with respect to x. Definition 4.3: (Derivative) For a function y = f(x), its derivative at x is the function f defined by f (x) = lim ℎ→0 𝑓(𝑥+ℎ)−𝑓(𝑥) ℎ , provided that the limit exists. If f (x) exists, then we say that f is differentiable at x. The process of finding the derivative of f is called differentiation. The definition gives us the three-step rule in calculating the derivative of a function f(x). Step 1. Write the difference quotient, 𝑓(𝑥+ℎ)−𝑓(𝑥) ℎ . Step 2. Simplify the difference quotient. Step 3. Find the limit as h approaches 0.
Page 3 of 12 Calculus 1 – Differential Calculus Professor: Dr. Clemente M. Aguinaldo Jr., PNU North Luzon Disclaimer: This learning material is the property of Dr. Clemente M. Aguinaldo Jr. of PNU North Luzon and is intended for use in his Calculus 1 class. Unauthorized reproduction, distribution, or use—including taking photos of any part of this material—is strictly prohibited. Example: For f(x) = x2, find f (x). Then, find f (4) and f (–3). Solution: We have: Step 1. 𝑓(𝑥+ℎ)−𝑓(𝑥) ℎ = . Step 2. (𝑥+ℎ)2−𝑥2 ℎ = ℎ = ℎ = ℎ = _________, h 0 Step 3. f (x) = lim ℎ→0 𝑓(𝑥+ℎ)−𝑓(𝑥) ℎ = lim ℎ→0(___________) = _____ = __ Using the fact that f (x) = ____, then f (4) = _______ = __ and f (____) = ______ = ____. This tells us that at x = 4, the tangent line has slope f (4) = 8, and at x = –3, the curve has a tangent line whose slope is f (–3) = –6. We can also say: • The tangent line to the curve at the point (4, 16) has slope 8. • The tangent line to the curve at the point (–3, 9) has slope –6. • The instantaneous rate of change at x = 4 is 8. • The instantaneous rate of change at x = –3 is –6. Example: Given f(x) = √𝑥. Use the three-step rule to find f (x). Also, answer the following: 1. Evaluate: f (16). 2. For what x-values is the function not differentiable? 3. Find an equation of the tangent line to the curve at x = 9. (Hint: Use the point-slope form y – y1 = m(x – x1).) Example: Given f(x) = 1 2𝑥+1. Use the three-step rule to find f (x). Also, answer the following: 1. Evaluate: f (3). 2. For what x-values is the function not differentiable? 3. Find an equation of the tangent line to the curve at x = 5.
Page 4 of 12 Calculus 1 – Differential Calculus Professor: Dr. Clemente M. Aguinaldo Jr., PNU North Luzon Disclaimer: This learning material is the property of Dr. Clemente M. Aguinaldo Jr. of PNU North Luzon and is intended for use in his Calculus 1 class. Unauthorized reproduction, distribution, or use—including taking photos of any part of this material—is strictly prohibited. In the previous example, note that since f(–1/2) does not exist for f(x) = 1/(2x + 1), we cannot evaluate the difference quotient 𝑓(0+ℎ)−𝑓(0) ℎ . Thus, f(–1/2) does not exist. We say that “f is not differentiable at 0.” In general, if a function is discontinuous at a point, it is not differentiable at that point. Example: Given f(x) = |𝑥|. Use the three-step rule to find f (x). Also, answer the following: 1. Evaluate: f (5) and f (0). 2. For what x-values is the function not differentiable? 3. Find an equation of the tangent line to the curve at x = 100. Sometimes, a function f is continuous at a point, but its derivative f is not defined at this point. The function f(x) = | x | above is an example. Observe that it is continuous at x = 0 since it meets all the requirements for continuity there. But what about a tangent line at this point? The graph of f(x) = | x | resembles a V-shape with the tip or “corner” at (0, 0). So, there could be many tangent lines at this “corner” and thus, many slopes. It follows that f (0) does not exist. In general, if a function has a “corner”, it will not have a derivative at that point. Practice Task 4.1: A. Given the functions f(x) below: a. 𝑓(𝑥) = 𝑥 − 1 𝑥2 b. 𝑓(𝑥) = (2𝑥 − 1)3 c. 𝑓(𝑥) = √3𝑥 − 6 Do the following: 1. Use the three-step rule to find f (x). 2. Find an equation of the tangent line to the curve at x = 6. 3. For what x-values is the function not differentiable? B. Consider the function h given by h(x) = | x – 3 | + 2. 1. For what value of x is this function not differentiable? 2. Evaluate h(0), h(1), h(4), h(10). Is there a shortcut you can use to find these slopes?
Page 5 of 12 Calculus 1 – Differential Calculus Professor: Dr. Clemente M. Aguinaldo Jr., PNU North Luzon Disclaimer: This learning material is the property of Dr. Clemente M. Aguinaldo Jr. of PNU North Luzon and is intended for use in his Calculus 1 class. Unauthorized reproduction, distribution, or use—including taking photos of any part of this material—is strictly prohibited. BASIC DIFFERENTIATION TECHNIQUES Before we discuss the techniques, note that there are symbols used to denote the derivative of y = f(x). A common way to express “the derivative of y with respect to x” is the notation 𝑑𝑦 𝑑𝑥, simply read as dy over dx, which was introduced by the German mathematician Gottfried Wilhelm von Leibniz. Together with Sir Isaac Newton, they are both credited with the invention of calculus though each performed his work independently. In practice, we often use y (y prime) to represent f (x) when there is no confusion as to which variables are involved. Other notations are Dxy, Dxf(x), and 𝑑 𝑑𝑥 𝑓(𝑥). Then if we wish to calculate a derivative at a number, say f (2), we can also write 𝑑𝑦 𝑑𝑥|𝑥=2. The following techniques in this section are stated as theorems. We use the three-step rule to prove them. Theorem 4.1: (The Constant Rule) The derivative of a constant function is zero. Proof: Let f(x) = c, where c is any constant. Then, f (x) = lim ℎ→0 𝑓(𝑥+ℎ)−𝑓(𝑥) ℎ = lim ℎ→0 𝑐−𝑐 ℎ = lim ℎ→0 0 = 0. ■ The result also follows from the fact that the graph of y = f(x) = c, for any constant c, is a horizontal line whose slope is zero. Theorem 4.2: (The Power Rule) For any real number n, if y = xn, then 𝑑𝑦 𝑑𝑥 = nxn – 1. Proof: Let y = xn, for any positive integer n. Then, 𝑑𝑦 𝑑𝑥 = lim ℎ→0 𝑓(𝑥+ℎ)−𝑓(𝑥) ℎ = lim ℎ→0 (𝑥+ℎ)𝑛−𝑥𝑛 ℎ = lim ℎ→0 (𝑥𝑛+𝑛𝑥𝑛−1ℎ+𝑛(𝑛−1) 2! 𝑥𝑛−2ℎ2+⋯+𝑛𝑥ℎ𝑛−1+ℎ𝑛)−𝑥𝑛 ℎ = lim ℎ→0 𝑛𝑥𝑛−1ℎ+𝑛(𝑛−1) 2! 𝑥𝑛−2ℎ2+⋯+𝑛𝑥ℎ𝑛−1+ℎ𝑛 ℎ = lim ℎ→0 ℎ(𝑛𝑥𝑛−1+𝑛(𝑛−1) 2! 𝑥𝑛−2ℎ+⋯+𝑛𝑥ℎ𝑛−2+ℎ𝑛−1) ℎ = lim ℎ→0 (𝑛𝑥𝑛−1 + 𝑛(𝑛−1) 2! 𝑥𝑛−2ℎ + ⋯ + 𝑛𝑥ℎ𝑛−2 + ℎ𝑛−1) = nxx – 1. ■ Note that (x + h)n was expanded using the binomial theorem in algebra. Using combination notation, the binomial expansion can also be written as (x + h)n = C(n, 0)xnh0 + C(n, 1)xn – 1h + C(n, 2)xn – 2h2 + ·· + C(n, n – 1)xhn – 1 + C(n, n)x0hn. Although we have proven the Power Rule only for the case where n is a positive integer, it is valid for all real numbers n. A shorter proof can be shown for any real number n using implicit differentiation and logarithmic differentiation. As a word of CAUTION, do not use the Power Rule when the exponent is not a number and/or the base is not a variable. As a special case of the power rule, when y = x, then 𝑑𝑦 𝑑𝑥 = 𝑑𝑥 𝑑𝑥 = 1. Example: For f(x) = x2 and g(x) = √𝑥 , find f (x) and g(x). Solution: We have f (x) = 2x2 – 1 = 2x, and since √𝑥 = 𝑥1 2, g(x) = ½x1/2 – 1 = ½x–1/2 = 1 2𝑥1/2 = 𝟏 𝟐√𝒙. (Memorize this form!)
Page 6 of 12 Calculus 1 – Differential Calculus Professor: Dr. Clemente M. Aguinaldo Jr., PNU North Luzon Disclaimer: This learning material is the property of Dr. Clemente M. Aguinaldo Jr. of PNU North Luzon and is intended for use in his Calculus 1 class. Unauthorized reproduction, distribution, or use—including taking photos of any part of this material—is strictly prohibited. Example: Find f (x) given the following functions: 1) f(x) = x2 2) f(x) = 1 √𝑥23 Theorem 4.3: (The Constant Multiple Rule) The derivative of a constant times a function is the constant times the derivative of the function. That is, if y = F(x) = cf(x), then𝑑𝑦 𝑑𝑥 = c f (x). Proof: Let F(x) = cf(x). Then, 𝑑𝑦 𝑑𝑥 = F (x) = lim ℎ→0 𝐹(𝑥+ℎ)−𝐹(𝑥) ℎ = lim ℎ→0 𝑐𝑓(𝑥+ℎ)−𝑐𝑓(𝑥) ℎ = clim ℎ→0 𝑓(𝑥+ℎ)−𝑓(𝑥) ℎ = c f (x). ■ Example: Find each of the following derivatives: 1) 𝑑 𝑑𝑥 4x3, 2) 𝑑 𝑑𝑥 (–5x4), and 3) 𝑑 𝑑𝑥 ( 1 6𝑥2). Example: Find the rate of change of volume V with respect to the radius r of the base of a right circular cylinder of height 5 inches. Solution: The formula for the volume of a right circular cylinder is V = f(r) = r2h. Then with h = 5, we have V = 5r2. So, dV/dr = 10r cubic inches with respect to its radius. Theorem 4.4: (The Sum-Difference Rule) The derivative of the sum of two functions is the sum of their derivatives. 𝑑 𝑑𝑥 [𝑓(𝑥) + 𝑔(𝑥)] = 𝑓(𝑥) + 𝑔(𝑥). The derivative of the difference of two functions is the difference of their derivatives. 𝑑 𝑑𝑥 [𝑓(𝑥) − 𝑔(𝑥)] = 𝑓(𝑥) − 𝑔(𝑥). Proof: The proof of the Sum Rule relies on the fact that the limit of a sum is the sum of the limits. Let F(x) = f(x) + g(x). Then, lim ℎ→0 𝐹(𝑥+ℎ)−𝐹(𝑥) ℎ = lim ℎ→0 [𝑓(𝑥+ℎ)+𝑔(𝑥+ℎ)]−[𝑓(𝑥)+𝑔(𝑥)] ℎ = lim ℎ→0 [𝑓(𝑥+ℎ)−𝑓(𝑥) ℎ + 𝑔(𝑥+ℎ)−𝑔(𝑥) ℎ ] = f (x) + g(x). ■ To prove the Difference Rule, we note that 𝑑 𝑑𝑥 [𝑓(𝑥) − 𝑔(𝑥)] = 𝑑 𝑑𝑥 [𝑓(𝑥) + (−1)𝑔(𝑥)] = 𝑑 𝑑𝑥 𝑓(𝑥) + 𝑑 𝑑𝑥 (−1)𝑔(𝑥) = 𝑓′(𝑥) + (−1) 𝑑 𝑑𝑥 𝑔(𝑥) = f (x) – g(x). ■ Example: For f(x) = 2𝑥3 − 4𝑥2 + 2√𝑥 − 8, find 𝑓′(𝑥).
Page 7 of 12 Calculus 1 – Differential Calculus Professor: Dr. Clemente M. Aguinaldo Jr., PNU North Luzon Disclaimer: This learning material is the property of Dr. Clemente M. Aguinaldo Jr. of PNU North Luzon and is intended for use in his Calculus 1 class. Unauthorized reproduction, distribution, or use—including taking photos of any part of this material—is strictly prohibited. Example: For f(x) = 15 𝑥3 − 3 𝑥2 + 5 𝑥 − 6 √𝑥, find 𝑓′(𝑥). Example: Find the vertex of the parabola y = x2 + 6x + 7. Solution: The vertex of the parabola lies at a point on the horizontal tangent line of the curve, that is, the slope is 0. Applying the Sum Rule, we have Theorem 4.5: (The Product Rule) Let F(x) = f(x)·g(x). Then, F (x) = 𝑑 𝑑𝑥 [𝑓(𝑥)𝑔(𝑥)] = f(x)[ 𝑑 𝑑𝑥 𝑔(𝑥)] + g(x)[ 𝑑 𝑑𝑥 𝑓(𝑥)]. The derivative of the product is the first function times the derivative of the second plus the second function times the derivative of the first. Proof: Let F(x) = f(x)·g(x). Then, 𝑑 𝑑𝑥 [𝑓(𝑥) ∙ 𝑔(𝑥)] = lim ℎ→0 𝑓(𝑥+ℎ)𝑔(𝑥+ℎ)−𝑓(𝑥)𝑔(𝑥) ℎ = lim ℎ→0 𝑓(𝑥+ℎ)𝑔(𝑥+ℎ)−𝒇(𝒙+𝒉)𝒈(𝒙)+𝒇(𝒙+𝒉)𝒈(𝒙)−𝑓(𝑥)𝑔(𝑥) ℎ = lim ℎ→0 𝑓(𝑥+ℎ)𝑔(𝑥+ℎ)−𝑓(𝑥+ℎ)𝑔(𝑥) ℎ + lim ℎ→0 𝑓(𝑥+ℎ)𝑔(𝑥)−𝑓(𝑥)𝑔(𝑥) ℎ = lim ℎ→0 𝑓(𝑥 + ℎ) 𝑔(𝑥+ℎ)−𝑔(𝑥) ℎ + lim ℎ→0 𝑔(𝑥) 𝑓(𝑥+ℎ)−𝑓(𝑥) ℎ = lim ℎ→0 𝑓(𝑥 + ℎ) lim ℎ→0 𝑔(𝑥+ℎ)−𝑔(𝑥) ℎ + lim ℎ→0 𝑔(𝑥) lim ℎ→0 𝑓(𝑥+ℎ)−𝑓(𝑥) ℎ = f(x)[ 𝑑 𝑑𝑥 𝑔(𝑥)] + g(x)[ 𝑑 𝑑𝑥 𝑓(𝑥)]. ■ A shorter way to memorize this rule is to let y = uv, where u and v are functions. Then, y = uv + vu. Using this form, a more elegant proof can be given by implicit differentiation and the derivatives of the natural logarithm. Example: For f(x) = (3x + 4)(x2 – 3x), find 𝑓′(𝑥). Consider y = (x2 + 4x – 11)(7x3 – √𝑥). Find y. Do not simplify. Solution:
Page 8 of 12 Calculus 1 – Differential Calculus Professor: Dr. Clemente M. Aguinaldo Jr., PNU North Luzon Disclaimer: This learning material is the property of Dr. Clemente M. Aguinaldo Jr. of PNU North Luzon and is intended for use in his Calculus 1 class. Unauthorized reproduction, distribution, or use—including taking photos of any part of this material—is strictly prohibited. Theorem 4.6: (The Quotient Rule) Let F(x) = 𝑓(𝑥) 𝑔(𝑥). Then, F (x) = 𝑑 𝑑𝑥 [𝑓(𝑥) 𝑔(𝑥)] = 𝑔(𝑥)𝑓′(𝑥)−𝑓(𝑥)𝑔′(𝑥) [𝑔(𝑥)]2 . The derivative of the quotient of two functions is the denominator times the derivative of the numerator minus the numerator times the derivative of the denominator all divided by the square of the denominator. The proof is similar to that of the product rule and will not be presented here anymore. However, an alternative, shorter proof can be given using logarithmic and implicit differentiation. A shorter way to memorize this rule is if we let y = u/v, where u and v are functions, then 𝑑 𝑑𝑥 (𝑢 𝑣) = y = 𝑣𝑢′−𝑢𝑣′ 𝑣2 . Example: For f(x) = 2𝑥+3 4𝑥−1, find 𝑓′(𝑥). Example: Find the derivative of y = 1−𝑥2 1+𝑥2. Solution: As a special case of the Quotient Rule, when u = c, where c is any constant, then 𝑑 𝑑𝑥 (𝑐 𝑣) = − 𝑐𝑣′ 𝑣2 . Example: Find the derivative of y = 4 1+3𝑥. Practice Task 4.2: A. Find the derivative of the following functions. Simplify your answers. 1. y = (4x – 3)(2x2 + 3x + 5) 2. y = (2x + 5)(3x2 – 4x + 1)
Page 9 of 12 Calculus 1 – Differential Calculus Professor: Dr. Clemente M. Aguinaldo Jr., PNU North Luzon Disclaimer: This learning material is the property of Dr. Clemente M. Aguinaldo Jr. of PNU North Luzon and is intended for use in his Calculus 1 class. Unauthorized reproduction, distribution, or use—including taking photos of any part of this material—is strictly prohibited. 3. y = −3 𝑥+2 4. y = 𝑥3−1 𝑥2+1 5. y = √𝑥 3 −7 √𝑥+2 B. Find how fast the (1) area and (2) circumference of a circle increase when its radius increases. C. Find the rate of change of the surface area 𝑆of a sphere with respect to its radius 𝑟. D. A cone has a height of 12cm. Find the rate of change of its volume 𝑉with respect to the radius 𝑟of its base. THE CHAIN RULE There are functions written in complicated form where differentiation formula will not immediately apply. Here is the role of the Chain Rule. But let us begin with a special case of the rule called the Extended or General Power Rule. We omit the proof here. Theorem 4.7: (The Extended Power Rule) Suppose that u is a differentiable function of x. Then, for any real number n, 𝑑 𝑑𝑥 (𝑢𝑛) = n𝑢𝑛−1 𝑑𝑢 𝑑𝑥. The Extended Power Rule allows us to differentiate functions such as y = (1 + x2)99 without having to expand the expression 1 + x2 to the 99th power (very time-consuming) and functions such as y = (1 + x6)1/4, for which expanding to the ¼ power is impractical. Let us show how these are done. Example: Differentiate: y = (1 + x2)99. Example: Differentiate: f(x) = (5x2 – 6)4. Example: Differentiate: y = (1 + x6)1/4
Page 10 of 12 Calculus 1 – Differential Calculus Professor: Dr. Clemente M. Aguinaldo Jr., PNU North Luzon Disclaimer: This learning material is the property of Dr. Clemente M. Aguinaldo Jr. of PNU North Luzon and is intended for use in his Calculus 1 class. Unauthorized reproduction, distribution, or use—including taking photos of any part of this material—is strictly prohibited. Example: Differentiate: f(x) = √2𝑥2 − 4𝑥 + 1. Example: Differentiate: f(x) = (3x – 5)4(7 – x)10. Solution: Here, we combine the Product Rule and the Extended Power Rule. Now, recall that the composition of f(x) with g(x) is written (f ∘g)(x) and is defined as (f ∘g)(x) = f(g(x)). How do we differentiate this kind of function? The next theorem tells us. Theorem 4.8: (The Chain Rule) The derivative of the composition f ∘g is given by 𝑑 𝑑𝑥 [(𝑓 ∘ 𝑔)(𝑥)] = 𝑑 𝑑𝑥 [𝑓(𝑔(𝑥))] = f (g(x)) g(x). Example: Suppose f(x) = 2x2 + x – 1and g(x) = 3x + 5. Find (𝑓 ∘ 𝑔)(𝑥) and then 𝑑 𝑑𝑥 [(𝑓 ∘ 𝑔)(𝑥)]. Also, use the Chain rule formula above. As we noted earlier, the Extended Power Rule is a special case of the Chain Rule. Consider f(x) = xn. For any other function g(x), we have (f ∘g)(x) = [g(x)]n, and the derivative of the composition is 𝑑 𝑑𝑥 [𝑔(𝑥)]𝑛 = n[𝑔(𝑥)]𝑛−1𝑔′(𝑥). The Chain Rule also appears in another form. Suppose that y = f(u) and u = g(x). Then, 𝑑𝑦 𝑑𝑥 = 𝑑𝑦 𝑑𝑢 ∙ 𝑑𝑢 𝑑𝑥.
Page 11 of 12 Calculus 1 – Differential Calculus Professor: Dr. Clemente M. Aguinaldo Jr., PNU North Luzon Disclaimer: This learning material is the property of Dr. Clemente M. Aguinaldo Jr. of PNU North Luzon and is intended for use in his Calculus 1 class. Unauthorized reproduction, distribution, or use—including taking photos of any part of this material—is strictly prohibited. Example: For y = 2 + √𝑢 and u = x3 + 1, find dy/du, du/dx, and dy/dx. Example: Find the derivative of f(x) = | x – 1 |. Solution: Note that | x – 1 | = √(𝑥 − 1)2. Now, let u = x – 1 and write f(x) = √𝑢2 = (𝑢2)1 2. Then, 𝑑𝑦 𝑑𝑢 = 1 2 (𝑢2)1 2−1(2𝑢) = and 𝑑𝑢 𝑑𝑥 = 𝑑(𝑥−1) 𝑑𝑥 = . By the Chain Rule, f (x) = 𝑑𝑦 𝑑𝑥 = 𝑑𝑦 𝑑𝑢 ∙ 𝑑𝑢 𝑑𝑥 = 𝑢 √𝑢2 (1) = We note that if x > 1, then | x – 1 | = x – 1 and f (x) = 1. But if x < 1, then | x – 1 | = –(x – 1) and f (x) = –1. Thus, f (x) does not exist at x = 1. HIGHER-ORDER DERIVATIVES Consider the function given by y = f(x) = x5 – 3x4 + x. Its derivative is given by y = f (x) = 5x4 – 12x3 + 1. The derivative function f (x) can also be differentiated. We can think of the derivative of f (x) as the rate of change of the slope of the tangent lines of f. We use the notation f for the derivative (f ). That is, f (x) = 𝑑 𝑑𝑥 𝑓′(𝑥). We call f the second derivative of f. For f(x) = x5 – 3x4 + x, the second derivative is given by y = f (x) = 20x3 – 36x2. The third derivative is y = f (x) = 60x2 – 72x. For the fourth derivative and higher, we use a number or n in parenthesis. Thus, f (n)(x) is the nth derivative. For the function above, f (4)(x) = 120x – 72 f (5)(x) = 120 f (6)(x) = 0, and f (n)(x) = 0, for any integer n > 6. Leibniz’ notation for the second derivative of a function f(x) is 𝑑2𝑦 𝑑𝑥2, or 𝑑 𝑑𝑥 (𝑑𝑦 𝑑𝑥), read “the second derivative of y with respect to x”. The 2’s in this notation are not exponents. Example: If f(x) = (x2 +10x)3, find f (x) and f (x).
Page 12 of 12 Calculus 1 – Differential Calculus Professor: Dr. Clemente M. Aguinaldo Jr., PNU North Luzon Disclaimer: This learning material is the property of Dr. Clemente M. Aguinaldo Jr. of PNU North Luzon and is intended for use in his Calculus 1 class. Unauthorized reproduction, distribution, or use—including taking photos of any part of this material—is strictly prohibited. Instead of factoring, we may simplify 2(6x + 30)(x2 +10x)(2x + 10) + 6(x2 +10x)2 to get f (x) = 30x4 + 600x3 + 3600x2 + 6000x. Both answers are acceptable. Continuing to get the derivative of f(x) = (x2 +10x)3, it will eventually be equal to zero. At what integer n do you think will f (n)(x) = 0? Practice Task 4.3: A. Find dy/dx by using the Chain Rule. 1. y = (𝑢2 − 𝑢 + 1)1/2 and u = √𝑥 2. y = (𝑥2+ 4 𝑥3−2 )3 3. y = | x2 + 2x | B. Find the second derivative of the following functions. Express your answers in factored form. 1. y = x3(2 – 3x4) 2. y = 𝑥3 𝑥−1 C. A point moves along the curve y = x3 – 3x + 6 so that x = t2 – 3, where t is time in minutes. At what rate is y changing when t = 2 minutes?