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Differentials: ∆푦 푑푦 When we first started to talk about derivatives, we said that becomes when the change in ∆푥 and ∆푦 in y ∆푥 푑푥 become very small.

⇒ dy can be considered a very small change in y. dx can be considered a very small change in x.

Let 푓(푥) be a differentiable function. The differential 푑푦 is:

푑푦 푑푦 = 푑푥 푑푥 푑푦 = 푓′(푥)푑푥

Linear approximation:

Very easy thing to do, and once you master it, you can impress all of your friends by calculating things like 3√70 in your head . . . . about 4.125! Impressed? I’ll teach you how. Recall that if a function 푓 (푥) is differentiable at = 푎 , we say it is locally linear at 푥 = 푎 . This means that as we zoom in closer and closer and closer and closer around 푥 = 푎 , the graph of 푓 (푥) , regardless of how curvy it is, will begin to look more and more and more and more like the tangent line at 푥 =a

This means that we can use the equation of the tangent line of 푓 (푥) at 푥 = 푎 to approximate 푓 (푥) for values close to 푥 = 푎.

Let’s take a look at ퟑ√ퟕퟎ and the figure below. 1/3 푓(푥) = (푥)

2 1 − 푓′(푥) = (푥) 3 3 2 1 − 1 푓′(64) = (64) 3 = 3 48 1 tangent line at (64,4): 푦 = 4 + (푥 − 64) 48 1 1 푓(70) ≈ 푦(70) = 4 + (70 − 64) = 4 + = 4.125 48 8

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Linear approximation of 풇(풙) at point 풙 around 풂 Tangent line approximation Linearization of f around a

1. step: Assuming that 푓(푥) is differentiable at 푥 = 푎:

푓(푥) = 푓(푎) + ∆푓 ≈ 푓(푎) + 푓′(푎)∆푥

Linear approximation of 푓(푥) at point 푥 around 푎

푓(푥) = 푓(푎) + 푓′(푎)(푥 − 푎) 퐹표푟 푡ℎ푒 푝표𝑖푛푡푠 푥 푐푙표푠푒 푡표 푎, 푡ℎ푒 푙𝑖푛푒푎푟 푎푝푟표푥𝑖푚푎푡𝑖표푛 표푓 푓푢푛푐푡𝑖표푛 푓(푥) 𝑖푠 푒푞푢푎푙 푡표 푡ℎ푒 푣푎푙푢푒 표푓 푡ℎ푒 푡푎푛푔푒푛푡 푙𝑖푛푒 푦(푥) 푎푡 푡ℎ푎푡 푝표𝑖푛푡 푥 푓(푥) = 푦(푥) 푅푒푚푎𝑖푛푑푒푟 푅 is small for the points close to point 푎

STEPS for linear approximation of a function 푓(푥) for points 푥 around 푎: 1. Find the equation of the tangent line 푦(푥) at the center (푎, 푓(푎)) in point-slope form. 푦(푥) = 푓(푎) + 푓′(푎)(푥 − 푎)) 2. For some 푥 = 푐 around : 푓(푐) ≈ 푦(푐) If asked, determine if 푦(푐) is an over-approximation or an under approximation by examining the concavity of 푓 (푥) at the center 푥 = 푎 a. If 푓′′(푎)<0, 푓 (푥) is concave down around 푎, tangent line is above the curve and 푦(푐) is an over-approximation

b. If 푓′′(푎) >0, 푓 (푥) is concave up around 푎 , tangent line is bellow the curve and 푦(푐) is an under-approximation

example: Find the linearization of the function 푓(푥) = √푥 + 3 at 푎 = 1, and use it to approximate the numbers √3.98 and √4.05 Are these approximations overestimates or underestimates? 1 푓′(푥) = 2√푥 + 3

′ 푓(푥) ≈ 푓(1) + [푓 (푥)]푥=1(푥 − 1)

1 7 푥 푓(푥) ≈ 2 + (푥 − 1) = + 4 4 4 7 푥 linear aproximation: √푥 + 3 ≈ + (when 푥 is near 1) 4 4 7 0.98 √3.98 ≈ + = 1.995 4 4 7 1.05 √4.05 ≈ + = 2.0125 4 4

푓′′(1) < 0 → 푓(푥) is concave down around 푥 = 1 Tangent line is above the curve, so estimated values are overestimation