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Lecture 4 Review of Math 131 with Trig Functions Math 13200 Trigonometry Derivative Rules Recall from last time: Dx sin x = cos x Dx cos x = − sin x Dx tan x = sec2 x Dx cot x = − csc2 x Dx sec x = sec x · tan x Dx csc x = − csc x · cot x Today, we will do a brief review of last quarter, but focusing on trig functions. Derivative Rules (with trig functions) For the following, let f, g be differentiable functions of x, k any constant, and r a rational number. Rules from last quarter: Constant Dx k = 0 Identity Dx x = 1 Constant Multiple Dx (k · f ) = k · Dx f Sum Dx (f + g) = Dx f + Dx g 1 Product Dx (f · g) = f · Dx g + g · Dx f Quotient f g · Dx f − f · Dx g Dx = g g2 Chain Dx (g ◦ f ) = (Dx g ◦ f ) · Dx f Dx (g(f (x)) = g 0 (f (x)) · f 0 (x) Absolute Value x Dx |x| = = sign x |x| Examples 1. What is the derivative of f (x) = sin(3x)? For this problem, we use the chain rule. The inside function is 3x. The derivative of 3x is 3 (using the constant multiple and identity rules). The outside function is sin x. Its derivative is cos x. We use the chain rule to compute the rest: Dx (sin 3x) = cos(3x) · (Dx 3x) = 3 cos 3x 2. Find the derivative f 0 of f (x) = x2 For this, we use the quotient rule and power rule. tan x tan x(2x) − x2 sec2 x tan2 x 2x tan x − x2 sec2 x = tan2 x f 0 (x) = 3. Find the derivative of f (x) = sin2 x + cos2 x without using trigonometric identities. Each term is a trigonometric function followed by raising to the 2nd power. Thus we use chain rule for each term. Dx (sin2 x + cos2 x) = 2 sin x(Dx sin x) + 2 cos x(Dx cos x) = 2 sin x · cos x + 2 cos x(− sin x) =0 Tangent Line of Function (with trig functions) Recall that the tangent line to the graph of f (x) at an x-value c is the line with slope f 0 (c) passing through the point (c, f (c)). 2 Example What is the tangent line to the function f (x) = sin x · tan x at the x-value π/6? First, we find the derivative using the product formula: f 0 (x) = sin x · Dx (tan x) + (Dx sin x) · tan x = sin x · sec2 x + cos x · tan x Now we find the slope of the tangent line: f 0 (π/6) = sin π6 · sec2 π6 + cos π6 · tan π6 2 √ 1 3 1 2 √ √ + = · 2 2 3 3 7 = 6 And to find the point (c, f (c)): √ f (π/6) = sin π6 tan π6 = 3 6 Now we use the point slope formula y − y1 = m(x − x1 ) to find the equation of the line: √ y− 7 π 3 = x− 6 6 6 Implicit Differentiation: Tangent Line of Curve (with trig functions) This is the same story as in the case of a tangent line to a curve, except now you use implicit differentiation to find the slope of the tangent line. Implicit differentiation will give you the slope dy dx in terms of x and y. Example Consider the curve C cut out by the equation 2 cos(4x) + 20 2 2 x y =5 π2 and the point P = (π/8, 4) on C. 3 What is the tangent line to C at P ? First we implicitly differentiate: 2(− sin 4x) · 4 + 20 (2xy 2 + 2x2 yy 0 ) = 0 π2 20 (2xy 2 + 2x2 yy 0 ) = 8 sin 4x π2 40 2 0 40 x yy = 8 sin 4x − 2 xy 2 π2 π 40x2 yy 0 = 8π 2 sin 4x − 40xy 2 π 2 sin 4x y − 5x2 y x dy π 2 sin 4x y = − dx 5x2 y x y0 = Now we evaluate dy dx at P π 2 sin π2 dy 4 32 2π − =− + ≈ −8.92928 = 2 dx P 5(π/8) · 4 π/8 π 5 Plugging in for point slope formula, we get the equation 32 2π π y−4= − + x− ≈ −8.92928(x − π/8) π 5 8 Related Rates (with trig functions) Blueprint for how to solve a related rate problem: (1) Draw a picture! (If you can, or if it hasn’t been made for you.) (2) What are all relevant, non-constant variables? (i.e., those things which are changing with time) (3) What rates are given? 4 (4) What is the deisred rate - i.e. the rate we want to know? And under what condition? (5) What are all possible relations between the variables from step 2? (6) Implicitly differentiate each relation from step 3 to obtain relations between the rates. (7) Box or circle the following things: • The given rate from step 3. • The condition(s) from step 4. • Each relation between variables from step 5. • Each relation between rates and variables from step 6. (8) The problem is now just basic (albeit complicated) algebra: solve for the desired rate. 5