Alex Green Avishek Khan Kristina Rowlett Weldon So Unit 3 Chapter Test Mark the solution in red 3.1)Superimpose the graph of the derivative f(x) = x2+4 @ x=1 Answer: 3.2) The derivative of the following functions fail to be differentiable. Tell why each function is fails. a.𝑦 = { 𝑡𝑎𝑛 −1 (𝑥), 𝑥≠0 1, 𝑥=0 } Discontinuity b. 𝑦 = 3𝑥 − 2|𝑥| − 1 Corner 5 c.𝑦 = √2 − 𝑥 Vertical Tangent d. 𝑓(𝑥) = 𝑥 2/3 Cusp 3.3) Find the horizontal tangents of the curve y=x4-4x2+1 y’=4x3-(4⋅2x)+0 y’=4x3-8x y’=4(x3⋅2x) y’=4x(x2-2) y’=0, ±√2 Find dy/dx (𝑥+1)(𝑥+2) y= (𝑥−1)(𝑥−2) Derivative of: x+1 is 1 x+2 is 1 x-1 is 1 x-2 is 1 By the Product Rule: (1)(𝑥+1)+(1)(𝑥+2) (1)(𝑥−1)+(1)(𝑥−2) 2𝑥+3 2𝑥−3 Derivative of: 2x+3 is 2 2x-3 is 2 By the Quotient Rule: (2)(2𝑥−3)−(2)(2𝑥+3) (2𝑥−3)2 4𝑥−6−4𝑥−6 (2𝑥−3)2 −12 y’= (2𝑥−3)2 3.4)Velocity and Other Rates of Change Solve the following parts of the problem using what you know about rates of change: A. Write the area A of a square as a function of the side length s. B. Find the (instantaneous) rate of change of the area A with respect to a side s. C. Evaluate the rate of change of A at s = 1 and s = 5. D. If s is measured in centimeters and A is measured in square centimeters, what units would be appropriate for 𝑑𝐴 ? 𝑑𝑠 A) Since the area of a square with respect to a side length is 𝑠 2 , then the area as a function with respect side s is 𝐴 = 𝑠 2 B) The instantaneous rate of change is the same as the first derivative. Thus, it is 𝐴′ = 2𝑠 C) 𝐴′ (1) = 2(1) = 2 ; 𝐴′ (5) = 2(5) = 10 D) 𝑐𝑚2 𝑐𝑚 3.5)Find the equation of the lines tangent and normal to the graph of y= sin x + 3 at x = π Answer: 3.6)Differentiate 𝑦 = 3 𝑡𝑎𝑛√𝑥. y’ = 3(sec2 √𝑥) t(√𝑥) y’ = 3(sec2 √𝑥) t(x1/2) y’ = 3(sec2 √𝑥) (½)(x-1/2) 1 ) 𝑥 1/2 y’ = 3(sec2 √𝑥) (½)( y’ = 3(𝑠𝑒𝑐2 √𝑥) 2√𝑥 3.7)Find dy/dx and find the slope of the curve at the indicated point x2+y2=13 @ (-2, 3) 2x+2y(dy/dx)=0 2x=-2y(dy/dx) -2x/2y=(dy/dx) -x/y=(dy/dx) -(-2)/(3)=(dy/dx) Slope=⅔ Find the lines that are (a) tangent and (b) normal to the curve at the given point x2+xy-y2=1 @ (2, 3) Derivative of xy: Using the product rule 𝑑𝑦 𝑑𝑥 ( )(x)+(1)(y) 𝑑𝑦 𝑑𝑥 𝑑𝑦 𝑑𝑥 𝑑𝑦 𝑑𝑦 2x+y=- x+ 2y 𝑑𝑥 𝑑𝑥 𝑑𝑦 2x+y=( )(-x+2y) 𝑑𝑥 𝑑𝑦 2𝑥+𝑦 = 𝑑𝑥 −𝑥+2𝑦 2x+ x+y-( )2y=0 Slope of Tangent Line: 2(2)+(3) 7 = −(2)+2(3) 4 Slope of Normal Line: - 4 7 Tangent Line: 7 4 y-3=( )(x-2) Normal Line: 4 y-3=(-7)(x-2) 3.8) Find the derivative of 𝑦 = 𝑐𝑜𝑠 −1 (8𝑥 4 )with respect to x The derivative of 𝑜𝑠 −1 (𝑥) = 𝑐𝑜𝑠 −1 (8𝑥 4 ) = −1 √1−(8𝑥 4 )2 √1−𝑥 2 3 x 32𝑥 3.9) FInd the derivative of Answer: −1 . Thus, the derivative with respect to x is: Chapter 1 Review)Solve for x 1 + 𝑠𝑖𝑛(𝑥) = 2𝑐𝑜𝑠 1+sinx = 2(1-sin2 x) 1+sinx = 2-2sin2 x 2sin2 -sinx -1=0 (2sinx-1)(sinx+1)=0 2sinx-1 =0 or sinx=-½ 𝜋 5𝜋 6 x= 6 , or x= 3𝜋 2 sinx+1=0 sinx=-1 2 (𝑥) Vectors) Solution on Mr. Wahbeh’s Website Solutions on Mr. Wahbeh’s website Chapter 2) Find the limit of: When you look at the graph of the function, there is a vertical asymptote at x=5. As the line approaches 5 from the left, the value heads towards negative infinity. Hence, the limit is negative infinity.