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# Calc 1: Practice Exam 2 Solutions Name: (1) Find ... - Westmont

Calc 1: Practice Exam 2 Solutions Name: (1) Find ... - Westmont

Answer: Use the quotient rule, the product rule, andthe chain rule.y ′ = x2 e 5x sec 2 ( √ x − 3√ x)( 1 2 x−1/2 − 1 3 x−2/3 ) − tan( √ x − 3√ x)(2xe 5x + 5x 2 e 5x )x 4 e 10x2(2) Consider the curve defined by the equation(a) Find dydxye x + xe y = e xyAnswer: Take the derivative with respect to x of bothsides, using implicit differentiation:ye x + e x dydx + ey + xe y dydx = exy (x dydx + y)Solve for dydxto find:dydx = yexy − ye x − e ye x + xe y − xe xy(b) Find the equation of the tangent line to the curve at thepoint (1,0).Answer: In the expression above, plug in x = 1 and y = 0−1to find that the slope of the tangent line is: . Thus theeequation of the tangent line is:y = − 1 (x − 1)e(3) Show all the steps for finding the derivative of:(a) y = arcsin(x)(b) y = arccos(x)

3(c) y = arctan(x)Answer: Rewrite y = arctan(x) as x = tan(y). Differentiateboth sides with respect to x to find that: 1 =sec 2 (y) dy . Thus:dxdydx = 1sec 2 (arctan(x)To simplify this make a right triangle which has an angley (since arctan(x) spits out the angle y). Since tangent isopposite over adjacent, label the side opposite the angle ywith x and the side adjacent with a 1. Thy hypotenusethen has length √ 1 + x 2 by Pythagoras’ theorem. Thefunction secant is defined as hypotenuse over adjacent, sosec(arctan(x)) = sec(y) = √ 1 + x 2 . Thus:.dydx = 11 + x 2(4) Suppose that f(t) is a solution of the DE y ′ = t 2 y + t and thatf(1) = 2. Find the equation of the tangent line to f(t) at thepoint (1,2).Answer: The slope of the tangent line is f ′ (t) when t = 1and y = 2. The fact that f(t) is a solution to the given DE,tells us that f ′ (t) = t 2 f(t) + t. So when t = 1 and y = 2 wehave: f ′ (1) = (1) 2 (2) + 1. In other words, the slope is 3. Thusthe equation of the tangent line is y = 3(t − 1) + 2.(5) Suppose that h(t) is a solution of the DE y ′ = e t y and thath(−2) = 1. Find an equation of the line tangent to h at thepoint (−2, 1).Answer: This works the same way as the previous one. Theslope is h ′ when t = −2 and h = 1. Thus the slope is e −2 (1) =e −2 . The equation of the tangent line to h(t) at the point (−2, 1)is:y = e −2 (t + 2) + 1

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