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Calculus I                                                         Quiz # 6                              October 24, 2003

Name____________________                   R.  Hammack                                          Score ______
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(1) (10 points) Find derivatives of the following functions.

(a) f(x) = tan((x^2 + 4)^(1/2))
f ' (x) = d/dx[tan((x^2 + 4)^(1/2))] = sec^2((x^2 + 4)^(1/2)) d/dx[(x^2 + 4)^(1/2)] = sec^2((x^2 + 4)^(1/2)) 1/2 (x^2 + 4)^(-1/2) 2x = (x sec^2((x^2 + 4)^(1/2)))/(x^2 + 4)^(1/2)


(b)  g(x) = sin(x) (3x + 4)
  g ' (x) = cos(x) (3x + 4) + 3sin(x)


(c) d/dx[4sec(x) + 1/x + 3^(1/2)] =4sec(x) tan(x) - 1/x^2



(2) (5 points) Consider the function f(x) = x^3 + 5. Find the equation of the line tangent to the graph of y = f(x) at the point (1, f(1)).

Slope at (x, f(x)) is f ' (x) = 3x^2
Slope at (1, f(1)) is f ' (1) = 3 (1)^2 = 3

Thus the slope of the line is 3 and it passes through (1, f(1)) = (1, 6).
Using the point-slope formula,
y - 6 = 3 (x - 1)
y = 3x + 3




(3)  (5 points) The graph of a function  f (x)  is sketched below.    If G(x) = (x + 2f(x))^(1/2),  find G ' (0)

[Graphics:HTMLFiles/quiz6sol_16.gif]

  G(x) = (x + 2f(x))^(1/2) = (x + 2f(x))^(1/2)
  
  G ' (x) = 1/2 (x + 2f(x))^(-1/2) (1 + 2f ' (x)) = (1 + 2f ' (x))/(2 (x + 2f(x))^(1/2))
  
  G ' (0) = (1 + 2f ' (0))/(2 (x + 2f(0))^(1/2)) = (1 + 2 (0))/(2 (0 + 2 (2))^(1/2)) = 1/(24^(1/2)) = 1/4