Скачать презентацию Calculus Ch 4 The Derivative Question 1 Скачать презентацию Calculus Ch 4 The Derivative Question 1

Tut_06_12.ppt

  • Количество слайдов: 14

Calculus Ch. 4. The Derivative Calculus Ch. 4. The Derivative

Question 1 (October 2011 Exam, Q 12). Complete the following definition of A function Question 1 (October 2011 Exam, Q 12). Complete the following definition of A function f (x) has a right derivative A at a point x if such that implies Answer: b)

Question 2 (October 2011 Exam, Q 20). Let Find an approximate value for using Question 2 (October 2011 Exam, Q 20). Let Find an approximate value for using differentials. Solution. Use the following approximation where x 0 = 1, and We have Hence Answer: a).

Tangent lines. 5 1 x Tangent lines. 5 1 x

Question 3 (November 2008 Exam, Q 14). At which of the following points is Question 3 (November 2008 Exam, Q 14). At which of the following points is the tangent line drawn to the graph of f (x) = x 3 + 4 x 2 - 10 parallel to the line y = 20 - 4 x? Solution. The equation of the tangent line at a point x = x 0: The two lines are parallel if We have The equation 3 x 2 + 8 x = -4 has solutions Answer: b.

Normal lines. 5 1 x Normal lines. 5 1 x

Question 4 (November 2008 Exam, Q 17). What is the equation of the line Question 4 (November 2008 Exam, Q 17). What is the equation of the line normal to the graph of f (x) = exp(x - 2) at the point x = 2? Solution. The equation of the normal line at a point x = x 0: We have Hence for x 0 = 2, and Answer: c.

Derivative of inverse functions f -1(x). Differentiate both sides of this equation using the Derivative of inverse functions f -1(x). Differentiate both sides of this equation using the chain rule. Hence

Question 5 (November 2009 Exam, Q 23). What it the derivative of the inverse Question 5 (November 2009 Exam, Q 23). What it the derivative of the inverse function at x = 3, if Solution. We have and It is necessary to find Since Hence we obtain

Derivative of implicit functions. We are given an equation If we solve this equation Derivative of implicit functions. We are given an equation If we solve this equation for y we obtain a function Can we find the derivative without solving the equation

Question 6 (November 2009 Exam, Q 22). Find if y(1) = 2 and y(x) Question 6 (November 2009 Exam, Q 22). Find if y(1) = 2 and y(x) is an implicit function given by Solution. We have and =1 Hence =1 =1 =1

The Lagrange (Mean Value) Theorem. Let f (x) be a function which satisfies the The Lagrange (Mean Value) Theorem. Let f (x) be a function which satisfies the following hypothesis: 1. f (x) is continuous on the closed interval [a, b]. 2. f (x) is differentiable on the open interval (a, b). Then there is a number c in (a, b) such that

Question 6. Suppose that f (0) = 1 and for all values of x. Question 6. Suppose that f (0) = 1 and for all values of x. How small f (2) can possibly be? Solution. According to the Lagrange theorem Hence Answer: The smallest possible value for f (2) is 3.

When the goings get tough the tough get going! When the goings get tough the tough get going!