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Engineering Mathematics: Differential Calculus

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Contents Concepts of Limits and Continuity Derivatives of functions Differentiation rules and Higher Derivatives Applications

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Differential Calculus Concepts of Limits and Continuity

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The idea of limits Consider a function The function is well-defined for all real values of x The following table shows some of the values: x2.92.99 33.0013.013.1 F(x)8.418.948.99499.0069.069.61

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The idea of limits

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Concept of Continuity E.g. is continuous at x=3? The following table shows some of the values: exists as and => f(x) is continuous at x=3! x2.92.99 33.0013.013.1 F(x)8.418.948.99499.0069.069.61

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Differential Calculus Derivatives of functions

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Derivative ( 導數 ) Given y=f(x), if variable x is given an increment x from x=x 0, then y would change to f(x 0 + x) y= f(x 0 + x) – f(x) y x is the slope ( 斜率 ) of triangular ABC x f(x 0 + x) f(x 0 ) x0+xx0+x x0x0 y xx yy Y=f(x) A B C

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Derivative What happen with y x as x tends to 0? It seems that y x will be close to the slope of the curve y=f(x) at x 0. We defined a new quantity as follows If the limit exists, we called this new quantity as the derivative ( 導數 ) of f(x). The process of finding derivative of f(x) is called differentiation ( 微分法 ).

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Derivative y f(x 0 + x) f(x 0 ) x0+xx0+x x0x0 x xx yy Y=f(x) A B C Derivative of f(x) at Xo = slope of f(x) at Xo

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Differentiation from first principle Find the derivative of with respect to (w.r.t.) x To obtain the derivative of a function by its definition is called differentiation of the function from first principles

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Differential Calculus Differentiation rules and Higher Derivatives

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Fundamental formulas for differentiation I Let f(x) and g(x) be differentiable functions and c be a constant. for any real number n

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Examples Differentiate and w.r.t. x

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Table of derivative (1) FunctionDerivative Constant k0 x1 kxk

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Table of Derivatives (2) Function Derivative Angles in radians

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Differential Calculus Product Rule, Quotient Rule and Chain Rule

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- The product rule and the quotient rule Form of product function Form of quotient function e.g.1 This is a _________ function with and

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e.g.2 This is a _________ function with and e.g.3 e.g.4 e.g.5

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The product rule Consider the function Using the product rule, e.g.1 Find where Solution:

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e.g.2 Find where Solution: e.g.3 Find where Solution:

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The quotient rule Consider the function. Applying the quotient rule, e.g.1 Find where Solution:

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e.g.2 Find where Solution: e.g.3 Find where Solution:

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More Example (1) Differentiate w.r.t. x

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More Example (2) Differentiate w.r.t. x

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Fundamental formulas for differentiation II Let f(x) and g(x) be differentiable functions

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Fundamental formulas for differentiation III ln(x) is called natural logarithm ( 自然對數 )

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Differentiation of composite functions To differentiate w.r.t. x, we may have problems as we don’t have a formula to do so. The problem can be simplified by considering composite function: Let so and we know derivative of y w.r.t. u (by formula): but still don ’ t know

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Chain Rule ( 鏈式法則 ) Chain Rule states that : given y=g(u), and u=f(x) So our problem and

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Example 1 Differentiate w.r.t. x Simplify y by letting so now By chain rule

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Example 2 Differentiate w.r.t. x Simplify y by letting so now By chain rule

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Example 3 Differentiate w.r.t. x Simplify y by letting so now By chain rule

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Higher Derivatives ( 高階導數 ) If the derivatives of y=f(x) is differentiable function of x, its derivative is called the second derivative ( 二階導數 ) of y=f(x) and is denoted by or f ’’(x). That is Similarly, the third derivative = the n-th derivative =

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Example Find if

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Differential Calculus Applications

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Slope of a curve Recall that the derivative of a curve evaluate at a point is the slope of the curve at that point. f(x 0 + x) f(x 0 ) x0+xx0+x x0x0 x xx yy Y=f(x) A B C Derivative of f(x) at Xo = slope of f(x) at Xo

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Slope of a curve Find the slope of y=2x+3 at x=0 To find the slope of a curve, we have to compute the derivative of y and then evaluate at a point The slope of y at x=0 equals 2 (y=mx+c now m=2)

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Slope of a curve Find the slope of at x=0, 2, -2 The slope of y = 2x The slope of y (at x=0) = 2(0) = 0 The slope of y (at x=-2) =2(-2) = -4 The slope of y (at x=2) =2(2) = 4 X=0 X=2X=-2

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Local maximum and minimum point For a continuous function, the point at which is called a stationary point. This gives the point local maximum or local minimum of the curve A B C D X1X1 X2X2

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First derivative test (Max pt.) Given a continuous function y=f(x) If dy/dx = 0 at x=x o & dy/dx changes from +ve to –ve through x 0, x=x 0 is a local maximum point x=x 0 local maximum point

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First derivative test (Min pt.) Given a continuous function y=f(x) If dy/dx = 0 at x=x o & dy/dx changes from -ve to +ve through x 0, x=x 0 is a local minimum point x=x 0 local minimum point

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Example 1 Determine the position of any local maximum and minimum of the function First, find all stationary point (i.e. find x such that dy/dx = 0), so when x=0 By first derivative test x=0 is a local minimum point

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Example 2 Find the local maximum and minimum of Find all stationary points first: x0x<1x=11

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Example 2 (con’d) By first derivative test, x=1 is the local maximum (+ve -> 0 -> -ve) x=3 is the local minimum (-ve -> 0 -> +ve)

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Second derivative test Second derivative test states: There is a local maximum point in y=f(x) at x=x0, if at x=x 0 and < 0 at x=x 0. There is a local minimum point in y=f(x) at x=x0, if at x=x 0 and > 0 at x=x 0. If dy/dx = 0 and =0 both at x=x 0, the second derivative test fails and we must return to the first derivative test.

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Example Find the local maximum and minimum point of Solution Find all stationary points first: By second derivative test, x=1 is max and x=3 is min. So, (1,2) is max. point and (3,-2) in min. point.

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Practical Examples 1 e.g.1 A rectangular block, with square base of side x mm, has a total surface area of 150 mm 2. Show that the volume of the block is given by. Hence find the maximum volume of the block. Solution:

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Practical Examples 2 A window frame is made in the shape of a rectangle with a semicircle on top. Given that the area is to be 8, show that the perimeter of the frame is. Find the minimum cost of producing the frame if 1 metre costs $75. Solution:

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