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1.4 The Derivatives of Some Basic Functions

1.4 The Derivatives of Some Basic Functions. The derivative function represents the slope of the tangent at each point on the graph of the function. (where it exists) Derivative of y=x. Derivative of y=x 2. Slope of tangent at x =2x y’=2x. Derivative of y=x 2. y’=2x.

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1.4 The Derivatives of Some Basic Functions

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  1. 1.4 The Derivatives of Some Basic Functions • The derivative function represents the slope of the tangent at each point on the graph of the function. (where it exists) • Derivative of y=x.

  2. Derivative of y=x2 • Slope of tangent at x =2x • y’=2x

  3. Derivative of y=x2 y’=2x • The derivative of y=x2 is y’=2x • See graph. • See rate triangle 4=2x 1 2=2x 1 The instantaneous rate of change of y with respect to x is 2x. Visualize the point moving from left to right along the graph of y=x2 The slope along the tangent is always double the x-coordinate of the point.

  4. Derivative of y=x3 • Slope of tangent at x is 3x2 • y’=3x2

  5. Derivative of y=x3 • The derivative of y=x3 is y’=3x2 • See graph. • See rate triangle 3=3x2 1 The instantaneous rate of change of y with respect to x is 3x2. Visualize the point moving from left to right along the graph of y=x3 . The slope along the tangent is always three time the square of the x-coordinate of the point.

  6. Graph of y=x3 and its derivative y ‘ = 3x2

  7. Example 1 • a) Find instantaneous rate of change of y=x2 at i) x=3 ii) x =-2 • b) Interpret graphically. • Solution i) • The derivative of y=x2 is y’=2x. • When x=3, y’=2(3)=6 • y is increasing 6 times as fast as x is increasing when x=3. m=6

  8. Example 1 • a) Find instantaneous rate of change of y=x2 at ii) x =-2 • b) Interpret graphically. • Solution ii) • The derivative of y=x2 is y’=2x. • When x=-2, y’=2(-2)=-4 • y is decreasing 4 times as fast as x is increasing when x=-2. m=-4

  9. Example 2: Determine the equation of the tangent line. • Determine the equation of the tangent to the graph of y=x3 at the point (-2,-8). • Illustrate the results on a graph. • Solution • The derivative of y=x3 is y’=3x2. • When x=-2, the slope of the tangent line is y’=3(-2)2 =12. • The slope of the tangent is 12. • The equation of the tangent line using y=m(x – p) + q is • y=12(x+2)-8, or y=12x+16.

  10. Example 2: Graph • Graph the function and the tangent line.

  11. Derivative of y=mx+b • y=x • y=mx • y=mx+b • a) y’=1 • b) y’=m • c) y’=m

  12. Vertical Translation Rule • When the graph of a function is translated vertically, the derivative is not affected. • The derivative of y=f(x)+c is y’=f ’(x), where c is any constant. • Example: Find the derivative of y=x2+4. • y’=2x.

  13. Vertical Translation Rule Example: Find the derivative of y=x2+4. • y’=2x.

  14. Vertical Stretch Rule • When the graph of a function is expanded or compressed vertically, the graph of the derivative is also expanded or compressed vertically. • The derivative of y=cf(x) is y’=cf’(x), where c is any constant.

  15. Vertical Stretch Rule • Example: Find the derivative of y=2x3 • f(x)=x3 , f’(x)=3x2 • y’=2f’(x) , so y’=2(3x2) • Or y’=6x2 • When the function is stretched by a factor of 2, the y-values for any x-value are doubled. • That means when the slope is calculated at any point the slope will be doubled.

  16. Practice • a) y’=3x2 • b) y’=-10x • c) y’=12x2 • d) y’=7 • e) y’=-3x2 • Find the derivatives of the following functions. • a) y=x3+ 23 • b) y=-5x2 • c) y=4x3-7 • d) y=7x • e) y=14 - x3 f(x)=x2 ;f ‘(x)=2x f(x)=x3 ;f ‘(x)=3x2 If y = f(x) +c, then y ‘= f ‘(x) If y = cf(x), then y ‘= cf ‘(x)

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