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2.3

2.3. Differentiation Formulas. Differentiation Formulas. Let’s start with the simplest of all functions, the constant function f ( x ) = c . The graph of this function is the horizontal line y = c , which has slope 0, so we must have f ( x ) = 0.

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2.3

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  1. 2.3 • Differentiation Formulas

  2. Differentiation Formulas • Let’s start with the simplest of all functions, the constant function f(x) = c. The graph of this function is the horizontal line y = c, which has slope 0, so we must have f(x) = 0. • The graph of f(x) = c is the line y = c, so f(x) = 0.

  3. Constant Rule • Using the formal definition of derivative:

  4. Power Rule • For functions f(x) = xn, where n is a positive integer:

  5. Proof by formal definition of derivative: • For n = 4 we find the derivative of f(x) = x4 as follows: • (x4) = 4x3

  6. Practice: Find each derivative • (a) If f(x) = x6 • (b) If y = x1000 • (c) If y = t4 • (d) (r3)

  7. Extended Power Rule

  8. PROPERTIES AND RULES OF DERIVATIVES

  9. Constant Multiple Rule

  10. Sum Rule • the derivative of a sum of functions is the sum of the derivatives.

  11. Difference Rule

  12. Example: • (x8 + 12x5 – 4x4 + 10x3 – 6x + 5) • = (x8)+12 (x5)–4 (x4)+10 (x3)–6 (x)+ (5) • = 8x7 + 12(5x4) – 4(4x3) + 10(3x2) – 6(1) + 0 • = 8x7+ 60x4 – 16x3 + 30x2 – 6

  13. Product Rule • the derivative of a product of two functions is the first function times the derivative of the second function plus the second function times the derivative of the first function. Or:

  14. Quotient Rule • the derivative of a quotient is the denominator times the derivative of the numerator minus the numerator times the derivative of the denominator, all divided by the square of the denominator. • Or: • Or:

  15. Example: • Let . Then

  16. Use quotient rule? • Don’t use the Quotient Rule every time you see a quotient. • Sometimes, when there is only ONE term in the quotient, it’s easier to rewrite the expression as a sum of power terms, then use the power rule. • Example: • f(x) = • We can use the quotient rule but it is much easier to perform the division first and write the function as: • f(x) = 3x + 2x –1 2

  17. Examples: • (a) If y = , then • = –x–2 • = • (b)

  18. Practice: • Differentiate the function f(t) =(a + bt). • Solution 1:Using the Product Rule, we have

  19. Practice – Solution2 • cont’d • If we first use the laws of exponents to rewrite f(t), then we can proceed directly without using the Product Rule.

  20. Normal line at a point: • The differentiation rules enable us to find tangent lines without having to resort to the definition of a derivative. • They also enable us to find normal lines. • The normal line to a curve C at point P is the line throughP that is perpendicular to the tangent line at P.

  21. Example: • Find equations of the tangent line and normal line to the curve • y = (1 + x2) at the point (1, ). • Solution: According to the Quotient Rule, we have

  22. Example – Solution • cont’d • So the slope of the tangent line at (1, ) is • We use the point-slope form to write an equation of the tangent line at (1, ): • y – = – (x – 1) or y =

  23. Example – Solution • cont’d • The slope of the normal line at (1, ) is the negative reciprocal of , namely 4, so an equation is • y – = 4(x – 1) or y = 4x – • The curve and its tangent and normal lines are graphed in Figure 5. • Figure 5

  24. Summary of Rules

  25. 2.4 • Derivatives of Trig Functions

  26. Example 1 • Differentiate y = x2 sin x. • Solution: • Using the Product Rule

  27. Example 2 • An object at the end of a vertical spring is stretched 4 cm beyond its rest position and released at time t = 0 (note that the downward direction is positive.) • Its position at time t is • s = f (t) = 4 cos t • Find the velocity and acceleration • at time t and use them to analyze • the motion of the object.

  28. Example 2 – Solution • The velocity and acceleration are

  29. Example 2 – Solution • cont’d The object oscillates from the lowest point (s = 4 cm) to the highest point (s = –4 cm). The period of the oscillation is 2, which is the period of cos t.

  30. Example 2 – Solution • cont’d • The speed is | v | = 4 | sin t |, which is greatest when | sin t | = 1, that is, when cos t =0. • So the object moves fastest as it passes through its equilibrium position (s = 0). Its speed is 0 when sin t =0, that is, at the high and low points. • The acceleration a = –4 cos t =0 when s = 0. It has greatest magnitude at the high and low points.

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