![Solution:
•Using quotient rule, you have
𝑑
𝑑𝑥
𝑥
ln 𝑥
=
(ln 𝑥)
𝑑
𝑑𝑥
𝑥 − 𝑥
𝑑
𝑑𝑥
[ln 𝑥]
(ln 𝑥)2
Quotient Rule
=
ln 𝑥 − 𝑥
1
𝑥
ln2𝑥
Recall: (ln 𝑥)2
= ln2
𝑥
=
ln 𝑥 − 1
ln2𝑥
Simplify](https://image.slidesharecdn.com/advanced-differentiation-rules-230423025608-3c1ed832/85/Advanced-Differentiation-Rules-pdf-48-320.jpg)
Advanced-Differentiation-Rules.pdf
- 3. Implicit Differentiation • Most of you are sometimes face with situations where you need to choose between expressing explicitly and expressing implicitly. Of course, you will choose the one that is clear and not ambiguous. The same situation happens also in mathematics, especially when dealing with functions. A function is written in explicit form when it is expressed as 𝑦 = 𝑓(𝑥). All functions that are dealt with in the previous modules are written in this form.
- 4. Implicit Differentiation •However, there are times when mathematical functions are written in more complicated form, wherein it is difficult (or sometimes impossible) to express 𝑦 explicitly in terms of 𝑥 . Such functions are expressed in implicit form. This section of the module will try to explain how these functions can be differentiated.
- 5. Implicit Differentiation •To start, consider the following equations describing two familiar curves: 𝑦 = 𝑥2 + 3 𝑥2 + 3𝑦 = 4
- 6. Implicit Differentiation •The first function defines 𝑦 as a function of 𝑥 explicitly, because for each 𝑥, the equation gives the explicit formula 𝑦 = 𝑓 𝑥 for finding the corresponding value of 𝑦. You can easily differentiate this function following the usual process by applying the differentiation rules: 𝑑𝑦 𝑑𝑥 = 2𝑥
- 7. Implicit Differentiation •The second function may still be differentiated easily, provided that it is possible to express it in explicit form. Because 𝑥2 + 3𝑦 = 4 , then by algebraic manipulation, 𝑦 = − 1 3 𝑥2 + 4 3 . You can now differentiate this function and yield 𝑑𝑦 𝑑𝑥 = − 1 3 2𝑥 = − 2 3 𝑥
- 8. Implicit Differentiation •However, there are cases when it is totally impossible to rewrite the given equation as function of 𝑥, such as 𝑥4 + 2𝑥𝑦 − 𝑦3 + 2 = 0. It is in these instances that implicit differentiation is used.
- 9. Implicit Differentiation • Because the equation is written in implicit form, the process of implicit differentiation involves differentiating each term of an equation on both sides, including those that contain the variably 𝑦. To totally understand how this works, remember that you are always differentiating in terms of 𝑥. Thus, you must differentiate the 𝑥 terms in the usual way. But for the 𝑦 terms, the chain rule must be applied because 𝑦 is defined implicitly as a function of 𝑥. For example, the expression 𝑥3 + 3𝑦2 has a derivative of 3𝑥2 + 6𝑦 𝑑𝑦 𝑑𝑥 .
- 10. Implicit Differentiation • To further guide you on the implicit differentiation process, follow these steps: • Differentiate both sides of the equation with respect to 𝑥. • Combine all terms containing 𝑑𝑦 𝑑𝑥 on the left side of the equation and all the other terms on the right side. • On the left side of the equation, factor out 𝑑𝑦 𝑑𝑥 . • Isolate 𝑑𝑦 𝑑𝑥 by dividing out the other factor on the left side.
- 11. Example 1: •Find 𝑑𝑦 𝑑𝑥 given that 𝑦3 + 𝑦 − 3𝑥2 = 2𝑥 + 5.
- 12. Solution: •Differentiate both sides of the equation with respect to 𝑥. 𝑑 𝑑𝑥 𝑦3 + 𝑦 − 3𝑥2 = 𝑑 𝑑𝑥 2𝑥 + 5 •Take derivative of each term with respect to 𝑥 and use chain rule. 3𝑦2 𝑑𝑦 𝑑𝑥 + 𝑑𝑦 𝑑𝑥 − 6𝑥 = 2
- 13. Solution: •Combine all terms with 𝑑𝑦 𝑑𝑥 on the left side. 3𝑦2 𝑑𝑦 𝑑𝑥 + 𝑑𝑦 𝑑𝑥 = 6𝑥 + 2 •Factor out 𝑑𝑦 𝑑𝑥 . 𝑑𝑦 𝑑𝑥 3𝑦2 + 1 = 6𝑥 + 2
- 14. Solution: •Divide both sides by 3𝑦2 + 1 to isolate 𝑑𝑦 𝑑𝑥 . 𝑑𝑦 𝑑𝑥 = 6𝑥 + 2 3𝑦2 + 1
- 15. Example 2: •Given the equation 𝑥2 𝑦 + 2𝑥𝑦2 = 𝑥4 + 𝑦4 , find 𝑑𝑦 𝑑𝑥 .
- 16. Solution: •Solution: Differentiate both sides of the equation with respect to 𝑥. 𝑑 𝑑𝑥 𝑥2 𝑦 + 2𝑥𝑦2 = 𝑑 𝑑𝑥 𝑥4 + 𝑦4
- 17. Solution: 𝑥2 𝑑𝑦 𝑑𝑥 + 2𝑥𝑦 + 4𝑥𝑦 𝑑𝑦 𝑑𝑥 + 2𝑦2 = 4𝑥3 + 4𝑦3 𝑑𝑦 𝑑𝑥 Differentiate with respect to 𝑥 𝑥2 𝑑𝑦 𝑑𝑥 + 4𝑥𝑦 𝑑𝑦 𝑑𝑥 − 4𝑦3 𝑑𝑦 𝑑𝑥 = 4𝑥3 − 2𝑥𝑦 − 2𝑦2 Combination of terms with 𝑑𝑦 𝑑𝑥 𝑑𝑦 𝑑𝑥 (𝑥2 + 4𝑥𝑦 − 4𝑦3) = 4𝑥3 − 2𝑥𝑦 − 2𝑦2 Factoring out 𝑑𝑦 𝑑𝑥 𝑑𝑦 𝑑𝑥 = 4𝑥3 − 2𝑥𝑦 − 2𝑦2 𝑥2 + 4𝑥𝑦 − 4𝑦3 Division of 𝑥2 + 4𝑥𝑦 − 4𝑦3
- 18. Solution: •The implicit differentiation can be used to find the slope of the tangent line to the curve, given that it is not a function, as in the next examples.
- 19. Example 3: •Find the slope of the curve described by 𝑥2 + 𝑦2 = 1 at 1 2 , 3 2
- 20. Solution: •The equation is that of unit circle. Thus, it is not a function. We can, however, restrict to apply implicit differentiation. •Differentiate both sides with respect to 𝑥. 𝑑 𝑑𝑥 𝑥2 + 𝑦2 = 𝑑𝑦 𝑑𝑥 1 𝑑 𝑑𝑥 𝑥2 + 𝑑 𝑑𝑥 𝑦2 = 0
- 21. Solution: •2𝑥 + 2𝑦 𝑑𝑦 𝑑𝑥 = 0 •2𝑦 𝑑𝑦 𝑑𝑥 = −2𝑥 • 𝑑𝑦 𝑑𝑥 = − 2𝑥 2𝑦 = − 𝑥 𝑦
- 22. Solution: •To find the slope at point at point 1 2 , 3 2 , substitute 1 2 for 𝑥 and 3 2 for 𝑦. You now obtain. • 𝑑𝑦 𝑑𝑥 = − 𝑥 𝑦 = − 1 2 3 2 = − 1 2 ∙ 2 3 = − 1 3
- 23. Example 4: •Find the equation of the tangent line to the curve 𝑥3 + 𝑦3 = 9 at the point 1, 2 .
- 24. Solution: •Sometimes, you can use 𝑦′ instead of 𝑑𝑦 𝑑𝑥 . Differentiating implicitly with respect to 𝑥, • 𝑑𝑦 𝑑𝑥 𝑥3 + 𝑦3 = 𝑑𝑦 𝑑𝑥 9 •3𝑥2 + 3𝑦2 𝑦′ = 0 •3𝑦2 𝑦′ = −3𝑥2 •𝑦′ = − 3𝑥2 3𝑦2 = − 𝑥2 𝑦2
- 25. Solution: •Now, at the point 1, 2 , the slope 𝑚 of the tangent line is 𝑚 = − 𝑥2 𝑦2 = − 1 2 2 2 = − 1 4 . Using the point-slope form of the equation of the line, you now have •𝑦 − 2 = − 1 4 𝑥 − 1 •𝑦 = − 1 4 𝑥 + 1 4 + 2 •𝑦 = − 1 4 𝑥 + 9 4
- 26. Example 5: • Consider the equation 𝑥2 + 𝑦2 = 9. a) Find 𝑦′ by implicit differentiation. b) Derive two functions that can be defined from the equation. c) Find the derivative of each functions obtained in (b) by explicit differentiation. d) Verify that the result obtained in (a) agrees with result in (c).
- 27. Solution: •Differentiating implicitly, you find that •2𝑥 + 2𝑦 𝑦′ = 0 •2𝑦 𝑦′ = −2𝑥 •𝑦′ = − 2𝑥 2𝑦 = − 𝑥 𝑦
- 28. Solution: •Solving the equation for 𝑦, you have 𝑦2 = 9 − 𝑥2 . Thus, •𝑦 = ± 9 − 𝑥2 •Let 𝑓 𝑥 = 9 − 𝑥2 and 𝑔 𝑥 = − 9 − 𝑥2.
- 29. Solution: •Because 𝑓 𝑥 = 9 − 𝑥2, by using chain rule, you now have •𝑓′ 𝑥 = 1 2 9 − 𝑥2 − 1 2 −2𝑥 = 1 2 −2𝑥 ∙ 1 9−𝑥2 = − 𝑥 9−𝑥2 •Similarly, you get 𝑔′ 𝑥 = 𝑥 9−𝑥2 .
- 30. Solution: •For 𝑦 = 𝑓 𝑥 where 𝑓 𝑥 = 9 − 𝑥2, you have from (c), •𝑓′ 𝑥 = − 𝑥 9−𝑥2 = − 𝑥 𝑦 •For 𝑦 = 𝑔 𝑥 where 𝑔 𝑥 = − 9 − 𝑥2, you have from (c), •𝑔′ 𝑥 = 𝑥 9−𝑥2 = 𝑥 − 9−𝑥2 = − 𝑥 𝑦
- 31. Solution: •The explicit differentiation for both functions agrees with the answer in (a).
- 32. Big Idea: •The implicit differentiation is performed on functions that are difficult to express directly. As a result, the differentiation process takes longer than usual and also a bit confusing when not properly carried out. This is the same for the things you cannot express directly. They may imply the right message, but the receiver might interpret them incorrectly when properly stated.
- 33. DERIVATIVE RULES FOR EXPONENTIAL AND LOGARITHMIC FUNCTIONS WITH BASE 𝒆
- 34. Derivative Rules for exponential and Logarithmic Functions with Base 𝒆 •You probably have now mastered how to differentiate algebraic functions based on the examples and exercises in the previous module. Not all functions, however, are algebraic in form, so you cannot apply the techniques for differentiation right away. In the case of exponential and logarithmic functions, they also have a set of rules that you must follows in finding the derivatives.
- 35. Derivative Rules for exponential and Logarithmic Functions with Base 𝒆 • Among all the exponential functions of the form 𝑦 = 𝑎𝑥 , natural exponential function is of great importance in terms of differentiation. Recall that the natural exponential function is written as 𝑦 = 𝑒𝑥 , where 𝑒 is the Euler number, an irrational number equivalent to 2.718281828 …. The most important thing about the natural exponential function is that its derivative is itself. Hence, 𝑑 𝑑𝑥 𝑒𝑥 = 𝑒𝑥 . To prove this, examine the value of 𝑒 once again. You may want to verify these values using your calculator.
- 36. Derivative Rules for exponential and Logarithmic Functions with Base 𝒆 Table 12.1. Values of 𝑒 as 𝑥 approaches 0 𝑥 1 + 𝑥 (1 + 𝑥) 1 𝑥 0.1 1.1 2.59374246 0.01 1.01 2.704813829 0.001 1.001 2.716923932 0.0001 1.0001 2.718145927 0.00001 1.00001 2.718268237 0.000001 1.000001 2.718280469 0.0000001 1.0000001 2.718281694 0.00000001 1.00000001 2.718281786
- 37. Derivative Rules for exponential and Logarithmic Functions with Base 𝒆 •From table 12.1, the value of (1 + 𝑥) 1 𝑥 approaches the number 𝑒 as 𝑥 approaches 0. Thus, lim 𝑥→0 (1 + 𝑥) 1 𝑥= 𝑒. This implies that as 𝑥 becomes sufficiently small, 𝑒 = (1 + 𝑥) 1 𝑥 or 𝑒𝑥 = 1 + 𝑥 . Now, suppose 𝑓 𝑥 = 𝑒𝑥 . Then,
- 38. Derivative Rules for exponential and Logarithmic Functions with Base 𝒆 𝑓′(𝑥) = lim ℎ→0 𝑒𝑥+ℎ − 𝑒𝑥 ℎ Definition of derivative = lim ℎ→0 𝑒𝑥 (𝑒ℎ − 1) ℎ Factoring out 𝑒𝑥 = lim ℎ→0 𝑒𝑥 (1 + ℎ − 1) ℎ Based on 𝑒ℎ = 1 + ℎ = lim ℎ→0 𝑒𝑥 (ℎ) ℎ Simplifying terms = lim ℎ→0 𝑒𝑥 = 𝑒𝑥
- 39. Derivative of Natural Exponential Functions •The natural exponential functions is its own derivative, that is, 𝑑 𝑑𝑥 𝑒𝑥 = 𝑒𝑥 •Generally, if 𝑢 is a differentiable function of 𝑥, then 𝑑 𝑑𝑥 𝑒𝑢 = 𝑒𝑢 𝑑𝑢
- 40. Example 6: •Differentiate each function. a) 𝑓 𝑥 = 𝑒3𝑥 b) 𝑔 𝑥 = 𝑒−2𝑥2 c) 𝑝 𝑥 = 𝑥𝑒2𝑥 d) ℎ 𝑥 = 𝑒2𝑥+1 𝑒𝑥−1
- 41. Solution: •Let 𝑢 = 3𝑥. Then, 𝑑 𝑑𝑥 𝑒3𝑥 = 𝑒3𝑥 ∙ 3 = 3𝑒3𝑥 •Let 𝑢 = −2𝑥2 . Then, 𝑑 𝑑𝑥 𝑒−2𝑥2 = 𝑒−2𝑥2 ∙ −4𝑥 = − 4𝑥𝑒−2𝑥2 •Using the product rule, 𝑝′ 𝑥 = 𝑥 ∙ 𝑑 𝑑𝑥 𝑒2𝑥 + 𝑒2𝑥 ∙ 𝑑 𝑑𝑥 𝑥 = 𝑥 2𝑒2𝑥 + 𝑒2𝑥 = 2𝑥𝑒2𝑥 + 𝑒2𝑥
- 42. Solution: •Let 𝑢 = 𝑒2𝑥 + 1 and 𝑣 = 𝑒𝑥 − 1 . Then, 𝑑𝑢 = 2𝑒2𝑥 and 𝑑𝑣 = 𝑒𝑥 . Applying quotient rule, ℎ′ 𝑥 = 𝑒𝑥 − 1 2𝑒2𝑥 − (𝑒2𝑥 + 1)(𝑒𝑥 ) (𝑒𝑥−1)2 = 2𝑒3𝑥 − 2𝑒2𝑥 − 𝑒3𝑥 − 𝑒𝑥 (𝑒𝑥−1)2 = 𝑒3𝑥 − 2𝑒2𝑥 − 𝑒𝑥 (𝑒𝑥−1)2
- 43. •Now that you know the derivative of the natural exponential function, you can now derive the rule for the derivative of its inverse, which is the natural logarithmic function. The derivative of the natural logarithmic function may be derived by using implicit differentiation. Suppose that 𝑦 = ln 𝑥.
- 44. 𝑦 = ln 𝑥 Given 𝑒𝑦 = 𝑥 Exponential notation 𝑑 𝑑𝑥 𝑒𝑦 = 𝑑 𝑑𝑥 𝑥 Differentiating both sides 𝑒𝑦 𝑑𝑦 𝑑𝑥 = 1 Implicit differentiation 𝑑𝑦 𝑑𝑥 = 1 𝑒𝑦 Isolating 𝑑𝑦 𝑑𝑥 𝑑𝑦 𝑑𝑥 = 1 𝑥 𝑒𝑦 = 𝑥
- 45. Derivative of Natural Logarithmic Function •The derivative of the natural logarithmic function is the reciprocal function, that is, 𝑑 𝑑𝑥 ln 𝑥 = 1 𝑥 •If 𝑢 is differentiable function of 𝑥, then 𝑑 𝑑𝑥 ln 𝑢 = 1 𝑢 𝑑𝑢
- 46. Example 7: •Find the derivative of each function. a) 𝑦 = ln 4𝑥 b) 𝑓 𝑥 = ln(5𝑥2 + 3) c) 𝑔 𝑥 = 𝑥 ln 𝑥 d) ℎ 𝑥 = 𝑥 ln 𝑥
- 47. Solution: •Let 𝑢 = 4𝑥 . Applying the derivative of natural logarithmic function, you have 𝑑 𝑑𝑥 ln 4𝑥 = 1 4𝑥 ∙ 𝑑 𝑑𝑥 4𝑥 = 4 4𝑥 = 1 𝑥 •Let 𝑢 = 5𝑥2 + 3. Then, 𝑑 𝑑𝑥 ln(5𝑥2 + 3) = 1 5𝑥2 + 3 ∙ 𝑑 𝑑𝑥 (5𝑥2 + 3) = 1 5𝑥2 + 3 (10𝑥) = 10𝑥 5𝑥2 + 3
- 48. Solution: •Using quotient rule, you have 𝑑 𝑑𝑥 𝑥 ln 𝑥 = (ln 𝑥) 𝑑 𝑑𝑥 𝑥 − 𝑥 𝑑 𝑑𝑥 [ln 𝑥] (ln 𝑥)2 Quotient Rule = ln 𝑥 − 𝑥 1 𝑥 ln2𝑥 Recall: (ln 𝑥)2 = ln2 𝑥 = ln 𝑥 − 1 ln2𝑥 Simplify
- 49. Solution: •Using product rule, you have ℎ′ 𝑥 = 𝑥 ∙ 𝑑 𝑑𝑥 ln 𝑥 + ln 𝑥 𝑑 𝑑𝑥 ∙ 𝑥 = 𝑥 ∙ 1 𝑥 + ln 𝑥 = 1 + ln 𝑥
- 50. Example 8: •Find the derivative of 𝑓 𝑥 = ln 4𝑥+2 𝑥−1 2
- 51. Solution: •Recall that the quotient rule of logarithms states that the logarithm of a quotient is equal to the difference of the logarithm of the numerator and the logarithm of the denominator. 𝑓 𝑥 = ln 4𝑥 + 2 𝑥 − 1 2 = ln 4𝑥 + 2 − ln 𝑥 − 1 2 •By power rule of logarithms, 𝑓 𝑥 = ln 4𝑥 + 2 − 2 ln(𝑥 − 1)
- 52. Solution: •Thus, 𝑓′ 𝑥 = 𝑑 𝑑𝑥 ln 4𝑥 + 2 − 2 𝑑 𝑑𝑥 ln(𝑥 − 1) = 1 4𝑥 + 2 ∙ 4 − 2 ∙ 1 𝑥 − 1 = 4 (4𝑥 + 2) − 2 𝑥 − 1 = −2𝑥 − 4 (2𝑥 + 1)(𝑥 − 1)
- 53. DERIVATIVE RULES FOR GENERAL EXPONENTIAL AND LOGARITHMIC FUNCTIONS
- 54. Derivative Rules for General Exponential and Logarithmic Functions •When the given exponential or logarithmic function expressed using a base other than the number 𝑒, the following rules may be applied. These rules can be derived through rewriting the function using base 𝑒.
- 55. Derivative with Other Bases • Let 𝑢 be a differentiable function of 𝑥 and 𝑎 is a real number such that 𝑎 > 0 and 𝑎 ≠ 1. Then, • 𝑑 𝑑𝑥 𝑎𝑥 = (ln 𝑎)𝑎𝑥 • 𝑑 𝑑𝑥 log𝑎 𝑥 = 1 ln 𝑎 1 𝑥 • 𝑑 𝑑𝑥 𝑎𝑢 = (ln 𝑎)𝑎𝑢 𝑑𝑢 • 𝑑 𝑑𝑥 log𝑎 𝑢 = 1 ln 𝑎 1 𝑢 𝑑𝑢
- 57. Proof: By definition, 𝑎𝑥 = 𝑒𝑥(ln 𝑎) . Therefore, 𝑑 𝑑𝑥 𝑎𝑥 = 𝑑 𝑑𝑥 𝑒𝑥(ln 𝑎) = 𝑒𝑥(ln 𝑎) ∙ 𝑑 𝑑𝑥 𝑥(ln 𝑎) = 𝑒𝑥(ln 𝑎) ∙ 𝑥 ∙ 𝑑 𝑑𝑥 (ln 𝑎) + ln 𝑎
- 58. Now, using constant rule, 𝑑 𝑑𝑥 (ln 𝑎) = 0. Thus, 𝑑 𝑑𝑥 𝑎𝑥 = 𝑒𝑥(ln 𝑎) ∙ 0 + ln 𝑎 = 𝑒𝑥(ln 𝑎) ∙ ln 𝑎 = (ln 𝑎)𝑎𝑥
- 59. Theorem: 𝑑 𝑑𝑥 log𝑎 𝑥 = 1 ln 𝑎 1 𝑥
- 60. Proof: Using the change-of-base formula, log𝑎 𝑥 = ln 𝑥 ln 𝑎 . Thus, 𝑑 𝑑𝑥 log𝑎 𝑥 = 𝑑 𝑑𝑥 ln 𝑥 ln 𝑎 = ln 𝑎 ∙ 𝑑 𝑑𝑥 ln 𝑥 − ln 𝑥 ∙ 𝑑 𝑑𝑥 ln 𝑎 ln2𝑎 = ln 𝑎 ∙ 1 𝑥 − 0 ln2𝑎 = 1 𝑥 ln 𝑎 ∙ 1 ln2𝑎 = 1 ln 𝑎 1 𝑥
- 61. Example 1: •Find the derivative of the following function: a) 𝑦 = 52𝑥 b) 𝑦 = 𝑒𝑥 + 8𝑥 c) 𝑦 = log5(3𝑥2 + 1) d) 𝑦 = log 𝑥 𝑥 − 2 3
- 62. Solution: •𝑦 = 52𝑥 𝑑 𝑑𝑥 52𝑥 = 52𝑥 ∙ ln 5 ∙ 𝑑 𝑑𝑥 2𝑥 = (2 ln 5)52𝑥 •𝑦 = 𝑒𝑥 + 8𝑥 𝑑 𝑑𝑥 𝑒𝑥 + 8𝑥 = 𝑑 𝑑𝑥 𝑒𝑥 + 𝑑 𝑑𝑥 8𝑥 = 𝑒𝑥 + ln 8 8𝑥
- 63. Solution: •𝑦 = log5 3𝑥2 + 1 𝑑 𝑑𝑥 log5 3𝑥2 + 1 = 1 ln 5 1 3𝑥2 + 1 ∙ 𝑑 𝑑𝑥 3𝑥2 + 1 = 1 ln 5 1 3𝑥2 + 1 6𝑥 = 6𝑥 (ln 5) 3𝑥2 + 1
- 64. Solution: •This function will be best differentiated if it will be simplified first using the product rule for logarithms, i.e., log 𝑢𝑣 = log 𝑢 + log 𝑣. Also, recall that this is a case of common logarithm in which the base, although not written, is understood to be 10. log 𝑥 𝑥 − 2 3 = log 𝑥 + log 𝑥 − 2 3 = log 𝑥 + 3 log(𝑥 − 2) 𝑑 𝑑𝑥 log 𝑥 𝑥 − 2 3 = 𝑑 𝑑𝑥 log 𝑥 + 3 log(𝑥 − 2)
- 65. Solution: = 𝑑 𝑑𝑥 log 𝑥 + 3 𝑑 𝑑𝑥 log(𝑥 − 2) = 1 ln 10 1 𝑥 + 3 1 ln 10 1 𝑥 − 2 ∙ 𝑑 𝑑𝑥 𝑥 − 2 = 1 𝑥 ln 10 + 3 𝑥 − 2 ln 10 = 4𝑥 − 2 (𝑥 ln 10)(𝑥 − 2)
- 67. Derivative Rules for Trigonometric Functions •You will deepen your understanding of trigonometric functions as you explore their derivatives. The trigonometric functions have many useful applications in several fields such as physics, astronomy, and biological sciences. Needed in the formulation of the rule for the derivatives of the six trigonometric functions are the two important trigonometric limits: lim 𝑥→0 sin 𝑥 𝑥 = 1 and lim 𝑥→0 1−cos 𝑥 𝑥 = 0.
- 68. Derivative Rules for Trigonometric Functions •The derivations of the differentiation formulas for trigonometric functions are based on these two limits. To start, let us derive the formula for 𝑑 𝑑𝑥 sin 𝑥 as follows. •By the definition of the derivative, 𝑑 𝑑𝑥 sin 𝑥 = lim ℎ→0 sin 𝑥 + ℎ − sin 𝑥 ℎ
- 69. Derivative Rules for Trigonometric Functions •Using the sum identity for sine, you have sin 𝑥 + ℎ = sin 𝑥 cos ℎ + cos 𝑥 sin ℎ. Then, 𝑑 𝑑𝑥 sin 𝑥 = lim ℎ→0 sin 𝑥 cos ℎ + cos 𝑥 sin ℎ − sin 𝑥 ℎ = lim ℎ→0 cos 𝑥 sin ℎ ℎ + sin 𝑥 cos ℎ − sin 𝑥 ℎ = lim ℎ→0 cos 𝑥 sin ℎ ℎ + lim ℎ→0 sin 𝑥 cos ℎ − sin 𝑥 ℎ
- 70. Derivative Rules for Trigonometric Functions = lim ℎ→0 cos 𝑥 ∙ sin ℎ ℎ + lim ℎ→0 − sin 𝑥 = cos 𝑥 lim ℎ→0 sin ℎ ℎ − sin 𝑥 lim ℎ→0 1 − cos ℎ ℎ = cos 𝑥 1 − sin 𝑥 (0) 𝑑 𝑑𝑥 sin 𝑥 = cos 𝑥
- 71. Derivative Rules for Trigonometric Functions •Thus, the derivative of the sine function is the cosine function. To derive the differentiation formula forcos 𝑥, you have 𝑑 𝑑𝑥 cos 𝑥 = lim ℎ→0 cos 𝑥 + ℎ − cos 𝑥 ℎ = lim ℎ→0 cos 𝑥 cos ℎ − sin 𝑥 sin ℎ − cos 𝑥 ℎ = lim ℎ→0 cos 𝑥 cos ℎ − cos 𝑥 − (sin 𝑥 sin ℎ) ℎ
- 72. Derivative Rules for Trigonometric Functions = lim ℎ→0 cos 𝑥 cos ℎ − cos 𝑥 ℎ = − cos 𝑥 lim ℎ→0 1 − cos ℎ ℎ = − cos 𝑥 0 − sin 𝑥 (1) 𝑑 𝑑𝑥 cos 𝑥 = − sin 𝑥
- 73. Derivative Rules for Trigonometric Functions •In general, for any function 𝑢 differentiable at any value of 𝑥 , 𝑑 𝑑𝑥 sin 𝑢 = cos 𝑢 𝑑𝑢 and 𝑑 𝑑𝑥 cos 𝑢 = − sin 𝑢 𝑑𝑢 . The following gives the differentiation formulas for the other functions. Their derivatives are left for you as an exercise.
- 74. Derivatives of the Trigonometric Functions • 𝑑 𝑑𝑥 sin 𝑢 = cos 𝑢 𝑑𝑢 • 𝑑 𝑑𝑥 cos 𝑢 = − sin 𝑢 𝑑𝑢 • 𝑑 𝑑𝑥 tan 𝑢 = sec2 𝑢 𝑑𝑢
- 75. Derivatives of the Trigonometric Functions • 𝑑 𝑑𝑥 cot 𝑢 = −csc2 𝑢 𝑑𝑢 • 𝑑 𝑑𝑥 sec 𝑢 = sec 𝑢 tan 𝑢 𝑑𝑢 • 𝑑 𝑑𝑥 csc 𝑢 = − csc 𝑢 cot 𝑢 𝑑𝑢
- 76. Example 1: •Differentiate the following functions: a) 𝑦 = cos 3𝑥 b) 𝑦 = sin 6𝑥2 − 1 c) 𝑦 = tan 5𝑥 4 d) 𝑦 = sec(6𝑥)
- 77. Solution: •Let 𝑢 = 3𝑥. Then, 𝑑 𝑑𝑥 cos 3𝑥 = − sin 3𝑥 ∙ 𝑑 𝑑𝑥 3𝑥 = −3 sin 3𝑥 •Let 𝑢 = 6𝑥2 − 1. Applying the derivative rule for sine, 𝑑 𝑑𝑥 sin 6𝑥2 − 1 = cos 6𝑥2 − 1 ∙ 𝑑𝑦 𝑑𝑥 6𝑥2 − 1 = 12𝑥 cos 6𝑥2 − 1
- 78. Solution: •Let 𝑢 = 5𝑥 4 . Therefore, 𝑑 𝑑𝑥 tan 5𝑥 4 = sec2 5𝑥 4 ∙ 𝑑 𝑑𝑥 5𝑥 4 = 5 4 sec2 5𝑥 4 •Let 𝑢 = 6𝑥. Then 𝑑𝑢 = 6 and you have 𝑑 𝑑𝑥 sec 6𝑥 = sec 6𝑥 tan 6𝑥 ∙ 6 = 6 sec 6𝑥 tan 6𝑥
- 79. Example 2: •Differentiate 𝑦 = (1 − cos 2𝑥)4
- 80. Solution: • Applying the chain rule, let 𝑢 = 1 − cos 2𝑥. 𝑑𝑢 = 𝑑 𝑑𝑥 1 − 𝑑 𝑑𝑥 cos 2𝑥 = 0 − − sin 2𝑥 𝑑 𝑑𝑥 2𝑥 = 2 sin 2𝑥 • Therefore, 𝑦′ = 4𝑢3 𝑑𝑢 = 4 1 − cos 2𝑥 3 (2 sin 2𝑥) = (8 sin 2𝑥) 1 − cos 2𝑥 3
- 81. Example 3: •Given that 𝑓 𝑥 = sin 𝑥 1−2 cos 𝑥 , find 𝑓′(𝑥).
- 82. Solution: • Let 𝑢 = sin 𝑥 and 𝑣 = 1 − 2 cos 𝑥 , giving 𝑑𝑢 = cos 𝑥 and 𝑑𝑣 = 2 sin 𝑥. Applying quotient rule, you have 𝑓′ 𝑥 = 1 − 2 cos 𝑥 cos 𝑥 − sin 𝑥 2 sin 𝑥 1 − 2 cos 𝑥 2 = cos 𝑥 − 2cos2 𝑥 − 2sin2 𝑥 1 − 2 cos 𝑥 2
- 83. Solution: = cos 𝑥 − 2(cos2 𝑥 + sin2 𝑥) 1 − 2 cos 𝑥 2 = cos 𝑥 − 2(1) 1 − 2 cos 𝑥 2 = cos 𝑥 − 2 1 − 2 cos 𝑥 2
- 84. Example 4: •Differentiate the following: 𝑦 = sin 3𝑥 4+5 cos 2𝑥
- 85. Solution: • Apply the quotient rule followed by chain rule. Then 𝑦′ = 4 + 5 cos 2𝑥 3 cos 3𝑥 − sin 3𝑥 −10 sin 2𝑥 4 + 5 cos 2𝑥 2 = 12 cos 3𝑥 + 15 cos 2𝑥 cos 3𝑥 + (10 sin 2𝑥 sin 3𝑥) 4 + 5 cos 2𝑥 2 = 12 cos 3𝑥 + 15 cos 2𝑥 cos 3𝑥 + 10 sin 2𝑥 sin 3𝑥 4 + 5 cos 2𝑥 2
- 86. Example 5: •Given that 𝑥 cos 𝑦 + 𝑦 cos 𝑥 = 1, find 𝑑𝑦 𝑑𝑥 .
- 87. Solution: • Differentiating implicitly with respect to 𝑥, you get 𝑥 ∙ − sin 𝑦 𝑑𝑦 𝑑𝑥 + cos 𝑦 1 + 𝑦 ∙ − sin 𝑥 + cos 𝑥 𝑑𝑦 𝑑𝑥 = 0 −𝑥 sin 𝑦 𝑑𝑦 𝑑𝑥 + cos 𝑦 − 𝑦 sin 𝑥 + cos 𝑥 𝑑𝑦 𝑑𝑥 = 0 − sin 𝑦 𝑑𝑦 𝑑𝑥 + cos 𝑥 𝑑𝑦 𝑑𝑥 = 𝑦 sin 𝑥 − cos 𝑦 𝑑𝑦 𝑑𝑥 cos 𝑥 − 𝑥 sin 𝑦 = 𝑦 sin 𝑥 − cos 𝑦 𝑑𝑦 𝑑𝑥 = 𝑦 sin 𝑥 − cos 𝑦 cos 𝑥 − 𝑥 sin 𝑦
- 88. Example 6: •Find the second the derivative of 𝑔 𝑥 = cot 3𝑥3 .
- 89. Solution: • To find the first derivative, use chain rule and the differentiation rule for sine: 𝑔′ 𝑥 = −csc2 3𝑥2 ∙ 9𝑥2 = −9𝑥2 csc2 3𝑥3 • Then, 𝑔′′ 𝑥 = 𝑑 𝑑𝑥 −9𝑥2 csc2 3𝑥3
- 90. Solution: = −9𝑥2 ∙ 𝑑 𝑑𝑥 csc2 3𝑥3 + csc2 3𝑥3 ∙ 𝑑 𝑑𝑥 −9𝑥2 = −9𝑥2 2 csc 3𝑥3 − csc 3𝑥3 cot 3𝑥3 9𝑥2 + csc2 3𝑥3 −18𝑥 = 162𝑥4 csc2 3𝑥3 cot 3𝑥3 − 18𝑥 csc2 3𝑥3 = 18𝑥 csc2 3𝑥3 9𝑥3 cot 3𝑥3 − 1
- 91. Example 7: •In a certain city, the normal average temperature (in degrees Fahrenheit) is modelled by the function 𝑓 𝑡 = 49 − 18 cos 𝜋(𝑡−32) 180 , where 𝑡 is the time in days so that 𝑡 = 1 corresponds to the first of January. What is the expected date of the warmest day and the coolest day?
- 92. Solution: • To solve this problem, you need to find the value of 𝑡 such that 𝑓′ 𝑡 = 0. 𝑓′ 𝑡 = 0 − 18 𝑑 𝑑𝑡 cos 𝜋 𝑡 − 32 180 = −18 − sin 𝜋 𝑡 − 32 180 ∙ 𝑑 𝑑𝑡 𝜋 𝑡 − 32 180 = 18 sin 𝜋(𝑡 − 32) 180 ∙ 𝜋 180 = 𝜋 10 sin 𝜋 𝑡 − 32 180
- 93. Solution: • Then, 𝜋 10 sin 𝜋 𝑡 − 32 180 = 0 sin 𝜋 𝑡 − 32 180 = 0
- 94. Solution: • In trigonometry, you have learned thatsin 𝜃 = 0 for all multiples of 𝜋. Thus, 𝜋 𝑡 − 32 180 = 0 𝜋 𝑡 − 32 = 0 𝑡 − 32 = 0 𝑡 = 32
- 95. Solution: 𝜋 𝑡 − 32 180 = 𝜋 𝜋 𝑡 − 32 = 180𝜋 𝑡 − 32 = 180 𝑡 = 212
- 96. Solution: 𝜋 𝑡 − 32 180 = 2𝜋 𝜋 𝑡 − 32 = 360𝜋 𝑡 − 32 = 360 𝑡 = 392
- 97. Solution: • For 𝜃 = 2𝜋 onward, 𝑡 is already out of the domain because a year has only 365 days. Substituting n the original function, if 𝑡 = 32, 𝑓 𝑡 = 49 − 18 cos 𝜋 32 − 32 180 = 49 − 18 cos 0 = 49 − 18 1 = 31℉ • If 𝑡 = 212, 𝑓 𝑡 = 49 − 18 cos 𝜋 212 − 32 180 = 49 − 18 cos 0 = 49 − 18 −1 = 67℉
- 98. Solution: • Therefore, the coolest day will happen when 𝑡 = 32, i.e., the first of February, and the warmest day will be when 𝑡 = 212, or 31st of July, provided that it is not leap year.