1. Packet #10 Proving Derivatives
Math 180
Packet #10
Proving Derivatives

2. Proof of Power Rule Suppose 𝑓 𝑥 = 𝑥 𝑛
Proof of Power Rule Suppose 𝑓 𝑥 = 𝑥 𝑛 . Then, 𝑓 ′ 𝑥 = lim ℎ→0 𝑓 𝑥+ℎ −𝑓 𝑥 ℎ = lim ℎ→0 𝑥+ℎ 𝑛 − 𝑥 𝑛 ℎ = lim ℎ→0 𝑥 𝑛 + 𝑛 1 𝑥 𝑛−1 ℎ+ 𝑛 2 𝑥 𝑛−2 ℎ 2 +…+ ℎ 𝑛 − 𝑥 𝑛 ℎ = lim ℎ→0 𝑛 𝑥 𝑛−1 ℎ+ 𝑛 2 𝑥 𝑛−2 ℎ 2 +…+ ℎ 𝑛 ℎ = lim ℎ→0 𝑛 𝑥 𝑛−1 + 𝑛 2 𝑥 𝑛−2 ℎ+…+ ℎ 𝑛−1 =𝑛 𝑥 𝑛−1 ∎

3. Proof of Power Rule Suppose 𝑓 𝑥 = 𝑥 𝑛
Proof of Power Rule Suppose 𝑓 𝑥 = 𝑥 𝑛 . Then, 𝑓 ′ 𝑥 = lim ℎ→0 𝑓 𝑥+ℎ −𝑓 𝑥 ℎ = lim ℎ→0 𝑥+ℎ 𝑛 − 𝑥 𝑛 ℎ = lim ℎ→0 𝑥 𝑛 + 𝑛 1 𝑥 𝑛−1 ℎ+ 𝑛 2 𝑥 𝑛−2 ℎ 2 +…+ ℎ 𝑛 − 𝑥 𝑛 ℎ = lim ℎ→0 𝑛 𝑥 𝑛−1 ℎ+ 𝑛 2 𝑥 𝑛−2 ℎ 2 +…+ ℎ 𝑛 ℎ = lim ℎ→0 𝑛 𝑥 𝑛−1 + 𝑛 2 𝑥 𝑛−2 ℎ+…+ ℎ 𝑛−1 =𝑛 𝑥 𝑛−1 ∎

4. Proof of Power Rule Suppose 𝑓 𝑥 = 𝑥 𝑛
Proof of Power Rule Suppose 𝑓 𝑥 = 𝑥 𝑛 . Then, 𝑓 ′ 𝑥 = lim ℎ→0 𝑓 𝑥+ℎ −𝑓 𝑥 ℎ = lim ℎ→0 𝑥+ℎ 𝑛 − 𝑥 𝑛 ℎ = lim ℎ→0 𝑥 𝑛 + 𝑛 1 𝑥 𝑛−1 ℎ+ 𝑛 2 𝑥 𝑛−2 ℎ 2 +…+ ℎ 𝑛 − 𝑥 𝑛 ℎ = lim ℎ→0 𝑛 𝑥 𝑛−1 ℎ+ 𝑛 2 𝑥 𝑛−2 ℎ 2 +…+ ℎ 𝑛 ℎ = lim ℎ→0 𝑛 𝑥 𝑛−1 + 𝑛 2 𝑥 𝑛−2 ℎ+…+ ℎ 𝑛−1 =𝑛 𝑥 𝑛−1 ∎

5. Proof of Power Rule Suppose 𝑓 𝑥 = 𝑥 𝑛
Proof of Power Rule Suppose 𝑓 𝑥 = 𝑥 𝑛 . Then, 𝑓 ′ 𝑥 = lim ℎ→0 𝑓 𝑥+ℎ −𝑓 𝑥 ℎ = lim ℎ→0 𝑥+ℎ 𝑛 − 𝑥 𝑛 ℎ = lim ℎ→0 𝑥 𝑛 + 𝑛 1 𝑥 𝑛−1 ℎ+ 𝑛 2 𝑥 𝑛−2 ℎ 2 +…+ ℎ 𝑛 − 𝑥 𝑛 ℎ = lim ℎ→0 𝑛 𝑥 𝑛−1 ℎ+ 𝑛 2 𝑥 𝑛−2 ℎ 2 +…+ ℎ 𝑛 ℎ = lim ℎ→0 𝑛 𝑥 𝑛−1 + 𝑛 2 𝑥 𝑛−2 ℎ+…+ ℎ 𝑛−1 =𝑛 𝑥 𝑛−1 ∎

6. Proof of Power Rule Suppose 𝑓 𝑥 = 𝑥 𝑛
Proof of Power Rule Suppose 𝑓 𝑥 = 𝑥 𝑛 . Then, 𝑓 ′ 𝑥 = lim ℎ→0 𝑓 𝑥+ℎ −𝑓 𝑥 ℎ = lim ℎ→0 𝑥+ℎ 𝑛 − 𝑥 𝑛 ℎ = lim ℎ→0 𝑥 𝑛 + 𝑛 1 𝑥 𝑛−1 ℎ+ 𝑛 2 𝑥 𝑛−2 ℎ 2 +…+ ℎ 𝑛 − 𝑥 𝑛 ℎ = lim ℎ→0 𝑛 𝑥 𝑛−1 ℎ+ 𝑛 2 𝑥 𝑛−2 ℎ 2 +…+ ℎ 𝑛 ℎ = lim ℎ→0 𝑛 𝑥 𝑛−1 + 𝑛 2 𝑥 𝑛−2 ℎ+…+ ℎ 𝑛−1 =𝑛 𝑥 𝑛−1 ∎

7. Proof of Power Rule Suppose 𝑓 𝑥 = 𝑥 𝑛
Proof of Power Rule Suppose 𝑓 𝑥 = 𝑥 𝑛 . Then, 𝑓 ′ 𝑥 = lim ℎ→0 𝑓 𝑥+ℎ −𝑓 𝑥 ℎ = lim ℎ→0 𝑥+ℎ 𝑛 − 𝑥 𝑛 ℎ = lim ℎ→0 𝑥 𝑛 + 𝑛 1 𝑥 𝑛−1 ℎ+ 𝑛 2 𝑥 𝑛−2 ℎ 2 +…+ ℎ 𝑛 − 𝑥 𝑛 ℎ = lim ℎ→0 𝑛 𝑥 𝑛−1 ℎ+ 𝑛 2 𝑥 𝑛−2 ℎ 2 +…+ ℎ 𝑛 ℎ = lim ℎ→0 𝑛 𝑥 𝑛−1 + 𝑛 2 𝑥 𝑛−2 ℎ+…+ ℎ 𝑛−1 =𝑛 𝑥 𝑛−1 ∎

8. Proof of Power Rule Suppose 𝑓 𝑥 = 𝑥 𝑛
Proof of Power Rule Suppose 𝑓 𝑥 = 𝑥 𝑛 . Then, 𝑓 ′ 𝑥 = lim ℎ→0 𝑓 𝑥+ℎ −𝑓 𝑥 ℎ = lim ℎ→0 𝑥+ℎ 𝑛 − 𝑥 𝑛 ℎ = lim ℎ→0 𝑥 𝑛 + 𝑛 1 𝑥 𝑛−1 ℎ+ 𝑛 2 𝑥 𝑛−2 ℎ 2 +…+ ℎ 𝑛 − 𝑥 𝑛 ℎ = lim ℎ→0 𝑛 𝑥 𝑛−1 ℎ+ 𝑛 2 𝑥 𝑛−2 ℎ 2 +…+ ℎ 𝑛 ℎ = lim ℎ→0 𝑛 𝑥 𝑛−1 + 𝑛 2 𝑥 𝑛−2 ℎ+…+ ℎ 𝑛−1 =𝑛 𝑥 𝑛−1 ∎

9. Trigonometric Functions
To figure out the derivatives of sin 𝑥 and cos 𝑥 by the definition, we need these two limits: lim 𝜃→0 sin 𝜃 𝜃 =1 and lim 𝜃→0 cos 𝜃 −1 𝜃 =0 These two limits can be proven using a geometric argument. Then we can find the derivatives of sin 𝑥 and cos 𝑥 as follows:

10. Trigonometric Functions
To figure out the derivatives of sin 𝑥 and cos 𝑥 by the definition, we need these two limits: lim 𝜃→0 sin 𝜃 𝜃 =1 and lim 𝜃→0 cos 𝜃 −1 𝜃 =0 These two limits can be proven using a geometric argument. Then we can find the derivatives of sin 𝑥 and cos 𝑥 as follows:

11. Trigonometric Functions
𝑑 𝑑𝑥 sin 𝑥 = lim ℎ→0 sin 𝑥+ℎ − sin 𝑥 ℎ = lim ℎ→0 sin 𝑥 cos ℎ + cos 𝑥 sin ℎ − sin 𝑥 ℎ = lim ℎ→0 sin 𝑥 cos ℎ −1 + cos 𝑥 sin ℎ ℎ = lim ℎ→0 sin 𝑥 ⋅ cos ℎ −1 ℎ + cos 𝑥 ⋅ sin ℎ ℎ = sin 𝑥 ⋅0+ cos 𝑥 ⋅1 = cos 𝑥

12. Trigonometric Functions
𝑑 𝑑𝑥 sin 𝑥 = lim ℎ→0 sin 𝑥+ℎ − sin 𝑥 ℎ = lim ℎ→0 sin 𝑥 cos ℎ + cos 𝑥 sin ℎ − sin 𝑥 ℎ = lim ℎ→0 sin 𝑥 cos ℎ −1 + cos 𝑥 sin ℎ ℎ = lim ℎ→0 sin 𝑥 ⋅ cos ℎ −1 ℎ + cos 𝑥 ⋅ sin ℎ ℎ = sin 𝑥 ⋅0+ cos 𝑥 ⋅1 = cos 𝑥

13. Trigonometric Functions
𝑑 𝑑𝑥 sin 𝑥 = lim ℎ→0 sin 𝑥+ℎ − sin 𝑥 ℎ = lim ℎ→0 sin 𝑥 cos ℎ + cos 𝑥 sin ℎ − sin 𝑥 ℎ = lim ℎ→0 sin 𝑥 cos ℎ −1 + cos 𝑥 sin ℎ ℎ = lim ℎ→0 sin 𝑥 ⋅ cos ℎ −1 ℎ + cos 𝑥 ⋅ sin ℎ ℎ = sin 𝑥 ⋅0+ cos 𝑥 ⋅1 = cos 𝑥

14. Trigonometric Functions
𝑑 𝑑𝑥 sin 𝑥 = lim ℎ→0 sin 𝑥+ℎ − sin 𝑥 ℎ = lim ℎ→0 sin 𝑥 cos ℎ + cos 𝑥 sin ℎ − sin 𝑥 ℎ = lim ℎ→0 sin 𝑥 cos ℎ −1 + cos 𝑥 sin ℎ ℎ = lim ℎ→0 sin 𝑥 ⋅ cos ℎ −1 ℎ + cos 𝑥 ⋅ sin ℎ ℎ = sin 𝑥 ⋅0+ cos 𝑥 ⋅1 = cos 𝑥

15. Trigonometric Functions
𝑑 𝑑𝑥 sin 𝑥 = lim ℎ→0 sin 𝑥+ℎ − sin 𝑥 ℎ = lim ℎ→0 sin 𝑥 cos ℎ + cos 𝑥 sin ℎ − sin 𝑥 ℎ = lim ℎ→0 sin 𝑥 cos ℎ −1 + cos 𝑥 sin ℎ ℎ = lim ℎ→0 sin 𝑥 ⋅ cos ℎ −1 ℎ + cos 𝑥 ⋅ sin ℎ ℎ = sin 𝑥 ⋅0+ cos 𝑥 ⋅1 = cos 𝑥

16. Trigonometric Functions
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 −1 − sin 𝑥 sin ℎ ℎ = lim ℎ→0 cos 𝑥 ⋅ cos ℎ −1 ℎ − sin 𝑥 ⋅ sin ℎ ℎ = cos 𝑥 ⋅0− sin 𝑥 ⋅1 =− sin 𝑥

17. Trigonometric Functions
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 −1 − sin 𝑥 sin ℎ ℎ = lim ℎ→0 cos 𝑥 ⋅ cos ℎ −1 ℎ − sin 𝑥 ⋅ sin ℎ ℎ = cos 𝑥 ⋅0− sin 𝑥 ⋅1 =− sin 𝑥

18. Trigonometric Functions
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 −1 − sin 𝑥 sin ℎ ℎ = lim ℎ→0 cos 𝑥 ⋅ cos ℎ −1 ℎ − sin 𝑥 ⋅ sin ℎ ℎ = cos 𝑥 ⋅0− sin 𝑥 ⋅1 =− sin 𝑥

19. Trigonometric Functions
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 −1 − sin 𝑥 sin ℎ ℎ = lim ℎ→0 cos 𝑥 ⋅ cos ℎ −1 ℎ − sin 𝑥 ⋅ sin ℎ ℎ = cos 𝑥 ⋅0− sin 𝑥 ⋅1 =− sin 𝑥

20. Trigonometric Functions
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 −1 − sin 𝑥 sin ℎ ℎ = lim ℎ→0 cos 𝑥 ⋅ cos ℎ −1 ℎ − sin 𝑥 ⋅ sin ℎ ℎ = cos 𝑥 ⋅0− sin 𝑥 ⋅1 =− sin 𝑥

21. Trigonometric Functions
But what about tan 𝑥 and the rest of the gang? Ex 1. Prove that the derivative of 𝑦= tan 𝑥 is 𝑑𝑦 𝑑𝑥 = sec 2 𝑥 by using the derivatives of sin 𝑥 and cos 𝑥 .

22. Exponential Functions
To find the derivative of 𝑒 𝑥 by the definition, we need the following limit: lim ℎ→0 𝑒 ℎ −1 ℎ =1 This limit is sometimes taken to just be definitional. That is, 𝑒 is the number where lim ℎ→0 𝑒 ℎ −1 ℎ =1.

23. Exponential Functions
To find the derivative of 𝑒 𝑥 by the definition, we need the following limit: lim ℎ→0 𝑒 ℎ −1 ℎ =1 This limit is sometimes taken to just be definitional. That is, 𝑒 is the number where lim ℎ→0 𝑒 ℎ −1 ℎ =1.

24. Exponential Functions
Then, 𝑑 𝑑𝑥 𝑒 𝑥 = lim ℎ→0 𝑒 𝑥+ℎ − 𝑒 𝑥 ℎ = lim ℎ→0 𝑒 𝑥 𝑒 ℎ − 𝑒 𝑥 ℎ = lim ℎ→0 𝑒 𝑥 𝑒 ℎ −1 ℎ = lim ℎ→0 𝑒 𝑥 ⋅ 𝑒 ℎ −1 ℎ = 𝑒 𝑥 ⋅1 = 𝑒 𝑥

25. Exponential Functions
Then, 𝑑 𝑑𝑥 𝑒 𝑥 = lim ℎ→0 𝑒 𝑥+ℎ − 𝑒 𝑥 ℎ = lim ℎ→0 𝑒 𝑥 𝑒 ℎ − 𝑒 𝑥 ℎ = lim ℎ→0 𝑒 𝑥 𝑒 ℎ −1 ℎ = lim ℎ→0 𝑒 𝑥 ⋅ 𝑒 ℎ −1 ℎ = 𝑒 𝑥 ⋅1 = 𝑒 𝑥

26. Exponential Functions
Then, 𝑑 𝑑𝑥 𝑒 𝑥 = lim ℎ→0 𝑒 𝑥+ℎ − 𝑒 𝑥 ℎ = lim ℎ→0 𝑒 𝑥 𝑒 ℎ − 𝑒 𝑥 ℎ = lim ℎ→0 𝑒 𝑥 𝑒 ℎ −1 ℎ = lim ℎ→0 𝑒 𝑥 ⋅ 𝑒 ℎ −1 ℎ = 𝑒 𝑥 ⋅1 = 𝑒 𝑥

27. Exponential Functions
Then, 𝑑 𝑑𝑥 𝑒 𝑥 = lim ℎ→0 𝑒 𝑥+ℎ − 𝑒 𝑥 ℎ = lim ℎ→0 𝑒 𝑥 𝑒 ℎ − 𝑒 𝑥 ℎ = lim ℎ→0 𝑒 𝑥 𝑒 ℎ −1 ℎ = lim ℎ→0 𝑒 𝑥 ⋅ 𝑒 ℎ −1 ℎ = 𝑒 𝑥 ⋅1 = 𝑒 𝑥

28. Exponential Functions
Then, 𝑑 𝑑𝑥 𝑒 𝑥 = lim ℎ→0 𝑒 𝑥+ℎ − 𝑒 𝑥 ℎ = lim ℎ→0 𝑒 𝑥 𝑒 ℎ − 𝑒 𝑥 ℎ = lim ℎ→0 𝑒 𝑥 𝑒 ℎ −1 ℎ = lim ℎ→0 𝑒 𝑥 ⋅ 𝑒 ℎ −1 ℎ = 𝑒 𝑥 ⋅1 = 𝑒 𝑥

29. Exponential Functions
Here’s how to get the derivative of 𝑦= 𝑎 𝑥 : 𝑑 𝑑𝑥 𝑎 𝑥 = 𝑑 𝑑𝑥 𝑒 ln 𝑎 𝑥 = 𝑑 𝑑𝑥 𝑒 𝑥 ln 𝑎 = 𝑒 𝑥 ln 𝑎 ln 𝑎 = 𝑎 𝑥 ln 𝑎

30. Exponential Functions
Here’s how to get the derivative of 𝑦= 𝑎 𝑥 : 𝑑 𝑑𝑥 𝑎 𝑥 = 𝑑 𝑑𝑥 𝑒 ln 𝑎 𝑥 = 𝑑 𝑑𝑥 𝑒 𝑥 ln 𝑎 = 𝑒 𝑥 ln 𝑎 ln 𝑎 = 𝑎 𝑥 ln 𝑎

31. Exponential Functions
Here’s how to get the derivative of 𝑦= 𝑎 𝑥 : 𝑑 𝑑𝑥 𝑎 𝑥 = 𝑑 𝑑𝑥 𝑒 ln 𝑎 𝑥 = 𝑑 𝑑𝑥 𝑒 𝑥 ln 𝑎 = 𝑒 𝑥 ln 𝑎 ln 𝑎 = 𝑎 𝑥 ln 𝑎

32. Exponential Functions
Here’s how to get the derivative of 𝑦= 𝑎 𝑥 : 𝑑 𝑑𝑥 𝑎 𝑥 = 𝑑 𝑑𝑥 𝑒 ln 𝑎 𝑥 = 𝑑 𝑑𝑥 𝑒 𝑥 ln 𝑎 = 𝑒 𝑥 ln 𝑎 ln 𝑎 = 𝑎 𝑥 ln 𝑎

33. Exponential Functions
Here’s how to get the derivative of 𝑦= 𝑎 𝑥 : 𝑑 𝑑𝑥 𝑎 𝑥 = 𝑑 𝑑𝑥 𝑒 ln 𝑎 𝑥 = 𝑑 𝑑𝑥 𝑒 𝑥 ln 𝑎 = 𝑒 𝑥 ln 𝑎 ln 𝑎 = 𝑎 𝑥 ln 𝑎

34. Logarithmic Functions
To find the derivative of logarithms, we can use implicit differentiation. Ex 2. Prove that the derivative of 𝑦= ln 𝑥 is 𝑑𝑦 𝑑𝑥 = 1 𝑥 .

35. Logarithmic Functions
Note: We can extend our results to include negative 𝑥-values: 𝑑 𝑑𝑥 ln 𝑥 = 1 𝑥 (𝑥≠0)

36. Logarithmic Functions
Here’s how to get the derivative of 𝑦= log 𝑎 𝑥 . 𝑑 𝑑𝑥 log 𝑎 𝑥 = 𝑑 𝑑𝑥 ln 𝑥 ln 𝑎 = 1 ln 𝑎 ⋅ 𝑑 𝑑𝑥 ln 𝑥 = 1 ln 𝑎 ⋅ 1 𝑥 = 1 𝑥 ln 𝑎

37. Logarithmic Functions
Here’s how to get the derivative of 𝑦= log 𝑎 𝑥 . 𝑑 𝑑𝑥 log 𝑎 𝑥 = 𝑑 𝑑𝑥 ln 𝑥 ln 𝑎 = 1 ln 𝑎 ⋅ 𝑑 𝑑𝑥 ln 𝑥 = 1 ln 𝑎 ⋅ 1 𝑥 = 1 𝑥 ln 𝑎

38. Logarithmic Functions
Here’s how to get the derivative of 𝑦= log 𝑎 𝑥 . 𝑑 𝑑𝑥 log 𝑎 𝑥 = 𝑑 𝑑𝑥 ln 𝑥 ln 𝑎 = 1 ln 𝑎 ⋅ 𝑑 𝑑𝑥 ln 𝑥 = 1 ln 𝑎 ⋅ 1 𝑥 = 1 𝑥 ln 𝑎

39. Logarithmic Functions
Here’s how to get the derivative of 𝑦= log 𝑎 𝑥 . 𝑑 𝑑𝑥 log 𝑎 𝑥 = 𝑑 𝑑𝑥 ln 𝑥 ln 𝑎 = 1 ln 𝑎 ⋅ 𝑑 𝑑𝑥 ln 𝑥 = 1 ln 𝑎 ⋅ 1 𝑥 = 1 𝑥 ln 𝑎

40. Logarithmic Functions
Here’s how to get the derivative of 𝑦= log 𝑎 𝑥 . 𝑑 𝑑𝑥 log 𝑎 𝑥 = 𝑑 𝑑𝑥 ln 𝑥 ln 𝑎 = 1 ln 𝑎 ⋅ 𝑑 𝑑𝑥 ln 𝑥 = 1 ln 𝑎 ⋅ 1 𝑥 = 1 𝑥 ln 𝑎

41. Inverse Trigonometric Functions
We can use implicit differentiation with inverse trig functions, too. Ex 3. Prove that the derivative of 𝑦= sin −1 𝑥 is 𝑑𝑦 𝑑𝑥 = 1 1− 𝑥 2 .

42. Hyperbolic Functions
Recall that sinh 𝑥 = 𝑒 𝑥 − 𝑒 −𝑥 2 and cosh 𝑥 = 𝑒 𝑥 + 𝑒 −𝑥 2 . Ex 4. Prove that the derivative of 𝑦= sinh 𝑥 is 𝑑𝑦 𝑑𝑥 = cosh 𝑥 .