If f(x)=x, then f^'(x)=1 ·x^0=1 If f(x)=x^2, ...

Key Concepts

Derivative

The derivative is a fundamental concept in calculus that measures the rate at which a function changes at any given point. It is defined as the limit of the difference quotient as the increment approaches zero, providing a precise mathematical formulation of instantaneous rate of change or slope of the tangent line to the function’s graph.

Power Rule

The power rule is a basic differentiation rule in calculus which states that for a function of the form f(x) = x^n, where n is a real number, the derivative is given by f'(x) = n * x^(n-1). This rule simplifies the process of finding derivatives of polynomial functions and is one of the first rules taught in differential calculus.

Mathematical Induction

Mathematical induction is a proof technique used to establish that a statement is true for all natural numbers. It consists of two steps: first, proving the base case is true, and second, proving that if the statement holds for an arbitrary natural number, it must also hold for the next one. This method is essential for proving properties that are defined recursively or that naturally extend over the set of positive integers.

Binomial Expansion

The binomial expansion, derived from the binomial theorem, provides a method for expanding expressions raised to a power, such as (x+h)^n. In the context of differentiation, it is used to expand functions so that the limit definition of the derivative can be applied term-by-term. This technique is particularly useful for algebraically simplifying the difference quotient when deriving the power rule.

Transcript

00:01 Equals x to the first f prime of x equals 1 times x to 0 if f of x equals x squared f prime of x equals 2 times x to the first if f of x equals x to the third f prime of x equals you can find this by taking the limit of the difference quotient if f of x equals x to the 3 f prime of x equals 3 times x to the 2 so when n is 1 2 2 and 3 we have f f f f f f f of x equaling x to the n and f prime of x equaling n times x to the n minus one.

00:38 When n is three, f prime of x is n times x to the n minus one.

00:47 So x to the n, x to the 3, f prime of x is n 3 times x to the n minus 1, 3 minus 1, which is 2.

00:57 So we know this is true for n is 1, 2, and 3.

01:05 When we have x to the n, when n is 1, 2, or 3, the derivative of x to the n will be n times x to the n minus 1.

01:16 Now, if we can show that if this is true for n, and we already know it's true for n equals 1, 2, and 3, if we can show that if this is true for n, then it's also going to be true for n plus 1, then by induction, we're going to be able to prove that f prime of x, if f of x equals n, then f prime of x equals n times x to the n minus 1 for all positive integers n.

01:49 In other words, okay, if this is true for n, and we already know, is true for n equals 1, 2, and 3.

01:58 But if this is true for n, and we're able to show that when, you know, when it's true for n, that it's also going to be true for n plus 1, then this will hold for all positive integers n.

02:10 Here's why.

02:12 We already know it's true when we have x cubed.

02:17 Okay? so in that case, n is 3.

02:20 Knowing that it already works for x cubed, if we can show that this formula works for x to the fourth, then you'll have it working for x to the fourth.

02:33 And then because for every n, you were able to show that it works for n plus one.

02:41 Okay, it works for x cubed.

02:43 If you're able to show that working for n means it works for n plus one, that means it's going to work for x to the fourth, working for x to the fourth.

02:51 If you already show it, it works for n and it works for n plus one.

02:54 You'll have it work for x to the fifth.

02:56 It works for x to the fifth.

02:58 You already know it's going to work for n plus 1.

03:00 It's going to work for x to the 6.

03:02 So a derivative of x to the 6 will be 6 times x to the 5th.

03:06 Derivative of x to 7 will be 7 times x to the 6.

03:10 Derivative of x to the end will be n times x to n minus 1.

03:14 So if we can show that when f of x equals x to the n, that the derivative, what we'll we want to show that this works for n plus 1.

03:27 So in other words, okay, we need to show, this is what we need to show.

03:41 We need to show if f of x equals x to the n plus 1, then f prime of x will equal n plus 1 times x to the n.

04:11 So for example, we know that it works for x to the third, n being three, n plus one, if n is three, n plus one would be four.

04:31 If we can show that, you know, when f of x is x to the fourth, then f prime of x will be four times x to the third.

04:44 This is what we're trying to show.

04:46 But instead of doing it for x to the fourth and x to the fifth and each one one at a time, if we can just show that when f of x equals x to the end, f prime of x is equaling n times x to the end minus one, we already know this is true for n equals 1, 2, and 3.

05:05 If this is working for x to the n, if we can show it works for the next integer up x to the n plus 1, if we can show that when f of x is x to the n plus 1, f prime of x will be the n plus 1 times x to the n, which is n plus 1 minus 1, then you'll be able to show that this formula works for all positive integers n.

05:34 Because we know it works for x to the first, x to the second, x to the third.

05:39 So if we're able to show this general formula, if we're able to prove this, if we're able to prove that knowing that this works for x to the n, we can prove that it works for x to the n plus one, then you know it's going to work for x to the fourth.

06:08 Because if it worked for x to the third and we're able to prove this, then you know it's going to be able to work for x to the fourth because that's just x a third plus one.

06:18 And if you know it works for x to the fourth, you know it's going to work for x to the fifth because that's x to the, you know, four plus one.

06:29 Basically, to prove that this works for all positive integers n, we're going to assume that it works for a certain n.

06:42 And we already know it works for n equals 1, 2, and 3.

06:44 So we're assuming this works for some positive integer n.

06:52 We already know it does for 1, 2, and 3.

06:55 Assuming that this works, and we already know it does, for some positive integer n, if we can prove that it also works for the next positive integer n plus 1, then you're going to know that it works for all positive integers.

07:20 Okay, so remember what we're doing.

07:25 We know that if f of x equals x to some positive, x to some positive integer n, then f prime of x equals n times x to the n minus 1.

07:36 We already know that's true when n is 1, 2, and 3.

07:39 Assuming this is true for x to the n, if we can show that is true for x to the n plus 1, that when f of x equals x to the n plus 1, that f prime of x comes out to be n plus 1.

07:51 Times x to the end, then you're going to know that this formula works for all positive integers n...

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