In each part, compute f^', f^'', f^''', and then state the formula for f^(n) . (a) f(x)=1 / x

In each part, compute f^', f^'', f^''', and then state the...

Key Concepts

Alternating Sign Patterns in Derivatives

Many functions, especially those involving negative exponents, exhibit alternating signs in their derivatives. This phenomenon can be succinctly represented using the term (-1)^n, where the exponent n determines the sign based on its parity. This concept is crucial in forming concise and accurate formulas for the nth derivative.

nth Derivative Formulas and Factorial Notation

The process of finding higher order (nth) derivatives frequently results in expressions that involve factorials. This occurs because each successive differentiation multiplies by a decrementing exponent, leading to a product of integers, which is naturally expressed using factorial notation. Recognizing this pattern allows one to state a general formula for the nth derivative of certain functions.

Power Rule for Differentiation

The power rule is a fundamental principle in calculus that allows the differentiation of functions of the form x^n. According to this rule, the derivative of x^n is given by n*x^(n-1). This rule is essential when handling functions with both positive and negative exponents, forming the basis for calculating successive derivatives.

Differentiation of Functions with Negative Exponents

When dealing with functions expressed as reciprocals, such as those of the form 1/x^k, it is often useful to rewrite them as x^(-k). This transformation allows for the direct application of the power rule, yielding derivatives that include increasing powers in the denominator and coefficients that can be expressed in factorial form. It also sets the stage for recognizing patterns in higher order derivatives.

Transcript

00:01 All right, so for this problem, there are two parts, a and b, and for each part, they want you to compute f -prime, f -double -prime, and f -triple -prime, and then they want you to write the general formula for the nth derivative of the function.

00:17 All right, so the way we're going to approach this problem is to write the first three derivatives and then find a pattern with those three derivatives in order to calculate the nth derivative.

00:29 All right so our first function is 1 over x and we can rewrite that as x to negative 1 power so we can use the power rule much more easily so if we were to take the first derivative f prime of x using the power rule it would be negative x to a negative 2 if we differentiated it again would get 2x to negative 3 and if we differentiated it a third time so i forgot prime for the second derivative.

01:02 If we differentiate it, a third time would get negative 6 x to the negative fourth power.

01:09 All right, so these are our first three derivatives of the function.

01:14 And now to calculate the nth derivative f of n of x, there's going to be three parts we want to talk about.

01:26 The first part is going to be the negative that's going to appear, then disappear, appear again with each consecutive through differentiation.

01:36 The next thing we're going to talk about is the coefficients in front of the x.

01:41 And then the final thing we're going to talk about are going to be the exponents, the exponents of the x variable.

01:48 All right.

01:48 So for our first tool for the negative sign, the hint that they give us is to use this expression, negative 1 to the n power, if negative 1, if there's a negative 1, if there's a negative.

02:02 Negative and n is odd and if it is positive for and is even and that is true for our differentiation.

02:09 If we took the first derivative, there's a negative symbol.

02:12 For our second, which is even, there is positive.

02:15 And then the third is a negative and so on and so on.

02:19 So we'll have that for our negative.

02:22 The next thing we want to talk about is the coefficients.

02:29 So first we have a one for a coefficient.

02:33 For the first derivative, then we have to, for our second derivative, we brought down the 2 from the exponent and multiplied it by 1.

02:42 And then for the third derivative, we brought down the negative 3 and multiply it by 2.

02:47 So you can imagine for the fourth derivative, we would have to bring down the negative 4 and multiply it by 6.

02:53 And the thing you can realize is that for each differentiation, we have to multiply the previous constant by the next integer number.

03:06 So if we were to take the fifth derivative, it would be the coefficient would be a multiplication of 1, 2, 3, 4, and 5.

03:16 Since to get to the 5th derivative, we would have to differentiate it 5 times before, 4 times before.

03:23 So generalizing this to the nth derivative, this would mean that we would have n factorial.

03:30 N factorial essentially means we're multiplying every integer that is n and smaller than n.

03:38 Alright so now that we got our n factorial, the last thing we need to discuss is the exponent...

Please give Ace some feedback