Problem

Prove Taylor's Inequality for $ n = 2, $ that is,…

05:59

Problem 82 Easy Difficulty

If $f(x)=\left(1+x^{3}\right)^{30},$ what is $f^{(5)}(0) ?$

Name: if fxleft1x3right30 what is f50
Uploaded: 2021-10-22
Duration: 3 min 41 s
Channel: Numerade
Description: If $f(x)=\left(1+x^{3}\right)^{30},$ what is $f^{(5)}(0) ?$

Answer

$$f^{58}(0)=0$$

Related Courses

Calculus 2 / BC

Calculus: Early Transcendentals

Chapter 11

Infinite Sequences and Series

Section 10

Taylor and Maclaurin Series

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Video Transcript

this next problem we are going to see how the use of the binomial expansion will vastly simplify taking a derivative. So if we consider F of x equals one plus x cubed To the 30th power. And we want to evaluate The 58th derivative of F ah At the value zero. Now naively we can look at F prime of X equals Well you would apply the power rule 30 times one plus X cubed to the 29 then the chain rule, you would need a three X squared in there. However, for the second derivative, you now need both the product rule and the chain rule And each subsequent derivative will have more and more applications of the product rule and more and more terms. So obviously this is less than ideal. However, instead we can write one plus x cubed to the 30th is equal to the sum from K equals 0 to 30 of 30, choose K of X cubed to the cath power. And now we see that d over dx of this simple some because each of these are now just polynomial is we can bring down a three K and then have 30 choose K X to the three K -1. And now we in fact notice that this cake will zero here is wrong because we need to be careful that D over dx of one is equal to zero. So now in fact what we need to do is we need to take this some From K. equals 1- 30. So the final step for getting the 58th derivative is we note that 19 times three is equal to 57. So all of those previous ones are going to be canceled out by the taking the derivative of the constant equals one. And so we can conclude That D over DX applied 58 times to the sum. Now from K equals 0 to 30 of 30 choose K. X. To the three K is equal to. We take the sum from K equals 20 to 30. And now we're going to have a three K times three K minus one, et cetera. All the way to three K minus 57 30, choose K, still X to the three K -58. And so each of these powers of X because now our largest K is our smallest K is 20 so the smallest power is going to be X squared are non constant, which means when we plug in X equals zero, Each of the terms goes to zero. And we conclude that The 58th derivative Of Exit zero is equal to zero.

Victoria C.

Rutgers, The State University of New Jersey

Related Courses

Calculus 2 / BC

Calculus: Early Transcendentals

Chapter 11

Infinite Sequences and Series

Section 10

Taylor and Maclaurin Series

Problem

Prove Taylor's Inequality for $ n = 2, $ that is,…

Problem 82 Easy Difficulty

If $f(x)=\left(1+x^{3}\right)^{30},$ what is $f^{(5)}(0) ?$

Answer

$$f^{58}(0)=0$$

Related Courses

Calculus: Early Transcendentals

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Discussion

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Watch More Solved Questions in Chapter 11

Video Transcript

Calculus: Early Transcendentals

Grace H.

Kayleah T.

Michael J.

Joseph L.

Integration Review - Intro

Definite vs Indefinite