Video: Use the Taylor Series Expansion to Find the Derivatives of a Function

The Taylor series representation of the function 𝑓 is given by 𝑓(π‘₯) = 2 βˆ’ 4(π‘₯ βˆ’ 7)Β² + 3 (π‘₯ βˆ’ 7)⁴ βˆ’ ((π‘₯ βˆ’ 7)⁢/6) + β‹…β‹…β‹…. Find the value of the fourth derivative of 𝑓 at π‘₯ = 7 . Find the value of the third derivative of 𝑓 at π‘₯ = 7.

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Video Transcript

The Taylor series representation of the function 𝑓 is given by 𝑓 of π‘₯ is equal to two minus four multiplied by π‘₯ minus seven squared plus three multiplied by π‘₯ minus seven to the fourth power minus π‘₯ minus seven to the sixth power divided by six. And the series keeps going and going. Find the value of the fourth derivative of 𝑓 at π‘₯ is equal to seven. Find the value of the third derivative of 𝑓 at π‘₯ is equal to seven.

The question wants us to find the fourth derivative of the function at π‘₯ is equal to seven and the third derivative of the function 𝑓 at π‘₯ is equal to seven. And to do this, it gives us a Taylor series representation of our function 𝑓 of π‘₯. We’ll recall that the Taylor series for a function 𝑓 about π‘Ž is equal to the sum from 𝑛 equals zero to infinity of the 𝑛th derivative of 𝑓 evaluated at π‘Ž divided by 𝑛 factorial multiplied by π‘₯ minus π‘Ž to the 𝑛th power. And this is for all values of π‘₯ where the series converges.

Since we know the representation of our function 𝑓 of π‘₯ is a Taylor series representation. And we can see that each term in this representation has π‘₯ minus seven raised to various powers. So this must be a Taylor series about seven. So that’s the π‘Ž in the definition of a Taylor series to be equal to seven.

So substituting the value of π‘Ž equals seven into our definition of a Taylor series gives us that 𝑓 of π‘₯ is equal to the sum from 𝑛 equals zero to infinity of the 𝑛th derivative of 𝑓 evaluated at seven divided by 𝑛 factorial multiplied by π‘₯ minus seven to the 𝑛th power. At this point, we can start writing out the terms of this series by substituting in each value of 𝑛 into our summand.

This starts with 𝑛 equals zero goes to 𝑛 equals one and carries on. Since the zeroth derivative of 𝑓 is just equal to 𝑓, zero factorial is equal to one, and raising π‘₯ minus seven to the zeroth power just gives us one. The first term is just 𝑓 evaluated at seven. Similarly, we can write the second term in our Taylor expansion as 𝑓 prime of seven multiplied by π‘₯ minus seven. We could do this for the third term. In fact, we could do this for as many terms as we would like.

What we want to notice from this Taylor expansion is the coefficient of π‘₯ minus seven to the 𝑛th power. What we want to notice from this is that the 𝑛th derivative evaluated at seven divided by 𝑛 factorial is equal to the coefficient of π‘₯ minus seven raised to the 𝑛th power. What this means in particular is that the fourth derivative of 𝑓 evaluated at seven divided by four factorial is equal to the coefficient of π‘₯ minus seven to the fourth power. And similarly, we will also have that the third derivative of 𝑓 evaluated at seven divided by three factorial is equal to the coefficient of π‘₯ minus three all cubed.

From the question, we can see that, in the Taylor series representation of our function 𝑓, the coefficient of π‘₯ minus seven to the fourth power is three. So we must have that the fourth derivative of 𝑓 evaluated at seven divided by four factorial is equal to three. Multiplying both sides of this equation by four factorial gives us that the fourth derivative of 𝑓 evaluated at seven is equal to three multiplied by four factorial, which is equal to 72.

Next, we know that the third derivative of 𝑓 evaluated at seven divided by three factorial will be equal to the coefficient of π‘₯ minus seven all cubed in our Taylor series representation. However, when we look at the question, we see that there is no π‘₯ minus seven all cubed term. If there is no π‘₯ minus seven all cubed term, this is the same as saying its coefficient is zero.

So we must have that the third derivative of 𝑓 evaluated at seven divided by three factorial is equal to zero. We can then multiply both sides of this equation by three factorial to get that the third derivative of 𝑓 evaluated at seven is equal to zero multiplied by three factorial, which is just equal to zero.

So what we’ve shown is that if the Taylor series representation of a function 𝑓 is given by two minus four multiplied by π‘₯ minus seven squared. Plus three multiplied by π‘₯ minus seven to the fourth power minus π‘₯ minus seven to the sixth power divided by six, et cetera. Then the fourth derivative of our function 𝑓 evaluated at seven is equal to 72. And the third derivative of our function 𝑓 evaluated at seven is equal to zero.

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