Video: Differentiating Trigonometric Functions Involving Trigonometric Ratios Using the Product Rule

Consider the series 𝑓(π‘₯) = 1/(1 + π‘₯) = βˆ‘_(𝑛 = 0)^(∞) (βˆ’1)^(𝑛) π‘₯^(𝑛). Differentiate the given series expansion of 𝑓 term by term to find the corresponding series expansion for the derivative of 𝑓. Use the result of the first part to evaluate the sum of the series βˆ‘_(𝑛 = 0)^(∞) ((βˆ’1)^(𝑛 + 1) (𝑛 + 1))/3^(𝑛).

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Video Transcript

Consider the series 𝑓 of π‘₯ equals one over one plus π‘₯, which equals the sum of negative one to the 𝑛th power times π‘₯ to the 𝑛th power, for values of 𝑛 between zero and ∞. Differentiate the given series expansion of 𝑓 term by term to find the corresponding series expansion for the derivative of 𝑓. Then use the result of the first part to evaluate the sum of the series the sum of negative one to the 𝑛 plus one times 𝑛 plus one over three to the 𝑛th power for values of 𝑛 between zero and ∞.

We’ve been given the series expansion of 𝑓. So let’s work out the first few terms. The first term is when 𝑛 is equal to zero. It’s negative one to the power of zero times π‘₯ to the power of zero, which is equal to one. The second term is when 𝑛 is equal to one. So it’s negative one to the power of one times π‘₯ to the power of one, which is negative π‘₯. We then add negative one squared times π‘₯ squared, which is π‘₯ squared. And we continue in this manner. And we see that 𝑓 of π‘₯ is equal to one minus π‘₯ plus π‘₯ squared minus π‘₯ cubed plus π‘₯ to the fourth power minus π‘₯ to the fifth power, and so on.

The question tells us to differentiate the given series expansion of 𝑓 that will give us 𝑓 prime. And of course we can simply do this term by term. The derivative of one is zero, and the derivative of negative π‘₯ is negative one. The derivative of π‘₯ squared is two π‘₯. And then our next term is negative three π‘₯ squared.

We continue in this way, multiplying each term by its exponent and then reducing that exponent by one. Without writing the zero, we see that we can write 𝑓 prime of π‘₯ as negative π‘₯ to the power of zero β€” remember, that’s just negative one β€” plus two π‘₯ to the power of one minus three π‘₯ squared plus four π‘₯ cubed, and so on.

Let’s think about how we can write this as a sum. We know that we start with a negative term, and then the sign alternates. To achieve this, we need negative one to the power of 𝑛 plus one. This will work for values of 𝑛 between zero and ∞. We then multiply each term by 𝑛 plus one. So the first term is when 𝑛 is equal to zero. And we’re multiplying that by one. The second term is when 𝑛 is equal to one. And we’re multiplying that by two, and so on. And this is all times π‘₯ to the 𝑛th power.

And so we found the series expansion for the derivative of 𝑓. It’s the sum of negative one to the power of 𝑛 plus one times 𝑛 plus one times π‘₯ to the 𝑛th power for values of 𝑛 between zero and ∞.

Part two says to use the result of the first part to evaluate the sum of the series the sum of negative one to the power of 𝑛 plus one times 𝑛 plus one over three to the 𝑛th power for values of 𝑛 between zero and ∞. Let’s compare this sum to the one we generated in the first part of this question. If we rewrite this as the sum of negative one to the power of 𝑛 plus one times 𝑛 plus one times a third to the 𝑛th power, then we see it’s of the same form. Just π‘₯ equals a third.

So what we’re going to do is we’re going to differentiate our original function one over one plus π‘₯ and evaluate that when π‘₯ is equal to one-third. Let’s write 𝑓 of π‘₯ as one plus π‘₯ to the power of negative one. And then we use the chain rule. When we do, we find its derivative to be equal to negative one over one plus π‘₯ squared. The sum of our series will be the value of the derivative when π‘₯ is equal to one-third. So that’s negative one over one plus one-third squared. That’s negative one over 16 over nine, which is simply negative nine sixteenths.

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