# Question Video: Differentiating Trigonometric Functions Involving Trigonometric Ratios Using the Product Rule Mathematics • Higher Education

Consider the series π(π₯) = 1/(1 + π₯) = β_(π = 0)^(β) (β1)^(π) π₯^(π). Differentiate the given series expansion of π term by term to find the corresponding series expansion for the derivative of π. Use the result of the first part to evaluate the sum of the series β_(π = 0)^(β) ((β1)^(π + 1) (π + 1))/3^(π).

03:17

### Video Transcript

Consider the series π of π₯ equals one over one plus π₯, which equals the sum of negative one to the πth power times π₯ to the πth power, for values of π between zero and β. Differentiate the given series expansion of π term by term to find the corresponding series expansion for the derivative of π. Then use the result of the first part to evaluate the sum of the series the sum of negative one to the π plus one times π plus one over three to the πth power for values of π between zero and β.

Weβve been given the series expansion of π. So letβs work out the first few terms. The first term is when π is equal to zero. Itβ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 π is equal to one. So itβ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 π 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 π that will give us π 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 π prime of π₯ as negative π₯ to the power of zero β remember, thatβs just negative one β plus two π₯ to the power of one minus three π₯ squared plus four π₯ cubed, and so on.

Letβ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 π plus one. This will work for values of π between zero and β. We then multiply each term by π plus one. So the first term is when π is equal to zero. And weβre multiplying that by one. The second term is when π is equal to one. And weβre multiplying that by two, and so on. And this is all times π₯ to the πth power.

And so we found the series expansion for the derivative of π. Itβs the sum of negative one to the power of π plus one times π plus one times π₯ to the πth power for values of π between zero and β.

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 π plus one times π plus one over three to the πth power for values of π between zero and β. Letβ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 π plus one times π plus one times a third to the πth power, then we see itβs of the same form. Just π₯ equals a third.

So what weβre going to do is weβre going to differentiate our original function one over one plus π₯ and evaluate that when π₯ is equal to one-third. Letβs write π 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 thatβs negative one over one plus one-third squared. Thatβs negative one over 16 over nine, which is simply negative nine sixteenths.