### Video Transcript

Consider the series π of π₯ equals
π₯ over one minus π₯ squared, which is equal to the sum of π₯ to the power of two π
plus one 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.

Letβs begin by using the power
series to write out the first few terms. The first term is when π is equal
to zero. So itβs π₯ to the power of two
times zero plus one, which is one. We then let π be equal to one. And we get two times one plus one,
which is three. Next, we let π be equal to
two. And we find the exponent to be
equal to five. And we can continue in this
manner. We find the first few terms to be
π₯ to the power of one plus π₯ cubed plus π₯ to the fifth power plus π₯ to the
seventh power, and so on.

Weβre told to differentiate the
series expansion of π term by term. And so weβre going to do just
that. The derivative of π₯ with respect
to π₯ is just one. And we recall that, for any
polynomial term, we differentiate by multiplying by the exponent and then reducing
that exponent by one. So the derivative of π₯ cubed is
three π₯ squared. The derivative of π₯ to the fifth
power is five π₯ to the fourth power. Differentiating π₯ to the seventh
power, and we get seven π₯ to the sixth power, and so on.

We need to find a way to write this
as a sum. So letβs have a look at whatβs
happened. Each time, weβve multiplied by the
exponent. And we saw that that exponent was
generated by using the expression two π plus one. Then we reduced each power by one,
while the original exponent was two π plus one. So our new exponent is two π plus
one minus one. The values for our sum remain the
same. Theyβre from π equals zero to
β. And of course two π plus one minus
one is simply two π. And so by differentiating term by
term, we found the corresponding series expansion for the derivative of π. Itβs the sum of two π plus one
times π₯ to the power of two π for values of π between zero and β.