Video: Finding the First Derivative of a Function Involving Negative Exponents Using the Power Rule

Find the Maclaurin series of sin 3π‘₯ = (𝑒^(3π‘₯) βˆ’ 𝑒^(βˆ’3π‘₯))/2.

03:35

Video Transcript

Find the Maclaurin series of the hyperbolic sin of three π‘₯ equals 𝑒 to the three π‘₯ power minus 𝑒 to the negative three π‘₯ power over two.

Let’s start by writing out the general form for the Maclaurin series expansion for a function 𝑓. To make things easier for ourselves, let’s start by finding the Maclaurin series expansion for the hyperbolic sin of π‘₯. We’ll then be able to substitute π‘₯ with three π‘₯ to get the Maclaurin series of the hyperbolic sin of three π‘₯. So, we’re going to need to take our function, 𝑓 of π‘₯, to be the hyperbolic sin of π‘₯, and we’re going to need to evaluate this at zero and its derivatives at zero. So, we’ll also have to remember that the derivative with respect to π‘₯ of the hyperbolic sin of π‘₯ is the hyperbolic cos of π‘₯. And the derivative of the hyperbolic cos of π‘₯ with respect to π‘₯ is the hyperbolic sin of π‘₯.

Let’s create a table for the derivatives of 𝑓 of π‘₯ equals the hyperbolic sin of π‘₯. So, when 𝑛 is zero, we just have 𝑓 of π‘₯, which is the hyperbolic sin of π‘₯. So, to evaluate this when π‘₯ is zero, we use the fact that the hyperbolic sin of π‘₯ is equal to 𝑒 to the π‘₯ power minus 𝑒 to the negative π‘₯ power over two. So, the hyperbolic sin of zero is 𝑒 to the zero power minus 𝑒 to the negative zero power over two. But as 𝑒 to the zero power is just one, this is one minus one over two, which is zero. When 𝑛 is equal to one, we have the first derivative of 𝑓, which we’ve seen already is the hyperbolic cos of π‘₯.

In order to evaluate this at zero, we recall that the hyperbolic cos of π‘₯ is equal to 𝑒 to the π‘₯ power add 𝑒 to the negative π‘₯ power over two. So, the hyperbolic cos of zero is equal to 𝑒 to the zero power add 𝑒 to the negative zero power over two. But we know that 𝑒 to the zero power is just one. So, this is one add one over two, which is just one. When 𝑛 is two, we’re looking for the second derivative of 𝑓 of π‘₯. We can get this by differentiating the first derivative, which was the hyperbolic cos of π‘₯. And because the hyperbolic cos of π‘₯ differentiates to give us the hyperbolic sin of π‘₯, this is the hyperbolic sin of π‘₯. We know what this is; when we evaluate it at zero, it just gives us zero.

And when 𝑛 is three, we want to find the third derivative of 𝑓, which is the derivative of the second derivative, which we found to be the hyperbolic sin. So, this is the hyperbolic cos of π‘₯, which we know when π‘₯ is zero, this is one. And We can start to see a bit of a pattern here. Each derivative of 𝑓 evaluated at zero alternates between zero and one. So, we’re going to use this table to write out the Maclaurin series for the hyperbolic sin of π‘₯. When we substitute in our values, we see that every other term is zero. We’re only left with the odd powers of π‘₯. We want to write this as a series, but we need to make sure we only end up with those odd powers of π‘₯.

So, our series is the sum from 𝑛 equals zero to ∞ of π‘₯ to the power of two 𝑛 add one over two 𝑛 add one factorial. We can use this now to find the Maclaurin series for the hyperbolic sin of three π‘₯. We do this by replacing π‘₯ with three π‘₯. And we can see that this gives us three π‘₯ over one factorial add three π‘₯ cubed over three factorial add three π‘₯ to the fifth power over five factorial, and so on. And we can see that we can write this as the sum from 𝑛 equals zero to ∞ of three π‘₯ raised to the power of two 𝑛 add one over two 𝑛 add one factorial.

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