# Question Video: Finding the First Derivative of a Function Involving Negative Exponents Using the Power Rule Mathematics • Higher Education

Find the Maclaurin series of sin 3π₯ = (π^(3π₯) β π^(β3π₯))/2.

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### Video Transcript

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

Letβs start by writing out the general form for the Maclaurin series expansion for a function π. To make things easier for ourselves, letβs start by finding the Maclaurin series expansion for the hyperbolic sin of π₯. Weβll then be able to substitute π₯ with three π₯ to get the Maclaurin series of the hyperbolic sin of three π₯. So, weβre going to need to take our function, π of π₯, to be the hyperbolic sin of π₯, and weβre going to need to evaluate this at zero and its derivatives at zero. So, weβ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₯.

Letβs create a table for the derivatives of π of π₯ equals the hyperbolic sin of π₯. So, when π is zero, we just have π 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 π to the π₯ power minus π to the negative π₯ power over two. So, the hyperbolic sin of zero is π to the zero power minus π to the negative zero power over two. But as π to the zero power is just one, this is one minus one over two, which is zero. When π is equal to one, we have the first derivative of π, which weβ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 π to the π₯ power add π to the negative π₯ power over two. So, the hyperbolic cos of zero is equal to π to the zero power add π to the negative zero power over two. But we know that π to the zero power is just one. So, this is one add one over two, which is just one. When π is two, weβre looking for the second derivative of π 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 π is three, we want to find the third derivative of π, 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 π evaluated at zero alternates between zero and one. So, weβ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. Weβ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 π equals zero to β of π₯ to the power of two π add one over two π 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 π equals zero to β of three π₯ raised to the power of two π add one over two π add one factorial.