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## Lesson 19: Differentiation Rules for Sinusoidal Functions

Sinusoidal functions and compound sinusoidal functions have a wide range of applications. For example, radio waves consist of electromagnetic fields of energy, which are made up of two parts which periodically alternate. These two parts are the signal wave and the carrier wave. They periodically alternate in a manner that can be modelled using sinusoidal functions.

In the last lesson you looked at simple rules of differentiation, such as the constant multiple rule. Note that all the other rules of differentiation which we have learnt also apply to sinusoidal functions. That is, the chain rule, the product rule, and the power rule all apply. Let us begin by looking at a problem where we'll have to use the chain rule:

Differentiate f(x)=5cos (πx).

We note here that the cosine function is acting on the internal function, πx. So we take the derivative of the cosine function with respect to the internal function and then multiply by the derivative of the internal function:

f(x)=5cos (πx)

f'(x)=-5sin (πx)×π

f'(x)=-5πsin (πx)

Next, let us run through a problem where we are utilizing the power of a function rule.

Differentiate f(x)=3sin³x-2cos²x.

Here we need to utilize the chain rule again. Note that sin³x=(sin x)³ and cos²x=(cos x)². But this time the internal function is the sinusoidal function itself and the external function is a power function, so we need to utilize the power function on the external function and multiply by the derivative of the sinusoidal function.

f(x)=3sin³x-2cos²x

f'(x)=3(3)sin²xcos x-2(2)cos x(-sin x) Apply the power rule and take the derivative of the inner function.

f'(x)=9 sin²xcos x +4 sin x cos x You can factor sin x cos x.

f'(x)=9 sin²xcos x +4 sin x cos x

f'(x)=sin x cos x (9 sin x+4)

Let us run through a more complicated problem where we will have to utilize the product and the chain rule. In dealing with such problems, it always helps to tackle the problem systematically, applying differentiation rules from outside inwards.

Differentiate f(t)= sin 2t cos³t.

Note that we have a product of two sinusoidal functions, sin 2t and cos³t, so we will have to utilize the product rule on these, taking the derivative of each function as required by the product rule. When we take the derivative of each function, we will also have to use other rules since both functions are composite functions. This means we will have to use the chain rule. For sin 2t, the external function is a sinusoidal and the internal function is 2t. For cos³t, the internal function is cos t and the external function is the power function (the internal function raised to the power of 3). So we will have to apply chain rule as we go.

f(t)= sin 2t cos³t

f'(t)= d/dt sin 2t cos³t+ sin 2t d/dt cos³t

f'(t)= (2) cos 2t cos³t+ sin 2t 3cos²t (-sin t)

f'(t)= 2 cos 2t cos³t- sin t sin 2t 3cos²t

f'(t)= 2 cos 2t cos³t- sin t sin 2t 3(1-sin²t) Note that we can farther simplify because cos²t+sin²t=1. Therefore cos²t=1-sin²t.

f'(t)= 2 cos 2t cos³t- sin 2t [3sint -3sin³t] Simplify. Note that sin t × sin²t=sin³t.

f'(t)= 2 cos 2t cos³t- 3sint sin 2t + 3sin³t sin 2t

That's all there is to know for this lesson. Just keep in mind that it helps to approach these problems from outside inwards. For example, in the last problem we applied the product rule first and then we applied the chain rule for the respective component functions as we went along.

In the last lesson you looked at simple rules of differentiation, such as the constant multiple rule. Note that all the other rules of differentiation which we have learnt also apply to sinusoidal functions. That is, the chain rule, the product rule, and the power rule all apply. Let us begin by looking at a problem where we'll have to use the chain rule:

Differentiate f(x)=5cos (πx).

We note here that the cosine function is acting on the internal function, πx. So we take the derivative of the cosine function with respect to the internal function and then multiply by the derivative of the internal function:

f(x)=5cos (πx)

f'(x)=-5sin (πx)×π

f'(x)=-5πsin (πx)

Next, let us run through a problem where we are utilizing the power of a function rule.

Differentiate f(x)=3sin³x-2cos²x.

Here we need to utilize the chain rule again. Note that sin³x=(sin x)³ and cos²x=(cos x)². But this time the internal function is the sinusoidal function itself and the external function is a power function, so we need to utilize the power function on the external function and multiply by the derivative of the sinusoidal function.

f(x)=3sin³x-2cos²x

f'(x)=3(3)sin²xcos x-2(2)cos x(-sin x) Apply the power rule and take the derivative of the inner function.

f'(x)=9 sin²xcos x +4 sin x cos x You can factor sin x cos x.

f'(x)=9 sin²xcos x +4 sin x cos x

f'(x)=sin x cos x (9 sin x+4)

Let us run through a more complicated problem where we will have to utilize the product and the chain rule. In dealing with such problems, it always helps to tackle the problem systematically, applying differentiation rules from outside inwards.

Differentiate f(t)= sin 2t cos³t.

Note that we have a product of two sinusoidal functions, sin 2t and cos³t, so we will have to utilize the product rule on these, taking the derivative of each function as required by the product rule. When we take the derivative of each function, we will also have to use other rules since both functions are composite functions. This means we will have to use the chain rule. For sin 2t, the external function is a sinusoidal and the internal function is 2t. For cos³t, the internal function is cos t and the external function is the power function (the internal function raised to the power of 3). So we will have to apply chain rule as we go.

f(t)= sin 2t cos³t

f'(t)= d/dt sin 2t cos³t+ sin 2t d/dt cos³t

f'(t)= (2) cos 2t cos³t+ sin 2t 3cos²t (-sin t)

f'(t)= 2 cos 2t cos³t- sin t sin 2t 3cos²t

f'(t)= 2 cos 2t cos³t- sin t sin 2t 3(1-sin²t) Note that we can farther simplify because cos²t+sin²t=1. Therefore cos²t=1-sin²t.

f'(t)= 2 cos 2t cos³t- sin 2t [3sint -3sin³t] Simplify. Note that sin t × sin²t=sin³t.

f'(t)= 2 cos 2t cos³t- 3sint sin 2t + 3sin³t sin 2t

That's all there is to know for this lesson. Just keep in mind that it helps to approach these problems from outside inwards. For example, in the last problem we applied the product rule first and then we applied the chain rule for the respective component functions as we went along.

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