# Question Video: Differentiating a Combination of Trigonometric Functions Mathematics • Higher Education

If π¦ = 6 cos 4π₯ + 2 sin 2π₯, find dπ¦/dπ₯.

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### Video Transcript

If π¦ equals six cos four π₯ plus two sin two π₯, find dπ¦ by dπ₯.

In this question, weβre asked to find the derivative of a sum of two trigonometric functions. So, we need to recall the rules for differentiating these. Our most basic rules tell us that the derivative of sin π₯ is cos π₯, the derivative of cos π₯ is negative sin π₯, the derivative of negative sin π₯ is negative cos π₯, and the derivative of negative cos π₯ is sin π₯. Although, we must remember that these rules are only true if the angle is measured in radians.

We can see, though, that the arguments in our function are not just π₯; we have four π₯ in the first term, and we have two π₯ in the second. So, we also need to recall how to differentiate sine and cosine functions of this type. We can quote further standard results. For a constant π, the derivative with respect to π₯ of sin ππ₯ is π cos π₯. And the derivative with respect to π₯ cos ππ₯ is negative π sin ππ₯. We just have an extra factor of π in our derivatives. These results could be proved using the chain rule if we wish.

Now, notice that we also have multiplicative constants in front of each term. We have a six in the first term and a two in the second. But we know that multiplying by a constant just means that the derivative will also be multiplied by the same constant. So, we can say that the derivative of π¦ with respect to π₯ will be equal to six multiplied by the derivative of cos four π₯ with respect to π₯ plus two multiplied by the derivative of sin two π₯ with respect to π₯. And we can use our standard results to find each of these derivatives.

Applying the second rule for the derivative with respect to π₯ of cos ππ₯, we have that the derivative with respect to π₯ of cos four π₯ is negative four sin four π₯. And then, applying our first rule for the derivative of sin ππ₯, we have that the derivative with respect to π₯ of sin two π₯ is two cos two π₯. So, dπ¦ by dπ₯ is equal to six multiplied by negative four sin four π₯ plus two multiplied by two cos two π₯. Simplifying the constants, and we have our answer. dπ¦ by dπ₯ is equal to negative 24 sin four π₯ plus four cos two π₯. And again, remember that this result is only true when π₯ is measured in radians.

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