Video: Finding the Third Derivative of Trigonometric Functions

Determine the third derivative of the function 𝑦 = βˆ’6 cos 2π‘₯.

02:52

Video Transcript

Determine the third derivative of the function 𝑦 is equal to negative six times the cos of two π‘₯.

We’re given 𝑦 which is a trigonometric function in π‘₯. And we’re asked to find the third derivative of 𝑦. This will be the third derivative of 𝑦 with respect to π‘₯. So we’ll start by finding the first derivative of 𝑦 with respect to π‘₯. That’s the derivative of negative six cos of two π‘₯ with respect to π‘₯. And to do this, we can recall one of our standard trigonometric derivative results. For any real constant π‘Ž, the derivative of negative cos of π‘Žπ‘₯ with respect to π‘₯ is equal to π‘Ž times the sin of π‘Žπ‘₯. And we can see our derivative is not exactly in this form. We’ll take the constant factor of six outside of our derivative.

So this gives us the following expression, and we can see this is exactly in this form with our value of π‘Ž set to be equal to two. So by setting our value of π‘Ž equal to two, we can differentiate negative the cos of two π‘₯ with respect to π‘₯ to get two times the sin of two π‘₯. Then we can simplify our coefficient in this expression. Six times two is equal to 12. So we’ve shown d𝑦 by dπ‘₯ is equal to 12 sin of two π‘₯. But remember, we need to find an expression for the third derivative of 𝑦 with respect to π‘₯. We need to differentiate this again. We differentiate to find d two 𝑦 by dπ‘₯ squared is equal to the derivative of 12 sin of two π‘₯ with respect to π‘₯. And once again, we’ll simplify this expression. We’ll take the constant factor of 12 outside of our derivative.

So to find our expression for d two 𝑦 by dπ‘₯ squared, we need to differentiate the sin of two π‘₯ with respect to π‘₯. And once again, we can do this by using a standard trigonometric derivative result. For any real constant π‘Ž, the derivative of the sin of π‘Žπ‘₯ with respect to π‘₯ is equal to π‘Ž times the cos of π‘Žπ‘₯. Once again, we can see the value of our constant π‘Ž is equal to two. So by using π‘Ž is equal to two, we get d two 𝑦 by dπ‘₯ squared is equal to 12 times two cos of two π‘₯. And we can simplify our coefficient. 12 times two is equal to 24. So we’ve shown the second derivative of 𝑦 with respect to π‘₯ is equal to 24 times the cos of two π‘₯.

To find our third derivative of 𝑦 with respect to π‘₯, we just need to differentiate this with respect to π‘₯ one more time. So we need to differentiate 24 cos of two π‘₯ with respect to π‘₯. And we can do this by using a trigonometric derivative result. For any real constant π‘Ž, the derivative of the cos of π‘Žπ‘₯ with respect to π‘₯ is equal to negative π‘Ž times the sin of π‘Žπ‘₯. Just as we did before, we’ll take the constant factor of 24 outside of our derivative and then evaluate our derivative by setting the value of π‘Ž equal to two. Doing this gives us d three 𝑦 by dπ‘₯ cubed is equal to 24 times negative two sin of two π‘₯.

And finally, by simplifying our coefficient, we get negative 48 times the sin of two π‘₯, which is our final answer. Therefore, we were able to show if 𝑦 is equal to negative six times the cos of two π‘₯, then the third derivative of 𝑦 with respect to π‘₯ will be equal to negative 48 times the sin of two π‘₯.

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