### Video Transcript

Find the third derivative of the function π¦ is equal to 44π₯ times the sin of two π₯.

The question is asking us to find the third derivative of our function. And since we want to find the third derivative of our function in terms of π₯, we need to differentiate this three times with respect to π₯. Letβs start by finding the first derivative of π¦ with respect to π₯. Thatβs the derivative of 44π₯ times the sin of two π₯ with respect to π₯. And we can see this is the derivative of the product of two functions. Itβs the product of 44π₯ and the sin of two π₯. So weβll differentiate this by using the product rule.

The product rule tells us the derivative of the product of two functions π’ and π£ with respect to π₯ is equal to π£ times dπ’ by dπ₯ plus π’ times dπ£ by dπ₯. So to evaluate our derivative, weβll set π’ to be 44π₯ and π£ to be the sin of two π₯. To apply the product rule, we need to find expressions for dπ’ by dπ₯ and dπ£ by dπ₯. Letβs start with dπ’ by dπ₯. Thatβs the derivative of 44π₯ with respect to π₯. And this is a linear function in terms of π₯. So its derivative is just the coefficient of π₯, which in this case is 44.

Letβs now find an expression for dπ£ by dπ₯. Thatβs the derivative of the sin of two π₯ with respect to π₯. And we know how to differentiate this using one of our standard trigonometric derivative results. For any real constant π, the derivative of the sin of ππ₯ with respect to π₯ is equal to π times the cos of ππ₯. Applying this, we get dπ£ by dπ₯ is equal to two times the cos of two π₯. Weβre now ready to apply the product rule to find the dπ¦ by dπ₯. Itβs equal to π£ times dπ’ by dπ₯ plus π’ times dπ£ by dπ₯.

Substituting in our expressions for π’, π£, dπ’ by dπ₯, and dπ£ by dπ₯, we get the sin of two π₯ times 44 plus 44π₯ times two times the cos of two π₯. Then we just simplify and rearrange this expression. We get 44 sin of two π₯ plus 88π₯ cos of two π₯. Remember, we want to find the third derivative of π¦ with respect to π₯ by differentiating our expression three times. So letβs now find our second derivative of π¦ with respect to π₯. Thatβs the derivative of our first derivative of π₯ with respect to π₯. We get the following expression. We see we can differentiate the first term by using our derivative rule for trigonometric functions.

However, the second term we need to differentiate is the product of two functions. So weβre going to need to use the product rule again. This time, we want π’ to be 88π₯ and we want π£ to be the cos of two π₯. We need to find expressions for dπ’ by dπ₯ and dπ£ by dπ₯. Letβs start with dπ’ by dπ₯. Thatβs the derivative of 88π₯ with respect to π₯. And again, this is a linear function. So its derivative is the coefficient of π₯, which in this case is 88.

Letβs now find an expression for dπ£ by dπ₯. Thatβs the derivative of the cos of two π₯ with respect to π₯. Again, we can evaluate this derivative by using one of our standard trigonometric derivative results. For any constant π, the derivative of the cos of ππ₯ with respect to π₯ is equal to negative π times the sin of ππ₯. Applying this gives us dπ£ by dπ₯ is equal to negative two times the sin of two π₯. Weβre now ready to find an expression for the second derivative of π¦ with respect to π₯.

First, weβll differentiate our first term, 44 times the sin of two π₯, to get 88 times the cos of two π₯. Next, we want to differentiate our second term by using the product rule. Itβs equal to π£ times dπ’ by dπ₯ plus π’ times dπ£ by dπ₯. Substituting in our expressions for π’, π£, dπ’ by dπ₯, and dπ£ by dπ₯. We get that the second derivative of π¦ with respect to π₯ is equal to 88 times the cos of two π₯ plus the cos of two π₯ times 88 plus 88π₯ times negative two times the sin of two π₯. And we can then simplify this expression to get 176 times the cos of two π₯ minus 176π₯ sin two π₯.

The last thing we need to do is differentiate this expression to find our third derivative of π¦ with respect to π₯. So letβs clear some space so we can find our third derivative of π¦ with respect to π₯. Thatβs the derivative of 176 cos of two π₯ minus 176π₯ sin of two π₯ with respect to π₯.

Again, we want to evaluate this derivative term by term. We can evaluate our first derivative by using our trigonometric derivative result. To evaluate our second derivative, we again notice itβs the product of two functions. So to evaluate the derivative of our second term, weβll use the product rule. Weβll set π’ to be 176π₯ and π£ to be the sin of two π₯. To use the product rule, we need expressions for dπ’ by dπ₯ and dπ£ by dπ₯. Letβs start with dπ’ by dπ₯. Thatβs the derivative of 176π₯ with respect to π₯. And again, this is a linear function. So its derivative is just the coefficient of π₯ which in this case is 176.

We now want to find an expression for dπ£ by dπ₯. Thatβs the derivative of the sin of two π₯ with respect to π₯. And we can do this by using our other standard trigonometric derivative result. Applying this gives us dπ£ by dπ₯ is equal to two times the cos of two π₯. Weβre now ready to find an expression for d three π¦ by dπ₯ cubed. First, letβs differentiate the first term. Differentiating this gives us negative two times 176 times the sin of two π₯, which is negative 352 sin of two π₯. We then want to subtract the derivative of our second term, which by the product rule is π£ times dπ’ by dπ₯ plus π’ times dπ£ by dπ₯.

Substituting in our expressions for π’, π£, dπ’ by dπ₯, and dπ£ by dπ₯, we have the third derivative of π¦ with respect to π₯ is equal to negative 352 sin of two π₯ minus the sin of two π₯ times 176 plus 176π₯ times two times the cos of two π₯. Then we just simplify and rearrange this expression. We get negative 528 sin of two π₯ minus 352π₯ times the cos of two π₯. And finally, weβll reorder our terms so that our powers of π₯ are descending.

Therefore, weβve shown the function π¦ is equal to 44π₯ times the sin of two π₯ has a third derivative equal to negative 352π₯ cos of two π₯ minus 528 sin two π₯.