# Question Video: Differentiating Functions Involving Trigonometric Ratios Using the Product Rule and the Chain Rule Mathematics • Higher Education

If π¦ = 6π₯Β³ sin (2π₯β΄ + 4 ), find dπ¦/dπ₯.

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### Video Transcript

If π¦ is equal to six π₯ cubed times the sin of two π₯ to the fourth power plus four, find dπ¦ by dπ₯.

The question wants us to find the first derivative of π¦ with respect to π₯, and we can see that π¦ is the product of two functions. Itβs the product of six π₯ cubed and the sin of two π₯ to the fourth power plus four. And to find the derivative of the product of two functions, weβll want to use the product rule for differentiation. The product rule tells us if we have π¦ is the product of two functions π’ of π₯ and π£ of π₯, then dπ¦ by dπ₯ is equal to π£ of π₯ times dπ’ by dπ₯ plus π’ of π₯ times dπ£ by dπ₯. So to use the product rule, weβll set our function π’ of π₯ to be six π₯ cubed and our function π£ of π₯ to be the sin of two π₯ to the fourth power plus four.

So to use the product rule, we need to find expressions for dπ’ by dπ₯ and dπ£ by dπ₯. Letβs start with finding dπ’ by dπ₯. Thatβs the derivative of six π₯ cubed with respect to π₯. And we can evaluate this by using the power rule for differentiation. We multiply by the exponent of π₯ and then reduce this exponent by one. That gives us 18π₯ squared. We now want to find an expression for dπ£ by dπ₯. Thatβs the derivative of the sin of two π₯ to the fourth power plus four with respect to π₯. However, we canβt evaluate this derivative directly since itβs not in a standard form, although we can notice that this is the derivative of the composition of two functions. Weβre taking the sin of two π₯ to the fourth power plus four.

And we know how to differentiate the composition of two functions by using the chain rule. We recall, the chain rule tells us if we have π£ is a function of π€ and π€ is a function of π₯, then dπ£ by dπ₯ is equal to dπ£ by dπ€ times dπ€ by dπ₯. In our case, we have that π£ is the composition of the sine function and a polynomial. So if we set π€ to be our inner function two π₯ to the fourth power plus four, then π£ of π₯ is equal to the sin of π€. So π£ is a function of π€, and π€ in turn is a function of π₯. This means we can evaluate the derivative of π£ with respect to π₯ by using the chain rule. We get that this is equal to dπ£ by dπ€ times dπ€ by dπ₯.

We can find expressions for both of these derivatives. First, dπ£ by dπ€ is the derivative of π£ with respect to π€. And we know that π£ is the sin of π€. So dπ£ by dπ€ is equal to the derivative of the sin of π€ with respect to π€. We can do something similar for dπ€ by dπ₯. Thatβs the derivative of π€ with respect to π₯. And π€ is two π₯ to the fourth power plus four. We can now evaluate both of these derivatives. First, we know the derivative of the sin of π€ with respect to π€ is equal to the cos of π€. Next, we can evaluate the derivative of two π₯ to the fourth power plus four with respect to π₯ by using the power rule for differentiation. We multiply by the exponent of π₯ and reduce this exponent by one. This gives us eight π₯ cubed. And of course, the derivative of the constant four is just equal to zero.

Finally, since this is an expression for dπ£ by dπ₯, we want our answer to be in terms of π₯. Weβll do this by using our substitution π€ is equal to two π₯ to the fourth power plus four. This gives us that dπ£ by dπ₯ is equal to eight π₯ cubed times the cos of two π₯ to the fourth power plus four. Weβre now ready to find an expression for dπ¦ by dπ₯. By the product rule, this is equal to π£ of π₯ times dπ’ by dπ₯ plus π’ of π₯ times dπ£ by dπ₯. Substituting in our expressions for π’ of π₯, π£ of π₯, dπ’ by dπ₯, and dπ£ by dπ₯, we get that dπ¦ by dπ₯ is equal to the sin of two π₯ to the fourth power plus four times 18π₯ squared plus six π₯ cubed multiplied by eight π₯ cubed times the cos of two π₯ to the fourth power plus four.

And by simplifying and rearranging this expression, weβve shown if π¦ is equal to six π₯ cubed times the sin of two π₯ to the fourth power plus four. Then, dπ¦ by dπ₯ is equal to 48π₯ to the sixth power times the cos of two π₯ to the fourth power plus four plus 18π₯ squared times the sin of two π₯ to the fourth power plus four.