### 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. Itβs 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, weβll 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, weβll
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₯. Letβs start with finding dπ’ by
dπ₯. Thatβs 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₯. Thatβs the derivative of the sin of
two π₯ to the fourth power plus four with respect to π₯. However, we canβt evaluate this
derivative directly since itβs not in a standard form, although we can notice that
this is the derivative of the composition of two functions. Weβre 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₯. Thatβs 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₯. Weβll 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. Weβre 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, weβve 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.