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If π¦ is equal to six π₯ cubed times the sin of two π₯ to the fourth power plus four, find dπ¦ by dπ₯.
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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.
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Itβs the product of six π₯ cubed and the sin of two π₯ to the fourth power plus four.
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And to find the derivative of the product of two functions, weβll want to use the product rule for differentiation.
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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₯.
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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.
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So to use the product rule, we need to find expressions for dπ’ by dπ₯ and dπ£ by dπ₯.
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Letβs start with finding dπ’ by dπ₯.
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Thatβs the derivative of six π₯ cubed with respect to π₯.
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And we can evaluate this by using the power rule for differentiation.
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We multiply by the exponent of π₯ and then reduce this exponent by one.
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That gives us 18π₯ squared.
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We now want to find an expression for dπ£ by dπ₯.
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Thatβs the derivative of the sin of two π₯ to the fourth power plus four with respect to π₯.
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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.
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Weβre taking the sin of two π₯ to the fourth power plus four.
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And we know how to differentiate the composition of two functions by using the chain rule.
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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₯.
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In our case, we have that π£ is the composition of the sine function and a polynomial.
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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€.
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So π£ is a function of π€, and π€ in turn is a function of π₯.
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This means we can evaluate the derivative of π£ with respect to π₯ by using the chain rule.
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We get that this is equal to dπ£ by dπ€ times dπ€ by dπ₯.
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We can find expressions for both of these derivatives.
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First, dπ£ by dπ€ is the derivative of π£ with respect to π€.
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And we know that π£ is the sin of π€.
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So dπ£ by dπ€ is equal to the derivative of the sin of π€ with respect to π€.
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We can do something similar for dπ€ by dπ₯.
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Thatβs the derivative of π€ with respect to π₯.
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And π€ is two π₯ to the fourth power plus four.
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We can now evaluate both of these derivatives.
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First, we know the derivative of the sin of π€ with respect to π€ is equal to the cos of π€.
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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.
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We multiply by the exponent of π₯ and reduce this exponent by one.
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This gives us eight π₯ cubed.
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And of course, the derivative of the constant four is just equal to zero.
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Finally, since this is an expression for dπ£ by dπ₯, we want our answer to be in terms of π₯.
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Weβll do this by using our substitution π€ is equal to two π₯ to the fourth power plus four.
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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.
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Weβre now ready to find an expression for dπ¦ by dπ₯.
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By the product rule, this is equal to π£ of π₯ times dπ’ by dπ₯ plus π’ of π₯ times dπ£ by dπ₯.
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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.
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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.
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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.