Video Transcript
If π¦ is equal to negative two times π₯ to the fourth power plus six times the sin of π₯ over two plus the cos of π by four, find the derivative of π¦ with respect to π₯.
We need to find the derivative of π¦ with respect to π₯, and we can see that π¦ is given as the sum of three terms. Itβs a polynomial term plus a trigonometric term plus another trigonometric term evaluated at a value of π₯. Since we know how to differentiate each of these three terms, we can find this by just differentiating each term separately. To start, remember, dπ¦ by dπ₯ is the derivative of π¦ with respect to π₯. In other words, this is equal to the derivative of negative two π₯ to the fourth power plus six times the sin of π₯ over two plus the cos of π by four with respect to π₯.
And now, instead of evaluating this derivative as a whole, weβll evaluate this derivative term by term. Letβs start with our first derivative of negative two π₯ to the fourth power with respect to π₯. This is the derivative of a polynomial, so we can do this by using the power rule for differentiation. We recall this tells us, for any real constants π and π, the derivative of ππ₯ to the πth power with respect to π₯ is equal to π times π times π₯ to the power of π minus one. We multiply by our exponent of π₯ and then reduce this exponent by one.
In our case, the exponent of π₯ is equal to four. So we need to multiply by this exponent of four and then reduce this exponent by one. This gives us negative two times four multiplied by π₯ to the power of four minus one. And of course, we can simplify this entire expression to negative eight π₯ cubed. To evaluate the derivative of six times the sin of π₯ over two with respect to π₯, we need to notice one small piece of manipulation. We can rewrite π₯ over two as one-half multiplied by π₯.
And in doing this, we can now see that this derivative is a standard trigonometric derivative result. We know for any real constants π and πΎ, the derivative of π times the sin of πΎπ₯ with respect to π₯ is equal to ππΎ times the cos of πΎπ₯. In our case, we can see that the coefficient of π₯ is equal to one-half, so weβll use this result with πΎ set to be equal to one-half. So by using our value of π equal to six and πΎ equal to one-half, we get six times one-half multiplied by the cos of one-half π₯. And then, of course, we can simplify and rewrite this as three times the cos of π₯ over two.
Now, all we have to do is evaluate the derivative of the cos of π by four with respect to π₯. But the cos of π by four is equal to a constant. In fact, we know itβs equal to the square root of two divided by two. And we know that constants donβt vary as our value of π₯ varies. In other words, the rate of change of this constant with respect to π₯ is equal to zero, so its derivative with respect to π₯ will also be equal to zero. And because this is equal to zero, we donβt need to include this. Therefore, given that π¦ is equal to negative two π₯ to the fourth power plus six times the sin of π₯ over two plus the cos of π by four, we were able to show that dπ¦ by dπ₯ is equal to negative eight π₯ cubed plus three times the cos of π₯ over two.