### Video Transcript

Evaluate the first derivative of the function π¦ equals four π₯ multiplied by the square root of π₯ plus six at the point with coordinates negative two, negative 16.

Letβs have a look at this function π¦. We can see that it consists of a product of terms. Itβs equal to four π₯ multiplied by the square root of π₯ plus six. And as the exponent in that second factor is a half, we canβt simply distribute the parentheses in order to turn this into a polynomial function.

Instead, weβre going to need to use the product rule of differentiation. This tells us that, for two differentiable functions π’ and π£, the derivative with respect to π₯ of their product π’π£ is equal to π’ times dπ£ by dπ₯ plus π£ times dπ’ by dπ₯. We multiply each function by the derivative of the other and add these together. So we define π’ to be one of our factors. Weβll let π’ equal four π₯. And we define π£ to be the other, the square root of π₯ plus six, which weβve already said can be expressed as π₯ plus six to the power of one-half.

We then need to find each of their individual derivatives. dπ’ by dπ₯ is straight forward. We can use the power rule of differentiation. We can think of four π₯ as four π₯ to the first power. So in order to differentiate, we multiply by the exponent of one and then reduce the exponent by one, giving four multiplied by one multiplied by π₯ to the power of zero. π₯ to the power of zero is simply one. So we have four multiplied by one multiplied by one, which is just four.

Now, dπ£ by dπ₯ is a little bit more complicated because we see that π£ is actually a composite function. Itβs a function of a function. We take the function π₯ plus six and then find its square root. In order to find this derivative then, weβre going to need to use the chain rule. The chain rule tells us that if π£ is a function of some other function π§, which is itself a function of π₯, then the derivative of π£ with respect to π₯ is equal to the derivative of π£ with respect to π§ multiplied by the derivative of π§ with respect to π₯. Iβve changed the letters from what you might usually see here. Often, we see it written in terms of π’, because we already have π’ defined to be another function in this question.

So, we can introduce this new variable π§. Weβll let π§ equal π₯ plus six. Thatβs the inner part of our composite function. π£ will then be equal to π§ to the power of one-half. So π£ is a function of π§. And π§ is a function of π₯. We then find each of their individual derivatives, both of which we can do using the power rule of differentiation. The derivative of π§ with respect to π₯ is simply one. And the derivative of π£ with respect to π§ is a half multiplied by π§ to the power of negative one-half.

Applying the chain rule then, we have that dπ£ by dπ₯ is equal to dπ£ by dπ§ multiplied by dπ§ by dπ₯. Thatβs one-half π§ to the power of negative one-half multiplied by one. But of course, multiplying by one has no effect. We can rewrite this as one over two π§ to the power of a half. And finally, we must remember to undo our substitution. So we need to give dπ£ by dπ₯ in terms of π₯. We find then that dπ£ by dπ₯ is equal to one over two multiplied by π₯ plus six to the power of a half. Weβve replaced π§ with π₯ plus six.

Now, you may also be able to find this derivative using the general power rule of differentiation rather than the chain rule. Which tells us that if we have some function π of π₯ to the πth power, then its derivative with respect to π₯ is equal to π multiplied by π prime of π₯ multiplied by π of π₯ to the power of π minus one. Itβs a lot like the power rule of differentiation. But we have this extra factor π prime of π₯. Thatβs the derivative of our function π of π₯.

For our function π£ then, the function π of π₯ would be π₯ plus six. So applying the general power rule, weβd find that dπ£ by dπ₯ is equal to π, thatβs one-half, multiplied by π prime of π₯, thatβs the derivative of π₯ plus six, which is simply one, multiplied by π of π₯ to the power of π minus one. Thatβs π₯ plus six to the power of negative one-half. This can of course be written as one over two multiplied by π₯ plus six to the power of a half, which is what weβve already found for our derivative of π£ with respect to π₯ using the chain rule. So either of these two methods would be absolutely fine.

In either case, now that we have both dπ’ by dπ₯ and dπ£ by dπ₯, we need to return to our formula for the product rule. We have that dπ¦ by dπ₯ is equal to π’ times dπ£ by dπ₯. Thatβs four π₯ multiplied by one over two π₯ plus six to the power of a half, which we can write as four π₯ over two π₯ plus six to the power of a half. Plus π£ times dπ’ by dπ₯. Thatβs four multiplied by π₯ plus six to the power of one-half. We can simplify by a factor of two in our first term. And we may also find it helpful to rewrite those exponents of one-half as square roots. So we have that dπ¦ by dπ₯ is equal to two π₯ over the square root of π₯ plus six plus four multiplied by the square root of π₯ plus six.

Finally, we recall that we werenβt asked to give a general expression for the first derivative. We were asked to evaluate it at a particular point. So the final step is to substitute the π₯-coordinate of this point, thatβs negative two, into our expression for dπ¦ by dπ₯. Doing so gives dπ¦ by dπ₯ is equal to two multiplied by negative two over the square root of negative two plus six plus four multiplied by the square root of negative two plus six. That simplifies to negative four over the square root of four plus four multiplied by the square root of four. Now, the square root of four is of course two. So we have negative four over two plus four multiplied by two. Thatβs negative two plus eight, which is equal to six.

So in this question, weβve had to use more than one rule of differentiation, either the product rule and the formal chain rule or the product rule and the general power rule. But in doing so, weβve found that the first derivative of the function π¦ equals four π₯ multiplied by the square root of π₯ plus six at the point with coordinates negative two, negative 16 is six.