### Video Transcript

Find the first derivative of the function π¦ is equal to four times π raised to the power seven π₯ over seven π₯ plus four.

Weβre asked to find the first derivative of a function π¦, which is a function of π₯. This means weβll need to differentiate π¦ with respect to π₯. Now, a function π¦ is a rational function of the form π’ over π£, where π’ and π£ are differentiable functions of π₯. There are a couple of different methods we could use to differentiate π¦. For example, since the denominator seven π₯ plus four is a simple polynomial, we could write our function as a product.

And we could use the general power rule to differentiate seven π₯ plus four to the negative one and then use the product rule. However, since our function is a fairly simple quotient, letβs use the quotient rule. And using the notation dπ by dπ₯ is equal to π prime, for a rational function, π of π₯ which is equal to π’ of π₯ over π£ of π₯, where π’ and π£ are differentiable functions of π₯, π prime of π₯ is equal to π£ of π₯ times π’ prime of π₯ minus π’ of π₯ times π£ prime of π₯ all over π£ squared.

Now, the function π¦, if we let π’ equal four π raised to the power seven π₯ where π is Eulerβs number and π£ equal to seven π₯ plus four, to use the quotient rule, we need to find π’ prime of π₯ and π£ prime of π₯. And to differentiate four π raised to the seven π₯ with respect to π₯, we can use the known result d by dπ₯ of π raised to the power ππ₯ is equal to ππ raised to the power ππ₯, where π is a constant. In our function π’, π is equal to seven. And bringing the seven down, dπ’ by dπ₯ is four times seven times π raised to the power seven π₯, which is 28π raised to the power seven π₯.

To find π£ prime of π₯, weβre going to use the power rule for differentiation, where we note that in π£, seven π₯ is actually seven π₯ raised to the power one. Remember that the power rule says for a function of the form π times π₯ raised to the power π, where π and π are constants, d by dπ₯ is equal to π times π times π₯ raised to the power π minus one. That is, we multiply by the exponent of π₯ and subtract one from the exponent. In our case, our exponent is one, so that the derivative of seven π₯ is seven times one times π₯ raised to the power one minus one. That is seven π₯ raised to the power zero. And we know that anything to the power zero is equal to one, so thatβs equal to seven.

The derivative of our constant four is equal to zero, since a constant has no π₯ dependence, so that π£ prime of π₯ is equal to seven. So now, using our results within the quotient rule, we have seven π₯ plus four which is π£ times 28 times π raised to the power seven π₯, which is π’ prime, minus four π raised to seven π₯, which is π’, times seven, which is π£ prime, all over π£ squared. Thatβs seven π₯ plus four squared. And since four times seven is 28, we can rewrite this as 28π raised to the seven π₯ times seven π₯ plus four minus 28π raised to the power seven π₯ all over seven π₯ plus four squared.

To simplify this, we can take the common factor of 28π raised to the power seven π₯ outside some parentheses, so that we have 28π raised to the power seven π₯ times seven π₯ plus four minus one all over seven π₯ plus four squared. And since four minus one is equal to three, inside our parentheses we have seven π₯ plus three. And if we now redistribute our parentheses, we have 28π raised to the power seven π₯ times seven π₯ plus 28π raised to the power seven π₯ times three.

The first derivative of the function π¦ is equal to four times π raised to the power seven π₯ over seven π₯ plus four is therefore 196π₯ times π raised to the power seven π₯ plus 84 times π raised to the power seven π₯ all over seven π₯ plus four squared.