### Video Transcript

Differentiate the function π of π₯ which is equal to five π₯ squared minus one over seven π₯ plus six.

The first thing that stands out to us in this question is that we have the quotient of two differentiable functions, the function five π₯ squared minus one in the numerator and the function seven π₯ plus six in the denominator. To differentiate functions of this type, we need to use the quotient rule. This tells us that, for two differentiable functions π’ and π£, the derivative of their quotient, π’ over π£, is equal to π£ multiplied by π’ prime minus π’ multiplied by π£ prime all over π£ squared.

Letβs see how to apply the quotient rule in this question. Remember, π’ is the function in the numerator of the quotient and π£ is the function in the denominator. So weβre going to let π’ equal the function in the numerator, five π₯ squared minus one, and π£ equal the function in the denominator, seven π₯ plus six. We then need to find each of their individual derivatives with respect to π₯, which we can do using the power rule of differentiation. π’ prime or dπ’ by dπ₯ is equal to five multiplied by two π₯, which is 10π₯. And remember, the derivative of a constant, in this case negative one, is just zero. π£ prime or dπ£ by dπ₯ is equal to seven. So we have both the derivatives of π’ and π£ with respect to π₯.

Weβre now ready to substitute into the quotient rule to find π prime of π₯. Itβs equal to π£ multiplied by π’ prime. Thatβs seven π₯ plus six multiplied by 10π₯ minus π’ multiplied by π£ prime. Thatβs five π₯ squared minus one multiplied by seven. And this is all divided by π£ squared. Thatβs seven π₯ plus six all squared. We now just need to distribute the parentheses in the numerator and simplify the result. We obtain 70π₯ squared plus 60π₯ minus 35π₯ squared plus seven all over seven π₯ plus six all squared. And then, collecting like terms in the numerator, we have 35π₯ squared plus 60π₯ plus seven all over seven π₯ plus six all squared.

By applying the quotient rule then, weβve differentiated our function π of π₯. Notice that it would also be possible to answer this question using the product rule. If we were to express our function π of π₯ as five π₯ squared minus one multiplied by seven π₯ plus six to the power of negative one. And if youβre interested, you could answer the question using the product rule and confirm it does indeed give the same result.