### Video Transcript

Suppose π of π₯ is equal to π₯ plus π over π₯ minus π and π prime of two is equal to negative two. Determine π.

So, in this question, weβve been given a function π of π₯ in terms of π₯ and an unknown quantity π. And weβve also been told the value of the first derivative, thatβs π prime of this function, when π₯ is equal to two. We need to use this information in order to determine the value of π. Well, first, weβre going to need to find a general expression for the first derivative of π of π₯.

We note that π of π₯ is a quotient of two differentiable functions, π₯ plus π and π₯ minus π. And therefore, in order to differentiate π of π₯, weβre going to need to apply the quotient rule. The quotient rule tells us that for two differentiable functions, π’ and π£, the derivative with respect to π₯ of their quotient, π’ over π£, is equal to π£ times dπ’ by dπ₯ minus π’ times dπ£ by dπ₯ all over π£ squared. We, therefore, define π’ to be the function in the numerator, thatβs π₯ plus π, and π£ to be the function in the denominator, thatβs π₯ minus π.

We now need to find each of their individual derivatives, which we can do by recalling the power rule of differentiation. This tells us that if weβre differentiating a power of π₯ with respect to π₯, then we multiply by the power and decrease the power by one. π₯ can be thought of as π₯ to the power of one. And therefore, its derivative with respect to π₯ is one multiplied by π₯ to the power of zero, which is just one. The derivative of a constant is zero. So, the derivatives of both π’ and π£ are both equal to one.

Next, we need to substitute the expressions for π’, π£, and their derivatives into the quotient rule. We have that π prime of π₯ is equal to π£ times dπ’ by dπ₯, thatβs π₯ minus π multiplied by one, minus π’ times dπ£ by dπ₯, thatβs π₯ plus π multiplied by one, all over π£ squared, thatβs π₯ minus π all squared. Distributing the parentheses in the numerator gives π₯ minus π minus π₯ minus π over π₯ minus π all squared, which simplifies to negative two π over π₯ minus π all squared.

Now, we can return to the other piece of information given in the question. Which tells us that if we evaluate this first derivative when π₯ is equal to two, we get negative two. So, substituting negative two for the value of π prime of π₯ and positive two for the value of π₯ gives the equation negative two equals negative two π over two minus π all squared.

We need to solve this equation for π. First, we divide both sides by negative two to give one equals π over two minus π all squared. Then, we multiply both sides by two minus π all squared, thatβs the denominator on the right-hand side, to give two minus π all squared is equal to π. Distributing the parentheses on the left-hand side gives four minus four π plus π squared is equal to π. And finally, subtracting π from each side gives π squared minus five π plus four is equal to zero. We now see that we have a quadratic equation in π, which we can solve by factorising.

The coefficient of π squared is just one, which means that in each of our brackets, the first term will be π. Weβre then looking for two numbers whose sum is the coefficient of π, thatβs negative five, and whose product is the constant term, which is positive four. Those two numbers are negative one and negative four. So, our quadratic factorises as π minus one multiplied by π minus four is equal to zero.

To solve, we recall that if the product of two things is equal to zero, then at least one of them must themselves be equal to zero. So, we set each bracket in turn equal to zero and solve the resulting linear equation. π minus one equals zero yields π equals one. And π minus four equals zero yields π equals four. So, we find that there are two possible values of π, π is equal to four or π is equal to one.

Remember that the key principle we applied in this question was the quotient rule. Which tells us that for two differentiable functions π’ and π£, the derivative with respect to π₯ of their quotient, π’ over π£, is equal to π£ times dπ’ by dπ₯ minus π’ times dπ£ by dπ₯ all over π£ squared.