### Video Transcript

Find the slope of the tangent to the curve five π₯ over two π¦ minus two π¦ over π₯ equals negative four at the point two, five.

We should recall that to find the slope of the tangent to a curve, we need to evaluate the derivative of that function at a given point. Now, weβre most used to evaluating derivatives of explicitly defined functions of π₯. In this case though, weβre looking at a function that is defined implicitly. And so, weβre going to use implicit differentiation to evaluate the derivative.

This is just a special case of the chain rule. And it says that the derivative of some function in π¦ with respect to π₯ is equal to the derivative of that function with respect to π¦ times dπ¦ by dπ₯. And so, letβs look at the equation of our curve. Itβs five π₯ over two π¦ minus two π¦ over π₯ equals negative four. And weβll just begin by differentiating each side of this equation with respect to π₯.

We recall that the derivative of the sum or difference of two or more functions is equal to the sum or difference of their respective derivatives. And so, weβre going to differentiate five π₯ over two π¦, two π¦ over π₯, and negative four. Well, in fact, the derivative of a constant is zero. So, the right-hand side here becomes zero. But how do we differentiate five π₯ over two π¦ and two π¦ over π₯?

Well, we use the quotient rule. This says that the derivative of the quotient of two differentiable functions π’ and π£ is π£ times dπ’ by dπ₯ minus π’ times dπ£ by dπ₯ all over π£ squared. Letβs begin by looking at the derivative of five π₯ over two π¦. Weβll let π’ be equal to five π₯, thatβs the numerator, and π£ be equal to two π¦, thatβs the denominator. Then, dπ’ by dπ₯, the first derivative of five π₯, is simply five. But what about dπ£ by dπ₯?

Well, according to the earlier formula, which was just a special case of the chain rule, we multiply the derivative of the function with respect to π¦, so thatβs two, by dπ¦ by dπ₯. And so, dπ£ by dπ₯ is simply two dπ¦ by dπ₯. Then, the derivative of five π₯ over two π¦ is π£ times dπ’ by dπ₯, so thatβs two π¦ times five, minus π’ times dπ£ by dπ₯, so thatβs five π₯ times two dπ¦ by dπ₯, all over π£ squared, so two π¦ squared. If we simplify a little, we get 10π¦ minus 10π₯ dπ¦ by dπ₯ all over four π¦ squared. And then, we should see that we can divide through by a common factor of two. And we found the derivative of the first part of our expression on the left-hand side. Itβs five π¦ minus five π₯ dπ¦ by dπ₯ all over π¦ squared.

Weβre now going to repeat this process for the derivative of two π¦ over π₯. This time, we let π’ be equal to two π¦ and π£ be equal to π₯. Of course, we know that the derivative of π₯ with respect to π₯ is just one. We already evaluated the derivative of two π¦. But as a reminder, itβs the derivative of two π¦ with respect to π¦ which is two times dπ¦ by dπ₯.

Letβs substitute everything we have into our formula for the quotient rule. And we get two π₯ dπ¦ by dπ₯ minus two π¦ over π₯ squared. And so, we find that when we differentiate both sides of our equation with respect to π₯, we get five π¦ minus five π₯ dπ¦ by dπ₯ over π¦ squared minus two π₯ dπ¦ by dπ₯ minus two π¦ over π₯ squared equals zero. But what on earth are we going to do with this?

Remember, weβre trying to find the slope of the tangent, so we need to evaluate dπ¦ by dπ₯ at our point two, five. The problem is, we havenβt currently got an expression for dπ¦ by dπ₯. So, letβs see if we can find a way to simplify a little. With our two fractions, weβre going to divide each bit of the numerator by the denominator. So, we divide five π¦ by π¦ squared to get five over π¦. And we divide negative five π₯ dπ¦ by dπ₯ by π¦ squared to get negative five π₯ over π¦ squared dπ¦ by dπ₯.

Similarly, we divide negative two π₯ dπ¦ by dπ₯ by π₯ squared to get negative two over π₯ dπ¦ by dπ₯ and negative negative two π¦ by π₯ squared to get positive two π¦ over π₯ squared. We now see that we can factor dπ¦ by dπ₯. And with those terms, we get dπ¦ by dπ₯ times negative five π₯ over π¦ squared minus two over π₯. At the same time, letβs subtract five over π¦ and two π¦ over π₯ squared from both sides. And so, we have this given expression.

We now should be able to see that we can divide through by negative five π₯ over π¦ squared minus two over π₯. We might also notice that every single term has a factor of negative one. So, weβre going to divide through by negative one at the same time. And so, we have a rather nasty-looking expression for the first derivative of π¦ with respect to π₯. Itβs five over π¦ plus two π¦ over π₯ squared all over five π₯ over π¦ squared plus two over π₯.

Normally, we might look to simplify this a little, but remember weβre trying to find the slope of the tangent at a given point. In fact, itβs the point where π₯ equals two and π¦ equals five. So, letβs substitute π₯ equals two and π¦ equals five into our expression for the derivative.

The numerator becomes five over five plus two times five over two squared, which simplifies to one plus five over two. Then, our denominator is five times two over five squared plus two over two, which is two-fifths plus one. One plus five over two is seven over two. And two-fifths plus one is seven-fifths. So, we have seven over two divided by seven-fifths.

And then, we might recall that to divide by a fraction, we multiply the reciprocal of that same fraction. So, itβs seven over two times five-sevenths. And then the sevens cancel, to get five over two. And so, weβre done. We found the slope of the tangent to our curve at the point two, five. Itβs five over two, or five-halves.