### Video Transcript

Find the first partial derivative of the function π of π₯, π¦ equals π₯π¦ plus one over π₯ plus π¦ with respect to π₯.

In this question, weβve been given a multivariable function in two variables, π₯ and π¦. And weβre being asked to find its first partial derivative with respect to π₯. Thatβs ππ ππ₯. And when weβre dealing with a function thatβs made up of more than one variable, we look to see how that function changes if we let just one of those variables change and hold all the others constant. In this case, since weβre finding the first partial derivative of our function with respect to π₯, weβre going to see what happens when the variable π₯ changes. And, therefore, we hold π¦ constant.

And this means we can differentiate our function with respect to π₯ and just imagine that π¦ is a constant. So letβs have a look at our function. Itβs π₯π¦ plus one over π₯ plus π¦. Itβs the quotient of two differentiable functions. And so letβs recall the quotient rule. This says that the derivative of the quotient of two differentiable functions, thatβs π’ over π£, is π£ times dπ’ by dπ₯ minus π’ times dπ£ by dπ₯ all over π£ squared. Now, of course, weβre dealing with partial derivatives. So letβs rephrase this slightly. Weβll let our function π§ be equal to π’ over π£.

Now, since weβre dealing with partial derivatives, letβs reword this slightly. Our function π is equal to the quotient of two multivariable functions. Thatβs π’ of π₯, π¦ over π£ of π₯, π¦. So we can say that the first partial derivative of π with respect to π₯ is equal to π£ of π₯, π¦ times the first partial derivative of π’ with respect to π₯ minus the function π’ times the first partial derivative of π£ with respect π₯ all over the function π£ squared.

Well, in this case, weβre going to let our function π’ of π₯, π¦ be equal to π₯π¦ plus one and π£ is π₯ plus π¦. Letβs begin by finding the first partial derivative of π’ with respect to π₯. Remember, weβre going to treat π¦ as a constant. We know that when we differentiate some constant times π₯ with respect to π₯, we get that constant. We also know that the derivative of a constant itself is zero. So the first partial derivative of the function π’ with respect to π₯ is π¦ plus zero or simply π¦.

Similarly, the first partial derivative of the function π£ with respect to π₯ is simply one. And thatβs because when we differentiate π₯ with respect to π₯, we get one. But the derivative of the constant π¦ is zero. And then ππ ππ₯ is equal to π£ times the first partial derivative of the function π’ with respect to π₯, so π₯ plus π¦ times π¦, minus π’ of π₯π¦ times ππ£ ππ₯, so π₯π¦ plus one times one. And thatβs all over the function of the denominator of our fraction squared, so π₯ plus π¦ all squared.

Weβre going to distribute the parentheses on the numerator of our fraction. And when we do, we get π₯π¦ plus π¦ squared minus π₯π¦ minus one. But of course π₯π¦ minus π₯π¦ is zero. So this simplifies to π¦ squared minus one over π₯ plus π¦ all squared. The first partial derivative of the function with respect to π₯ is π¦ squared minus one over π₯ plus π¦ squared.