Video Transcript
Find the first partial derivative of the function 𝑓 of 𝑥, 𝑦 equals 𝑥𝑦 plus one over 𝑥 plus 𝑦 with respect to 𝑥.
In this question, we’ve been given a multivariable function in two variables, 𝑥 and 𝑦. And we’re being asked to find its first partial derivative with respect to 𝑥. That’s 𝜕𝑓 𝜕𝑥. And when we’re dealing with a function that’s 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 we’re finding the first partial derivative of our function with respect to 𝑥, we’re going to see what happens when the variable 𝑥 changes. And, therefore, we hold 𝑦 constant.
And this means we can differentiate our function with respect to 𝑥 and just imagine that 𝑦 is a constant. So let’s have a look at our function. It’s 𝑥𝑦 plus one over 𝑥 plus 𝑦. It’s the quotient of two differentiable functions. And so let’s recall the quotient rule. This says that the derivative of the quotient of two differentiable functions, that’s 𝑢 over 𝑣, is 𝑣 times d𝑢 by d𝑥 minus 𝑢 times d𝑣 by d𝑥 all over 𝑣 squared. Now, of course, we’re dealing with partial derivatives. So let’s rephrase this slightly. We’ll let our function 𝑧 be equal to 𝑢 over 𝑣.
Now, since we’re dealing with partial derivatives, let’s reword this slightly. Our function 𝑓 is equal to the quotient of two multivariable functions. That’s 𝑢 of 𝑥, 𝑦 over 𝑣 of 𝑥, 𝑦. So we can say that the first partial derivative of 𝑓 with respect to 𝑥 is equal to 𝑣 of 𝑥, 𝑦 times the first partial derivative of 𝑢 with respect to 𝑥 minus the function 𝑢 times the first partial derivative of 𝑣 with respect 𝑥 all over the function 𝑣 squared.
Well, in this case, we’re going to let our function 𝑢 of 𝑥, 𝑦 be equal to 𝑥𝑦 plus one and 𝑣 is 𝑥 plus 𝑦. Let’s begin by finding the first partial derivative of 𝑢 with respect to 𝑥. Remember, we’re going to treat 𝑦 as a constant. We know that when we differentiate some constant times 𝑥 with respect to 𝑥, we get that constant. We also know that the derivative of a constant itself is zero. So the first partial derivative of the function 𝑢 with respect to 𝑥 is 𝑦 plus zero or simply 𝑦.
Similarly, the first partial derivative of the function 𝑣 with respect to 𝑥 is simply one. And that’s because when we differentiate 𝑥 with respect to 𝑥, we get one. But the derivative of the constant 𝑦 is zero. And then 𝜕𝑓 𝜕𝑥 is equal to 𝑣 times the first partial derivative of the function 𝑢 with respect to 𝑥, so 𝑥 plus 𝑦 times 𝑦, minus 𝑢 of 𝑥𝑦 times 𝜕𝑣 𝜕𝑥, so 𝑥𝑦 plus one times one. And that’s all over the function of the denominator of our fraction squared, so 𝑥 plus 𝑦 all squared.
We’re going to distribute the parentheses on the numerator of our fraction. And when we do, we get 𝑥𝑦 plus 𝑦 squared minus 𝑥𝑦 minus one. But of course 𝑥𝑦 minus 𝑥𝑦 is zero. So this simplifies to 𝑦 squared minus one over 𝑥 plus 𝑦 all squared. The first partial derivative of the function with respect to 𝑥 is 𝑦 squared minus one over 𝑥 plus 𝑦 squared.
