### Video Transcript

Find the first partial derivative of the function π of π₯, π¦ is the cube root of π₯ squared plus π¦ minus four with respect to π₯.

In this question, weβve been given a multivariable function, a function in two variables: π₯ and π¦. Weβre asked to find the first partial derivative of this function with respect to π₯. So, what does that mean? Well, in this case, it means we treat any other variable apart from π₯ as a constant. So, we treat π¦ here as a constant. We represent this partial derivative as shown. And so, to find the partial derivative of our function with respect to π₯, weβre going to differentiate it and treat π¦ as a constant. We are going to rewrite it slightly as π₯ squared plus π¦ minus four to the power of one-third.

And then we see, with respect to π₯, we have a function of a function or a composite function. And so, we could use the chain rule or the general power rule to differentiate it. Letβs use the chain rule. The chain rule says that if π¦ is some differentiable function of π’ and π’ itself is some differentiable function of π₯, then the derivative of π¦ with respect to π₯ is equal to the product of the derivatives of each of our functions, dπ¦ by dπ’ times dπ’ by dπ₯.

Now, weβre going to need to be a little bit careful here since our function is in terms of π₯ and π¦. So, weβre going to let π’ be equal to the inner part of our function, to π₯ squared plus π¦ minus four. Then, π of π’ is equal to π’ to the power of one-third. The first partial derivative of our function with respect to π₯ is the product of the derivatives of each of these functions. When we find the partial derivative of π’, we get two π₯. Remember, weβre treating π¦ like a constant, and the derivative of a constant is zero. Then, we differentiate our function in π’ with respect to π’. So, we get a third times π’ to the power of negative two-thirds.

Remember, we multiply the entire term by the exponent and then reduce that exponent by one. But of course, our partial derivative is with respect to π₯. So, we go back to our original substitution. We replace π’ with π₯ squared plus π¦ minus four. And our first partial derivative is two π₯ times a third times π₯ squared plus π¦ minus four to the power of negative two-thirds. Now, a negative exponent tells us to find the reciprocal. So, we can write this as two π₯ over three times π₯ squared plus π¦ minus four to the power of two-thirds.

By using one of our laws of exponents, the one that says that π₯ to the power of π to the power of π is the same as π₯ to the power of π times π, we can split our exponent up. And we can write π₯ squared plus π¦ minus four to the power of two-thirds as π₯ squared plus π¦ minus four squared to the power of one-third. But of course, we already saw that the power of one-third is the same as finding the cube root of that number. And so, we found the first partial derivative of the function with respect to π₯. Itβs two π₯ over three times the cube root of π₯ squared plus π¦ minus four squared.