# Video: Finding the First Partial Derivative of a Multivariable Root Function of Two Variables

Find the first partial derivative of the function 𝑓(𝑥, 𝑦) = ∛(𝑥² + 𝑦 − 4) with respect to 𝑥.

02:59

### Video Transcript

Find the first partial derivative of the function 𝑓 of 𝑥, 𝑦 is the cube root of 𝑥 squared plus 𝑦 minus four with respect to 𝑥.

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

And then we see, with respect to 𝑥, 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. Let’s use the chain rule. The chain rule says that if 𝑦 is some differentiable function of 𝑢 and 𝑢 itself is some differentiable function of 𝑥, then the derivative of 𝑦 with respect to 𝑥 is equal to the product of the derivatives of each of our functions, d𝑦 by d𝑢 times d𝑢 by d𝑥.

Now, we’re going to need to be a little bit careful here since our function is in terms of 𝑥 and 𝑦. So, we’re going to let 𝑢 be equal to the inner part of our function, to 𝑥 squared plus 𝑦 minus four. Then, 𝑓 of 𝑢 is equal to 𝑢 to the power of one-third. The first partial derivative of our function with respect to 𝑥 is the product of the derivatives of each of these functions. When we find the partial derivative of 𝑢, we get two 𝑥. Remember, we’re treating 𝑦 like a constant, and the derivative of a constant is zero. Then, we differentiate our function in 𝑢 with respect to 𝑢. So, we get a third times 𝑢 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 𝑥. So, we go back to our original substitution. We replace 𝑢 with 𝑥 squared plus 𝑦 minus four. And our first partial derivative is two 𝑥 times a third times 𝑥 squared plus 𝑦 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 𝑥 over three times 𝑥 squared plus 𝑦 minus four to the power of two-thirds.

By using one of our laws of exponents, the one that says that 𝑥 to the power of 𝑎 to the power of 𝑏 is the same as 𝑥 to the power of 𝑎 times 𝑏, we can split our exponent up. And we can write 𝑥 squared plus 𝑦 minus four to the power of two-thirds as 𝑥 squared plus 𝑦 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 𝑥. It’s two 𝑥 over three times the cube root of 𝑥 squared plus 𝑦 minus four squared.