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

Find the first partial derivative of the function 𝑓(π‘₯, 𝑦) = βˆ›(π‘₯Β² + 𝑦 βˆ’ 4) with respect to π‘₯.

02:59

Video Transcript

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

In this question, we’ve been given a multivariable function, a function in two variables: π‘₯ and 𝑦. We’re 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 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 π‘₯, we’re going to differentiate it and treat 𝑦 as a constant. We are going to rewrite it slightly as π‘₯ squared plus 𝑦 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. Let’s use the chain rule. The chain rule says that if 𝑦 is some differentiable function of 𝑒 and 𝑒 itself is some differentiable function of π‘₯, then the derivative of 𝑦 with respect to π‘₯ is equal to the product of the derivatives of each of our functions, d𝑦 by d𝑒 times d𝑒 by dπ‘₯.

Now, we’re going to need to be a little bit careful here since our function is in terms of π‘₯ and 𝑦. So, we’re going to let 𝑒 be equal to the inner part of our function, to π‘₯ squared plus 𝑦 minus four. Then, 𝑓 of 𝑒 is equal to 𝑒 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 𝑒, we get two π‘₯. Remember, we’re treating 𝑦 like a constant, and the derivative of a constant is zero. Then, we differentiate our function in 𝑒 with respect to 𝑒. So, we get a third times 𝑒 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 𝑒 with π‘₯ squared plus 𝑦 minus four. And our first partial derivative is two π‘₯ times a third times π‘₯ 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 π‘₯ over three times π‘₯ squared plus 𝑦 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‘Ž to the power of 𝑏 is the same as π‘₯ to the power of π‘Ž times 𝑏, we can split our exponent up. And we can write π‘₯ squared plus 𝑦 minus four to the power of two-thirds as π‘₯ 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 π‘₯. It’s two π‘₯ over three times the cube root of π‘₯ squared plus 𝑦 minus four squared.

Nagwa uses cookies to ensure you get the best experience on our website. Learn more about our Privacy Policy.