Video: Finding the First Partial Derivative of a Three-Variable Function

Find the first partial derivative with respect to 𝑦 of 𝑓(π‘₯, 𝑦, 𝑧) = π‘₯³𝑦𝑧² + 2𝑦𝑧.

02:24

Video Transcript

Find the first partial derivative with respect to 𝑦 of 𝑓 of π‘₯, 𝑦, 𝑧 equals π‘₯ cubed 𝑦𝑧 squared plus two 𝑦𝑧.

In this question, we’ve been given a multivariable function. That’s a function in terms of more than one variable. Here, those are π‘₯, 𝑦, and 𝑧. We’re looking to find the first partial derivative of our function with respect to 𝑦. And we denote this as shown. Essentially, what we’re now doing is seeing how the function changes as we let just one of our variables change. In this case, that’s 𝑦. And to do so, we hold all the other variables constant.

And so, when we find the first partial derivative of π‘₯ cubed 𝑦𝑧 squared plus two 𝑦𝑧, we’re going to treat π‘₯ and 𝑧 both as constants. Then, we’ll perform this process term by term. It might be helpful to write π‘₯ cubed 𝑦𝑧 squared as π‘₯ cubed 𝑧 squared 𝑦. And then, treating π‘₯ cubed 𝑧 squared as a constant and differentiating with respect to 𝑦, we see we’re just left with the constant part, with π‘₯ cubed 𝑧 squared. Similarly, when we differentiate two 𝑦𝑧 with respect to 𝑦 and we treat 𝑧 as a constant, we’re simply left with two 𝑧. And so we found the first partial derivative with respect to 𝑦 of our function. It’s π‘₯ cubed 𝑧 squared plus two 𝑧.

And of course, had we chosen to do so, we could’ve found the first partial derivative with respect to π‘₯ and with respect to 𝑧. Let’s see what that might’ve looked like. When we differentiate π‘₯ cubed with respect to π‘₯, we get three π‘₯ squared. Now, we treat 𝑦𝑧 squared as a constant here, and we get three π‘₯ squared 𝑦𝑧 squared. Then, when we’re finding the first partial derivative with respect to π‘₯, we treat 𝑦 and 𝑧 as a constant. So the derivative of two 𝑦𝑧 is zero. And then, the first partial derivative with respect to π‘₯ of our function is three π‘₯ squared 𝑦𝑧 squared.

Let’s perform this process one more time, this time finding the first partial derivative with respect to 𝑧. Now, if we were to differentiate 𝑧 squared with respect to 𝑧, we just get two 𝑧. So we treat π‘₯ cubed and 𝑦 like a constant, and we find that the first term becomes two π‘₯ cubed 𝑦𝑧. Then, in our second term, we’re treating the 𝑦 as a constant. So we’re left with two 𝑦. And the first partial derivative with respect to 𝑧 is two π‘₯ cubed 𝑦𝑧 plus two 𝑦.

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