# Video: The Chain Rule for Multivariate Functions

Let 𝑧 = 𝑓(𝑥, 𝑦) over the curve 𝑥 = 𝑡² + 1, 𝑦 = 𝑡² − 1. Write an expression for d𝑧/d𝑡, giving your answer in terms of the partial derivatives of 𝑓.

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### Video Transcript

Let 𝑧 be equal to 𝑓 of 𝑥, 𝑦 over the curve 𝑥 equals 𝑡 squared plus one, 𝑦 equals 𝑡 squared minus one. Write an expression for d𝑧 by d𝑡, giving your answer in terms of the partial derivatives of 𝑓.

We see that 𝑧 is a multivariate function. It’s a function in both 𝑥 and 𝑦. And we’re asked to find an expression for d𝑧 by d𝑡 giving our answers in terms of the partial derivatives of 𝑓. Those are 𝜕𝑓 𝜕𝑥 ⁠— that’s the partial derivative of 𝑓 with respect to 𝑥 ⁠— and 𝜕𝑓 𝜕𝑦 — the partial derivative of 𝑓 with respect to 𝑦. Now, our functions for 𝑥 and 𝑦 are themselves differentiable functions of 𝑡. And so we can quote the chain rule for one independent variable. Here, that’s 𝑡.

This says that suppose 𝑥 and 𝑦 are differentiable functions of 𝑡 and 𝑧 is a function in 𝑥 and 𝑦 as a differentiable function. Then this means 𝑧 is 𝑓 of 𝑥 of 𝑡, 𝑦 of 𝑡 is a differentiable function of 𝑡. Such that d𝑧 by d𝑡, the derivative of 𝑧 with respect to 𝑡 is equal to the partial derivative of 𝑧 with respect to 𝑥 times the derivative of 𝑥 with respect to 𝑡 plus the partial derivative of 𝑧 with respect to 𝑦 times the ordinary derivative of 𝑦 with respect to 𝑡. Well, in this case, 𝑥 is the function 𝑡 squared plus one and 𝑦 is the function 𝑡 squared minus one. These are both polynomials in 𝑡, and we know polynomials are differentiable over their entire domain. So 𝑥 and 𝑦 are both differentiable functions.

We’re going to evaluate d𝑥 by d𝑡, the ordinary derivative of 𝑥 with respect to 𝑡, and d𝑦 by d𝑡, the ordinary derivative of 𝑦 with respect to 𝑡. Now, we know that, to differentiate a power term, we multiply the entire term by the exponent, then reduce that exponent by one. So in both cases, the derivative of 𝑡 squared with respect to 𝑡 is two 𝑡. But we also know that the derivative of a constant is zero. So both one and negative one differentiate to zero. And we see that d𝑥 by d𝑡 and d𝑦 by d𝑡 are both equal to two 𝑡.

The question tells us to give our answer in terms of the partial derivatives of 𝑓. Those are 𝜕𝑧 𝜕𝑥 and 𝜕𝑧 𝜕𝑦. And so we can say that the derivative of 𝑧 with respect to 𝑡 is equal to 𝜕𝑧 𝜕𝑥 times two 𝑡 plus 𝜕𝑧 𝜕𝑦 times two 𝑡. We simplify a little, and we can write this as two 𝑡 𝜕𝑧 𝜕𝑥 plus two 𝑡 𝜕𝑧 𝜕𝑦, where we said 𝜕𝑧 𝜕𝑥 is the first partial derivative of 𝑧 with respect to 𝑥 and 𝜕𝑧 𝜕𝑦 is the first partial derivative of 𝑧 with respect to 𝑦.

But remember, 𝑧 is a function in 𝑥 and 𝑦, where 𝑥 is 𝑡 squared plus one and 𝑦 is 𝑡 squared minus one. So we can rewrite 𝜕𝑧 𝜕𝑥 as shown and 𝜕𝑧 𝜕𝑦 as shown. And so we found an expression for d𝑧 by d𝑡 in terms of the partial derivatives of 𝑓. It’s two 𝑡 times the first partial derivative of the function in 𝑡 squared plus one and 𝑡 squared minus one with respect to 𝑥 plus two 𝑡 times the first partial derivative of the function in 𝑡 squared plus one and 𝑡 squared minus one with respect to 𝑦.