Video: The Chain Rule for Multivariate Functions

Let 𝑧 = 𝑓(π‘₯, 𝑦) over the curve π‘₯ = 𝑑² + 1, 𝑦 = 𝑑² βˆ’ 1. Write an expression for d𝑧/d𝑑, giving your answer in terms of the partial derivatives of 𝑓.

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Video Transcript

Let 𝑧 be equal to 𝑓 of π‘₯, 𝑦 over the curve π‘₯ equals 𝑑 squared plus one, 𝑦 equals 𝑑 squared minus one. Write an expression for d𝑧 by d𝑑, 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 π‘₯ and 𝑦. And we’re asked to find an expression for d𝑧 by d𝑑 giving our answers in terms of the partial derivatives of 𝑓. Those are πœ•π‘“ πœ•π‘₯ ⁠— that’s the partial derivative of 𝑓 with respect to π‘₯ ⁠— and πœ•π‘“ πœ•π‘¦ β€” the partial derivative of 𝑓 with respect to 𝑦. Now, our functions for π‘₯ and 𝑦 are themselves differentiable functions of 𝑑. And so we can quote the chain rule for one independent variable. Here, that’s 𝑑.

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

We’re going to evaluate dπ‘₯ by d𝑑, the ordinary derivative of π‘₯ with respect to 𝑑, and d𝑦 by d𝑑, the ordinary derivative of 𝑦 with respect to 𝑑. 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 𝑑 squared with respect to 𝑑 is two 𝑑. 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 dπ‘₯ by d𝑑 and d𝑦 by d𝑑 are both equal to two 𝑑.

The question tells us to give our answer in terms of the partial derivatives of 𝑓. Those are πœ•π‘§ πœ•π‘₯ and πœ•π‘§ πœ•π‘¦. And so we can say that the derivative of 𝑧 with respect to 𝑑 is equal to πœ•π‘§ πœ•π‘₯ times two 𝑑 plus πœ•π‘§ πœ•π‘¦ times two 𝑑. We simplify a little, and we can write this as two 𝑑 πœ•π‘§ πœ•π‘₯ plus two 𝑑 πœ•π‘§ πœ•π‘¦, where we said πœ•π‘§ πœ•π‘₯ is the first partial derivative of 𝑧 with respect to π‘₯ and πœ•π‘§ πœ•π‘¦ is the first partial derivative of 𝑧 with respect to 𝑦.

But remember, 𝑧 is a function in π‘₯ and 𝑦, where π‘₯ is 𝑑 squared plus one and 𝑦 is 𝑑 squared minus one. So we can rewrite πœ•π‘§ πœ•π‘₯ as shown and πœ•π‘§ πœ•π‘¦ as shown. And so we found an expression for d𝑧 by d𝑑 in terms of the partial derivatives of 𝑓. It’s two 𝑑 times the first partial derivative of the function in 𝑑 squared plus one and 𝑑 squared minus one with respect to π‘₯ plus two 𝑑 times the first partial derivative of the function in 𝑑 squared plus one and 𝑑 squared minus one with respect to 𝑦.

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