Video: The Chain Rule for Multivariate Functions

Let π§ = π(π₯, π¦) over the curve π₯ = π‘Β² + 1, π¦ = π‘Β² β 1. Write an expression for dπ§/dπ‘, giving your answer in terms of the partial derivatives of π.

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Video Transcript

Let π§ be equal to π of π₯, π¦ over the curve π₯ equals π‘ squared plus one, π¦ equals π‘ squared minus one. Write an expression for dπ§ by dπ‘, giving your answer in terms of the partial derivatives of π.

We see that π§ is a multivariate function. Itβs a function in both π₯ and π¦. And weβre asked to find an expression for dπ§ by dπ‘ giving our answers in terms of the partial derivatives of π. Those are ππ ππ₯ β β thatβs the partial derivative of π with respect to π₯ β β and ππ ππ¦ β the partial derivative of π with respect to π¦. Now, our functions for π₯ and π¦ are themselves differentiable functions of π‘. And so we can quote the chain rule for one independent variable. Here, thatβs π‘.

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

Weβre going to evaluate dπ₯ by dπ‘, the ordinary derivative of π₯ with respect to π‘, and dπ¦ by dπ‘, the ordinary derivative of π¦ with respect to π‘. 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 π‘ squared with respect to π‘ is two π‘. 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 dπ‘ and dπ¦ by dπ‘ are both equal to two π‘.

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

But remember, π§ is a function in π₯ and π¦, where π₯ is π‘ squared plus one and π¦ is π‘ squared minus one. So we can rewrite ππ§ ππ₯ as shown and ππ§ ππ¦ as shown. And so we found an expression for dπ§ by dπ‘ in terms of the partial derivatives of π. Itβs two π‘ times the first partial derivative of the function in π‘ squared plus one and π‘ squared minus one with respect to π₯ plus two π‘ times the first partial derivative of the function in π‘ squared plus one and π‘ squared minus one with respect to π¦.

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