Lecture Note
University
Massachusetts Institute of TechnologyCourse
Multivariable CalculusPages
2
Academic year
2023
Sporkz
Views
99
Theoretical Application. Product Rule From Chain Rule So an application of this is to justify the product and quotient rules. For example, wecan show using this idea that a function of two variables u and v that's just theproduct uv is in fact equal to u multiplied by v plus u times v divided by 2. π = π’π£ Let us suppose that u and v are actually functions of one variable t. π’ = π’ π‘ ( ), π£ = π£(π‘) Then, well, the derivative of uv with respect to t is equal to the product of f and itsderivative with respect to u times f and its derivative with respect to v times thederivative of v with respect to t. π(π’π£) ππ‘ = π π’ ππ’ ππ‘ + π π£ ππ£ ππ‘ Now, what is the partial derivative of f with respect to u? It's v. So that's v dt. And thepartial derivative of f with respect to v is going to be just u. π(π’π£) ππ‘ = π π’ ππ’ ππ‘ + π π£ ππ£ ππ‘ = π£ ππ’ ππ‘ + π’ ππ£ ππ‘ The product rule states that the derivative of the product of two functions equals thederivative of one function times the other function, times an extra term. This is aslightly complicated way of deriving it. However, it is valid and useful to understandhow to take the derivative of a product by thinking of the product first as a function oftwo variables, which are u and v, and then saying, oh, but u and v were actuallyfunctions of a variable t. Then you do the differentiation in two stages using chainrule. Similarly, you can apply the quotient rule. Let's do it just for practice. π = π’π£ When we consider the function g equal to u over v, we see that u and v are actuallyfunctions of t. π’ = π’ π‘ ( ), π£ = π£(π‘) To determine the partial derivatives of, we must first differentiate the equation withrespect to v. Then, it is going to be partial g, partial u. How much is that? 1 over vtimes du dt plus - next, we need to have partial g over partial v. What's the derivativeof this with respect to v? It's minus u over v squared times dv dt. π π’π£ ( ) ππ‘ = 1π£ ππ’ ππ‘ + βπ’ π£ 2 ( ) ππ£ ππ‘
And that is actually the usual quotient rule written in a different way. In other words, ifyou clear denominators to put v squared in there, then you will see that it is just utimes v minus u times v prime. π π’π£ ( ) ππ‘ = 1π£ ππ’ ππ‘ + βπ’ π£ 2 ( ) ππ£ ππ‘ = π’ πΌ π£βπ£ πΌ π’ π£ 2
Theoretical Application. Product Rule From Chain Rule
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