Taking second derivatives in Calc 1 wasn't so bad. Here it can be rather tedious and challenging depending on the number of variables and parameters. To take the simplest case, suppose we have an equation in two variables f(x, y) = 0, which defines y as an implicit function of x, twice differentiable. (a) Starting with the first derivative, d²y_2fxfyfry - ffxx - fzfyy f3 dx² dy dx = fx fy where fr = apply the quotient rule and chain rule to show: af მე 2 af = მყ " fxy = Ꭷ f მყმუ and so on. (b) Use the above formula (whether or not you figured it out) to find d²y for the equation x²+y² = 4, dx² even though you probably did this in Calc 1 without such a monstrous formula.
Taking second derivatives in Calc 1 wasn't so bad. Here it can be rather tedious and challenging depending on the number of variables and parameters. To take the simplest case, suppose we have an equation in two variables f(x, y) = 0, which defines y as an implicit function of x, twice differentiable. (a) Starting with the first derivative, d²y_2fxfyfry - ffxx - fzfyy f3 dx² dy dx = fx fy where fr = apply the quotient rule and chain rule to show: af მე 2 af = მყ " fxy = Ꭷ f მყმუ and so on. (b) Use the above formula (whether or not you figured it out) to find d²y for the equation x²+y² = 4, dx² even though you probably did this in Calc 1 without such a monstrous formula.
Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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