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7) Suppose w=rt+rs' and r= tan-' (2x² +3y), s=x²y' +3y and t = x²y² .Use the chain rule (from calculus III) to find ду (You can leave your answer in terms of r, s, t, x, and y.)

7) Suppose w=rt+rs' and r= tan-' (2x² +3y), s=x²y' +3y and t = x²y² .Use the chain rule (from calculus III) to find ду (You can leave your answer in terms of r, s, t, x, and y.)

Calculus: Early Transcendentals
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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Please show steps and i will rate :D

**Problem 7:** Given the functions: - \( w = r^2 t + rs^3 \) - \( r = \tan^{-1}(2x^2 + 3y) \) - \( s = x^2 y^3 + 3y \) - \( t = x^2 y^2 \) Use the chain rule from calculus to find the partial derivative \(\frac{\partial w}{\partial y}\). (Note: You can leave your answer in terms of \( r, s, t, x, \) and \( y \).)
Transcribed Image Text:**Problem 7:** Given the functions: - \( w = r^2 t + rs^3 \) - \( r = \tan^{-1}(2x^2 + 3y) \) - \( s = x^2 y^3 + 3y \) - \( t = x^2 y^2 \) Use the chain rule from calculus to find the partial derivative \(\frac{\partial w}{\partial y}\). (Note: You can leave your answer in terms of \( r, s, t, x, \) and \( y \).)
Expert Solution
Step 1

Given w=r2t+rs3r=tan−1(2x2+3y)s=x2y3+3yt=x2y2

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