How Do You Tackle Partial Differentiation with Trigonometric Functions?

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Homework Help Overview

The problem involves finding the first and second partial derivatives of a function defined by the equations x = yz and y = 2sin(y+z), specifically dx/dy and d^2x/dy^2. The context includes partial differentiation with trigonometric functions.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • The original poster expresses confusion about how to start the problem and questions the relationship between variables. Some participants suggest finding dz in terms of dy and substituting it into the equation for dx. Others discuss the appropriateness of using inverse trigonometric functions in this context.

Discussion Status

The discussion is ongoing, with participants exploring different approaches to the problem. Some guidance has been offered regarding the manipulation of equations, but there is no explicit consensus on the best method to proceed.

Contextual Notes

There is mention of confusion regarding the setup of the problem and the relationships between the variables, as well as a potential typo in the original poster's equations. The participants are navigating these constraints as they discuss the problem.

phrygian
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Homework Statement



If x = yz and y = 2sin(y+z), find dx/dy and d^2x/dy^2

Homework Equations





The Attempt at a Solution



I am beyond confused at how to even start this one, the problem is not like any of the examples in my book.

I know x = ydz + zdy but don't know how to deal with z and dz?

Thanks for the help
 
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Find dz in terms of dy from the second equation and plug it into the first: dx=ydz+zdy.
 
Is it correct to use inverse trig function for that?
 
I can see no reason to. The second equation is y= 2sin(y+z) so dy= 2cos(y+z)(dy+ dz). From dx= zdy+ ydz, (NOT the "x= zdy+ ydz" you give. I assume that was a typo.), we get ydz= dx- zdy so dz= (1/y)dx- (z/y)dy. Putting that into the second equation, dy= 2cos(y+z)(dy+ (1/y)dx- (z/y)dy)= 2cos(y+z)(1- (z/y)dy+ (2/y)cos(y+z)dx. Solve that for dx and divide both sides by dy.
 
You can do anything you want that doesn't violate the laws of math!
 

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