Implicit Differentiation problem

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Homework Help Overview

The problem involves finding dy/dx using implicit differentiation for the equation y^2 + xsiny = 4.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the original poster's attempt involving dy/dt instead of dy/dx, with suggestions to clarify the differentiation process. There is also a focus on the importance of parentheses in the final expression for dy/dx.

Discussion Status

Some participants have provided guidance on correcting the differentiation approach and emphasized the need for clarity in the expression. Multiple interpretations of the final expression for dy/dx are being explored, but no consensus has been reached.

Contextual Notes

There is a mention of potential confusion regarding the differentiation variable and the need for careful notation in the expressions derived.

suxatphysix
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Homework Statement



Find dy/dx by implicit differentiation when it is known that y^2 + xsiny = 4

Homework Equations





The Attempt at a Solution



2y dy/dt + xcosy dy/dt + siny = 0

2y dy/dt + xcosy dy/dt = -siny

dy/dt + dy/dt = -siny/2y/xcosy

I'm sure I'm doing it wrong so I stopped right there. What am I doing it wrong and how do I solve it?

Thanks
 
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you're looking for dy/dx not dy/dt o.o, but assuming by dt you mean dx then in the 2nd line of your attempt you should pull out the dy/dt and the answer will be easy from there.
 
Is this the answer?

dy/dx = -siny/2y+xcosy
 
It would be if you would parenthesize the denominator.
 
dy/dx = -siny/(2y+xcosy) ? What's that's do
 
It distinguishes between -(siny/(2y))+(xcosy) and -siny/(2y+xcosy). Which are two quite different expressions. But look the same if you omit the parentheses.
 
Last edited:
1. Homework Statement

Find dy/dx by implicit differentiation when it is known that y^2 + xsiny = 4

[tex]\frac{d}{dx}(y^2+x\sin(y)) = 0[/tex]

[tex]=2y\left(\frac{dy}{dx}\right)+\left[\frac{d}{dx}x\cdot\sin(y)+\frac{d}{dx}\sin(y)\cdot x\right]=0[/tex]

[tex]= 2y\left(\frac{dy}{dx}\right) + \left[\sin(y) + x\cdot\cos(y)\left(\frac{dy}{dx}\right)\right]= 2y\left(\frac{dy}{dx}\right) + \sin(y) + x\cos(y)\left(\frac{dy}{dx}\right) = 2y\left(\frac{dy}{dx}\right) + x\cos(y)\left(\frac{dy}{dx}\right) + \sin(y) = 0[/tex]

[tex]= \left(\frac{dy}{dx}\right)(2y+x\cos(y)) + \sin(y) = 0[/tex]

[tex]= \left(\frac{dy}{dx}\right)(2y+x\cos(y)) = -\sin(y)[/tex]

[tex]\left(\frac{dy}{dx}\right) = \frac{-\sin(y)}{(2y+x\cos(y))}[/tex]

Edit: I'm late by about 25 minutes. :biggrin:
 
cool thanks
 

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