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Q: find dy/dx by implicit differentiation ( Answered ,   0 Comments )
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 Subject: find dy/dx by implicit differentiation Category: Reference, Education and News > Homework Help Asked by: skirock-ga List Price: \$4.00 Posted: 14 Jul 2005 13:20 PDT Expires: 13 Aug 2005 13:20 PDT Question ID: 543587
 ```this problem seems to not work out right for me. Please show all steps. Thanks. find dy/dx by implicit differentiation x^2 cos y + sin 2y = xy```
 Subject: Re: find dy/dx by implicit differentiation Answered By: elmarto-ga on 14 Jul 2005 18:39 PDT Rated:
 ```Hello skirock, In the following link you will find a tutorial on how to solve problems involving implicit differentiation. http://www.howardcc.edu/math/ma145/2.7/2.7.htm Basically, we must start by differentiating both sides of the equation with respect to x, taking in account that y also depends on x. So what we have here is: d[x^2 cos y + sin 2y]/dx = d[xy]/dx I'll differentiate each term separately here: d[x^2 cos y]/dx =2x*cos(y) + x^2*(-sin(y))*dy/dx =2x*cos(y) - x^2*sin(y)*dy/dx (notice that I used the product rule and the chain rule) The next term: d[sin 2y]/dx =cos(2y)*2*dy/dx Finally, the right hand side of the equation: d[xy]/dx =y + x*dy/dx So what we have is: 2x*cos(y) - x^2*sin(y)*dy/dx + cos(2y)*2*dy/dx = y + x*dy/dx In order to find dy/dx we must simply isolate it from this equation. Let's group all terms with dy/dx together: - x^2*sin(y)*dy/dx + cos(2y)*2*dy/dx - x*dy/dx = y - 2x*cos(y) Taking dy/dx as common factor: dy/dx*[-x^2*sin(y) + 2cos(2y) - x] = y - 2x*cos(y) And now we easily isolate dy/dx: dy/dx = [y - 2x*cos(y)] / [-x^2*sin(y) + 2cos(2y) - x] I hope this helps! Best wishes, elmarto```
 skirock-ga rated this answer: `perfect!!`

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