

Subject:
find dy/dx by implicit differentiation
Category: Reference, Education and News > Homework Help Asked by: skirockga 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: elmartoga 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 
skirockga
rated this answer:
perfect!! 

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