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Q: find dy/dx by implicit differentiation ( Answered 5 out of 5 stars,   0 Comments )
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:5 out of 5 stars
Hello skirock,
In the following link you will find a tutorial on how to solve
problems involving implicit differentiation.

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 

Finally, the right hand side of the equation:

=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,
skirock-ga rated this answer:5 out of 5 stars

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