What is the Implicit Differentiation of (x + y)sin(xy) = 1?

[james.farrow]

I have a question thus:

If (x + y)sin(xy) = 1 find dy/dx.

It looks to me as I should use the product rule.

d/dx(x + y) I get 1 + dy/dx

Now this is where it gets kinda tricky.

d/dx sin(xy) its the fuction of a function I think!

I get (eventually) cos(xy)xdy/dx + y

Now putting it all together and using the prod rule as 1st stated my final effort is


dy/dx = -y -sin(xy)/(x +y)cos(xy)x + sin(xy)


Can any shed some light on my work so far!

Cheers

Jimmy

 

[Hurkyl]

I think you understand the relevant calculus here!


Alas, your arithmetic needs some work -- some of the equations you've written are wrong. The one I looked at could be just that you forgot to use parentheses. But it could also be the result of a different arithmetic error, I can't tell which.

 

[Samuelb88]

You're right, you do have to use the product rule, but remember the product rule is

(uv)` = uv` + vu`

It's a bit hard to follow your work, but it looks like you differentiated both "products" and then multiplied them together which would be wrong. Also, looking at your derivatives, perhaps you just forgot to write them out, but don't forget when you use the chain rule when finding d/dx(sinxy) you must distribute cos(xy) to the inner quantity. Sometimes implicit differentiation can get a bit complicated and messy... It might make it a bit more clear if you let u = x+y, and v = sin(xy), find u` and v` and plug then back into the formula above.

 

[james.farrow]

Cheers eveyone, I have taken heed of your advice and have had another go at it, and yes my final answer was a little different...

I will post my efforts when I get 5 minutes...

Thanks for your help

James