dy/dx of a polynomial in x and y

• Apr 30th 2013, 07:58 PM
kingsolomonsgrave
dy/dx of a polynomial in x and y
what is dy/dx of x^2 + sin(xy) +y^2 = 0?

I would guess the answer is

2x + cos (xy)* y = 0

Is this correct?

MH
• Apr 30th 2013, 08:24 PM
ibdutt
Re: dy/dx of a polynomial in x and y
The derivation would be: 2x + cos (xy) [ 1 + x dy/dx ] + 2y dy/dx = 0
that is: 2x + cos ( xy ) + x cos(xy) dy/dx + 2y dy/dx = 0
now you can find dy/dx
• Apr 30th 2013, 09:21 PM
kingsolomonsgrave
Re: dy/dx of a polynomial in x and y
Thanks! Why is it dy/dx of sin(xy) = cos(xy)[1+ x dy/dx ]? I would have thought cos(xy)[y + x dy/dx]
• Apr 30th 2013, 09:47 PM
Prove It
Re: dy/dx of a polynomial in x and y
Quote:

Originally Posted by kingsolomonsgrave
what is dy/dx of x^2 + sin(xy) +y^2 = 0?

I would guess the answer is

2x + cos (xy)* y = 0

Is this correct?

MH

Are we assuming that y is a function of x?
• May 1st 2013, 01:31 AM
ibdutt
Re: dy/dx of a polynomial in x and y
You are right it was a typing error. On differentiating wrt x we get
2x + sin (xy) [ y + x dy/dx] + 2y dy/dx]

We foffow chain rule for sin(xy): d/dx(sin (xy) ) = cos ( xy) d/dx(xy) = cos xy [ y + x dy/dx ]
THANKS
• May 1st 2013, 07:10 AM
kingsolomonsgrave
Re: dy/dx of a polynomial in x and y
yes I think we are to assume y is a function of x
• May 1st 2013, 07:26 AM
HallsofIvy
Re: dy/dx of a polynomial in x and y
Quote:

Originally Posted by kingsolomonsgrave
Thanks! Why is it dy/dx of sin(xy) = cos(xy)[1+ x dy/dx ]? I would have thought cos(xy)[y + x dy/dx]

First, please stop saying "dy/dx of" something. You are asking for dy/dx itself, where y is (implicitely) defined by that equation.

However yes, the derivative of sin(u) with respect to x is cos(u)(du/dx). And the derivative of u= xy, with respect to x, is (dx/dx)y+ x(dy/dx)= y+ xdy/dx so that the derivative of sin(xy), with respect to x, is cos(xy)(y+ x dy/dx).
• May 1st 2013, 07:53 AM
kingsolomonsgrave
Re: dy/dx of a polynomial in x and y
thanks HallsofIvey. Why wouldn't I say 'dy/dx of'? I thought dy/dx = the derivative?

Or is it that dy/dx is read 'the derivative of y wrt x' and so contains 'the derivative' in its definition?
• May 1st 2013, 12:26 PM
topsquark
Re: dy/dx of a polynomial in x and y
The derivative is an expression, ie a function. What you are referring to (in English) is taking the derivative of an expression. The two are vastly different.

-Dan
• May 1st 2013, 12:32 PM
HallsofIvy
Re: dy/dx of a polynomial in x and y
You could say "$\displaystyle \frac{d}{dx}$ of a function" (or, more generally, of an expression or equation). Note that there is no "y" in that. It is the function that takes the place of the 'y'.
• May 1st 2013, 05:08 PM
Prove It
Re: dy/dx of a polynomial in x and y
This is because \displaystyle \displaystyle \begin{align*} \frac{d}{dx} \end{align*} is the OPERATOR, while \displaystyle \displaystyle \begin{align*} \frac{dy}{dx} = \frac{d}{dx} \left( y \right) \end{align*}, the RESULT after the operator has operated on the function y.