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

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- April 30th 2013, 07:58 PMkingsolomonsgravedy/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 - April 30th 2013, 08:24 PMibduttRe: 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 - April 30th 2013, 09:21 PMkingsolomonsgraveRe: 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]

- April 30th 2013, 09:47 PMProve ItRe: dy/dx of a polynomial in x and y
- May 1st 2013, 01:31 AMibduttRe: 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 AMkingsolomonsgraveRe: 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 AMHallsofIvyRe: dy/dx of a polynomial in x and y
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 AMkingsolomonsgraveRe: 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 PMtopsquarkRe: 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 PMHallsofIvyRe: dy/dx of a polynomial in x and y
You could say " 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 PMProve ItRe: dy/dx of a polynomial in x and y
This is because is the OPERATOR, while , the RESULT after the operator has operated on the function y.