MaplePrimes - answers and comments on Question, implicit differentiation

a way

jakubi — Sun, 03 Jan 2010 23:39:29 Z

For instance:

implicitdiff(x*sin(x*y)=x,y,x):
simplify(%,{sin(x*y)=1});
                                - y/x

eq:=x*sin(x*y)=x:
implicitdiff(eq,y,x):
simplify(%,{eq/x});

                                - y/x

perfect

longrob — Mon, 04 Jan 2010 05:45:29 Z

thanks !

comment

Joe Riel — Mon, 04 Jan 2010 09:27:03 Z

Your original premise is flawed. That is, the additional assumption that cos(x*y) <> 0 implies sin(x*y) <> 1. So the only solution to the equation is x=0.

strange

longrob — Mon, 04 Jan 2010 10:03:00 Z

I realised a little while ago that the question is quite strange. Adding to the strangeness is the page where I saw the question posed: http://www.analyzemath.com/calculus/Differentiation/implicit.html Note the given answer at the bottom

x=0 ? Surely not.....?

longrob — Mon, 04 Jan 2010 10:21:25 Z

Surely if x=0 then sin(x*y)=0 which contradicts the original relation sin(x*y)=1 From the original relationship, sin(x*y) = 1, provided that x is *not* equal to zero. Thus, xy = pi/2 ± k.2pi; ie xy = const. This gives dy/dx = - y/x immediately. In the second method, before dividing by x^2cos(x*y), we have the expression x{cos(xy)[y + x.dy/dx]} + sin(xy) = 1. But sin(xy) = 1, xy = Pi/2 ±k*2*Pi, so cos(xy) = 0

Look at the curve

Alex Smith — Mon, 04 Jan 2010 21:52:27 Z

The "curve" defined by the equation x*sin(x*y)=x consists of hyperbolas y=c/x, where c=(2*k+1)*Pi/2, along with the y-axis, from the solution x=0.

So when x=0, dy/dx is undefined.