Finding the instantaneous rate of change of y with respect to x?

Question

I need help finding the instantaneous rate of change of y with respect to x for the curve of

xe^y+x=xy

All help is greatly appreciated.

Answer 1

What you are looking for is the derivative dy/dx

Equations where the function y can't be isolated in terms of x need something called implicit differentation.

For the purposes of illustration I won't factor out the x which would make things a bit easier.

xe^y + x = xy

The derivative operator is linear so

d[f + g]/dx = df/dx + fg/dx

Using this, take the derivative of each term with respect to x.

d(xe^y)/dx + dx/dx = d(xy)/dx

For the first term of the left hand side and the term on the right hand side, you use the power rule and chain rule.

Per the product rule, d(fg)/dx = gdf/dx + fdg/dx
Per the chain rule, df(y)/dx = df/dy dy/dx

Using this the individual terms of the expression are

d(xe^y) = xd(e^y)/dx + e^ydx/dx = xe^y dy/dx + e^y
dx/dx = 1
d(xy)/dx = xdy/dx + ydx/dx = xdy/dx + y

Substituing all this into

d(xe^y)/dx + dx/dx = d(xy)/dx

we get

xe^y dy/dx + e^y + 1 = xdy/dx + y

Now we isolate dy/dx

xe^y dy/dx - xdy/dx =

xdy/dx(e^y - 1) = y - 1 - e^y

xdy/dx(1 - e^y) = y - 1 - e^y

dy/dx = [(e^y + 1) - y] / [x(1 - e^y)]

Answer 2

Following dobbyisbored, I agree that there must be a typo; and as it stands, the only possible gradient is zero.

I wonder whether parentheses have been omitted:
x e^(y + x) = xy . Even then we can factor out an x; but the result is at least no longer trivial. Then
e^(y + x) = y. Two methods to proceed:
(a)
e^(y + x) * (dy/dx +1) = dy/dx
So dy/dx * (1 - e^(y + x) ) = e^(y + x) ;
and dy/dx = e^(y + x) / (1 - e^(y + x) ), = 1 / (e^-(y + x) - 1)
or
(b)
y + x = ln y, so x = ln y - y
so dx/dy = 1/y - 1, = (1 - y) / y
so dy/dx = y / 1 - y

Answer 3

are you sure this is typed in correctly?

the x's cancel leaving e^y +1 = y
which is not a curve any more as it is independant of x... so you get straight horizontal lines thro any solutions of the above on the y axis... ie lines y = m

so i guess there is a typo... otherwise answer is 0!


PS looking at answers above maybe im completely wrong... it has been a long day!

Answer 4

Do an implicit differentation then solve for y':

(xe^y)' + (x)' = (xy)'

1.e^y + xe^y*y' + 1 = 1y + xy'. Now collect all terms with y' to the left hand side

(xe^y-x)y' = (y-e^y - 1)

y' = (y-e^y - 1)/[(e^y-1)*x]

Answer 5

The easiest way is to differentiate each term by parts:
[ (xe^y)dy + (e^y)dx ] + [ dx ] = [ x dy + y dx ]

Now collect the terms:
(xe^y - x) dy = (y - e^y - 1) dx

dy/dx = (y - e^y - 1) / (xe^y - 1)

Answer 6

This is dy/dx=y´
e^y +xe^y *y´+1 =y +xy´
so
y´= (e^y-y+1)/(x-x*e^y)

Answer 7

Taking deriv. of both sides
e^y+xe^y(dy/dx)+1=y+xy(dy/dx)
dy/dx=[y-1-e^y]/[xe^y-xy]

Answer 8

x.e^y(dy/dx)+e^y.1+1= x.(dy/dx)+y.1
~ (x.e^y-x).dy/dx = y-e^y-1
~ dy/dx = (y-e^y-1)/(x.e^y-x)