Question

What is the implicit derivative of `25=sin(xy)/x-3xy`?

Answer 1

Long time since tried this but I am having a go!
`color(blue)((dy)/dx =(sin(xy)-yxcos(xy))/(x^2cos(xy)-3x^3)+(3y)/(cos(xy)-3x))`
`color(red)("You will need to check this!!")`

Explanation:

`d/(dx)( (sin(xy))/x -3xy) =d/dx(25)`

`d/dx((sin(xy))/x) - d/dx(3xy)=0`

Using std forms `d/dx(uv) =v(du)/dx + u(dv)/dx`

`color(white)(xxxxxxxxxx)d/dx(u/v) = (v(du)/dx - u(dv)/dx )/v^2`
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Let `color(white)(xx)w =uv -> xy`
then `color(green)((dw)/dx ->(dy)/dx= y +x(dy)/dx ...........................(1))_"confirmed"`

Let `t=sin(xy) = sin(w)`
then `color(green)( (dt)/dx->(dy)/dx = (y+x(dy)/dx)cos(xy) .............(2))_"confirmed"`

`color(green)(d/dx(3xy) = 3(y+x(dy)/dx) "from (1) "................(3))_"confirmed"`
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Consider: `d/dx((sin(xy))/x) -> u/v`

`(x{ycos(xy)+x(dy/dx)cos(xy)} - sin(xy))/x^2 ......._color(red)("corrected y-> ycos(xy)")`

`color(green)((ycos(xy))/x +(dy)/dx cos(xy) -sin(xy)/x^2.........(4))_color(red)(("corrected" y->ycos(xy)))`

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
`color(blue)("Combining (3) and (4)")`

`{y/x cos(xy)+(dy)/dx cos(xy)-sin(xy)/x^2}-{3y+3x(dy)/dx}=0_("corrected " y->ycos(xy))`

Collecting like terms `color(red)("(rebuilt calculations)")`

`(dy)/dx cos(xy) -3x(dy)/dx color(red)( =) (sin(xy))/x^2-y/xcos(xy)+3y_(color(red)(" corrected " y->ycos(xy)))`
`color(white)(x)`

`(dy)/dx color(red)( =) 1/(cos(xy)-3x) ((sin(xy))/x^2 -y/xcos(xy)+3y)`

`(dy)/dxcolor(red)( =) 1/(cos(xy)-3x)((sin(xy)-yxcos(xy))/x^2+3y)`

`color(blue)((dy)/dx =(sin(xy)-yxcos(xy))/(x^2cos(xy)-3x^3)+(3y)/(cos(xy)-3x))`

Answer 2

By working step-wise taking derivatives of individual components you should end up with
`(dy)/(dx) = (xycos(xy)-sin(xy)-3x^2y)/(3x^3-x^2cos(xy)`

Explanation:

This is complex so check carefully before assuming what follows is valid:

Breaking the equation up into small pieces and working on the derivatives for each:
Part 1: The `color(red)(xy)` component of `25=sin(color(red)(xy))/x-3xy`
Finding the derivative of `xy` is of the form for finding the derivative of `uv` and we know `(d(uv))/(dx) = v(du)/(dx)+u(dv)/(dx)`
so
`color(white)("XXX")(d(xy))/(dx) = y(cancel(dx))/(cancel(dx)) + x(dy)/(dx)`
`color(white)("XXXXXX")=y+x(dy)/(dx)`

Part 2: The `color(red)(sin(xy))` component of `25=color(red)(sin(xy))/x-3xy`
Finding the derivative of `sin(xy)` is of the form for finding the derivative of `f(g(u))` and we know
`color(white)("XXX")(df(g(x)))/(dx)=(df(g(x)))/(d(g(x))) * (dg(x))/(dx)`
with `f(x) = sin(x)` and `g(x)=xy`
so
`color(white)("XXX")(dsin(xy))/(dx) = cos(xy)*(y+x(dy)/(dx))`

Part 3: The `color(red)(sin(xy)/x)` component of `25= color(red)(sin(xy)/x)-3xy`
Finding the derivative of `sin(xy)/x` is of the form for finding the derivative of `u/v` and we know
`color(white)("XXX")d(u/v)/dx = (v(du)/(dx) -u(dv)/(dx))/(v^2)`
with `v=x` and `u=sin(xy)`
so
`color(white)("XXX")(d(sin(xy))/x)/(dx)= (x*(cos(xy)*(y+x(dy)/(dx)))-sin(xy)(cancel(dx))/(cancel(dx)))/(x^2)`

`color(white)("XXXXXXXX")=(cos(xy)(y+x(dy)/(dx)))/x - (sin(xy))/(x^2)`

Part 4: The `color(red)(xy)` component of `25 = sin(xy)/x - 3color(red)(xy)`
from Part 1 we already have
`color(white)("XXX")`(d(xy))/(dx) = y+x(dy)/(dx)#

Putting the pieces together
Take the derivative of both sides
`d(25)/(dx) = (d(sin(x)/x-3xy))/dx`

`0 = (dsin(x)/x)/dx-3((d(xy))/(dx))`

`(cos(xy)(y+x(dy)/(dx)))/x - (sin(xy))/(x^2)-3(y+x(dy)/(dx))=0`

`(ycos(xy))/x +cos(xy)(dy)/(dx)-sin(xy)/(x^2)-3y-3x(dy)/(dx)=0`

`(ycos(xy))/x-sin(xy)/(x^2)-3y = (3x-cos(xy))(dy)/(dx)`

`(dy)/(dx) = (xycos(xy)-sin(xy)-3x^2y)/(3x^3-x^2cos(xy)`