Question

How do I find the solution to the following problem using Implicit Differentiation?

`x sin(y) +y sin (x) =2`

Answer 1

`dy/dx=-(siny+ycosx)/(xcosy+sinx)`

Explanation:

`"assuming you require "dy/dx`

`color(blue)"differentiate implicitly with respect to x"`

`"differentiate "xsiny" and "ysinx" using the "color(blue)"product rule"`

`[xcosydy/dx+siny]+[ycosx+sinxdy/dx]=0`

`rArrdy/dx(xcosy+sinx)=-siny-ycosx`

`rArrdy/dx=-(siny+ycosx)/(xcosy+sinx)`

Answer 2

Given: `x sin(y) +y sin (x) =2`

Differentiate each term with respect to x:

`(d(x sin(y)))/dx +(d(y sin (x)))/dx =(d(2))/dx`

Use the product rule on the first term:

`(d(uv))/dx= (du)/dxv+u(dv)/dx`

where `u = x` and `v = sin(y)`, then `(du)/dx = 1` but we must use the chain rule for `(dv)/dx = (d(sin(y)))/dy dy/dx = cos(y)dy/dx`

Substituting into the product rule:

`(d(xsin(y)))/dx= sin(y)+xcos(y)dy/dx`

Substituting this into the equation:

`sin(y)+xcos(y)dy/dx+(d(y sin (x)))/dx =(d(2))/dx`

Use the product rule on the next term:

`(d(uv))/dx= (du)/dxv+u(dv)/dx`

where `u = y` and `v = sin(x)`, then `(du)/dx = dy/dx` and `(dv)/dx = cos(x)`

Substituting into the product rule:

`(d(ysin(x)))/dx= (dy)/dxsin(x)+ycos(x)`

Substituting this into the equation:

`sin(y)+xcos(y)dy/dx+(dy)/dxsin(x)+ycos(x) =(d(2))/dx`

The derivative of a constant is 0:

`sin(y)+xcos(y)dy/dx+(dy)/dxsin(x)+ycos(x) =0`

Move all of the terms that do not contain `dy/dx` to the right:

`xcos(y)dy/dx+(dy)/dxsin(x)=-(ycos(x)+sin(y))`

Factor out `dy/dx`:

`(xcos(y)+sin(x))(dy)/dx=-(ycos(x)+sin(y))`

Divide both sides by the leading coefficient:

`dy/dx=-(ycos(x)+sin(y))/(xcos(y)+sin(x))`