Question

How do you find `dy/dx` by implicit differentiation given `y=cos(x+y)`?

Answer 1

`dy/dx = -sin(x + y)/(1 + sin(x + y))`

Explanation:

Method 1: Expand and use the product rule

We expand using the formula `cos(A + B) = cosAcosB - sinAsinB`.

`y = cosxcosy - sinxsiny`

`d/dx(y) = d/dx(cosxcosy - sinxsiny)`

`d/dx(y) = d/dx(cosxcosy) - d/dx(sinxsiny)`

Remember that we are differentiating with respect to `x` here.

`1(dy/dx) = -sinx(cosy) - siny(dy/dx)(cosx) - (cosx(siny) + cosy(sinx)dy/dx)`

Solve for `dy/dx`:

`1(dy/dx) + sinycosx(dy/dx) + cosysinx(dy/dx) = -sinxcosy - cosxsiny`

`dy/dx(1 + sinycosx + cosysinx) = -(sinxcosy + cosxsiny)`

`dy/dx = -(sinxcosy + cosxsiny)/(1 + sinycosx + cosysinx)`

`dy/dx= -(sin(x + y))/(1 + sin(x + y))`

Method 2: Use the chain rule

Let `y= cosu` and `u = x + y`. Then `dy/(du) = -sinu` and `(du)/dx = 1 + dy/dx`.

The derivative is given by `color(magenta)(dy/dx= dy/(du) xx (du)/dx`.

`dy/dx = -sinu xx (1 + dy/dx)`

`dy/dx = -sinu - dy/dxsinu`

`dy/dx +dy/dxsinu = -sinu`

`dy/dx(1 +sinu) = -sinu`

`dy/dx = -sinu/(1+ sinu)`

Since `u = x+ y`:

`dy/dx = -sin(x + y)/(1 + sin(x + y))`

Hopefully this helps!