Question

How do you implicitly differentiate `-3=cos(y-x)/x`?

Answer 1

`frac{"d"y}{"d"x} = 1 - cot(y-x)/x`

Explanation:

Differentiate both sides w.r.t. `x`. Use the quotient rule and the chain rule along the way.

`frac{"d"}{"d"x}(3) = frac{"d"}{"d"x}( cos(y-x)/x )`

`0 = frac{ xfrac{"d"}{"d"x}( cos(y-x) ) - cos(y-x)frac{"d"}{"d"x}(x) }{x^2}`

`0 = x(-sin(y-x))frac{"d"}{"d"x}(y-x) - cos(y-x)`

`0 = xsin(y-x)(frac{"d"y}{"d"x}-1) + cos(y-x)`

Now, we just have to make `frac{"d"y}{"d"x}` the subject of formula. Begin by bringing all the terms containing `frac{"d"y}{"d"x}` to one side.

`xsin(y-x) - cos(y-x) = xsin(y-x)frac{"d"y}{"d"x}`

`frac{"d"y}{"d"x} = frac{xsin(y-x) - cos(y-x)}{xsin(y-x)}`

`= 1 - cot(y-x)/x`