Question

What is the implicit derivative of `1=xy-cosy`?

Answer 1

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

Explanation:

Implicit differentiation is basically an application of the chain rule.

`frac{"d"}{"d"x}(f(y)) = frac{"d"}{"d"y}(f(y)) frac{"d"y}{"d"x}`

`= f'(y) frac{"d"y}{"d"x}`

So in this problem, we take the derivative of `x` on both sides.

`frac{"d"}{"d"x}(1) = frac{"d"}{"d"x}(xy - cos(y))`

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

`= [yfrac{"d"}{"d"x}(x)+xfrac{"d"}{"d"x}(y)] - frac{"d"}{"d"y}(cos(y))frac{"d"y}{"d"x}`

`= y + x frac{"d"y}{"d"x} - (-sin(y))frac{"d"y}{"d"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.

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

`= (x + sin(y))frac{"d"y}{"d"x}`

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