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

Mar 15, 2016

$\frac{\text{d"y}{"d} x}{=} - \frac{y}{x + \sin \left(y\right)}$

#### 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.

$\frac{\text{d"}{"d"x}(1) = frac{"d"}{"d} x}{x y - \cos \left(y\right)}$

$0 = \frac{\text{d"}{"d"x}(xy) - frac{"d"}{"d} x}{\cos \left(y\right)}$

= [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}

$\frac{\text{d"y}{"d} x}{=} - \frac{y}{x + \sin \left(y\right)}$