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

Mar 16, 2016

$\frac{\text{d"y}{"d} x}{=} 1 - \cot \frac{y - x}{x}$

#### Explanation:

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

$\frac{\text{d"}{"d"x}(3) = frac{"d"}{"d} x}{\cos \frac{y - x}{x}}$

$0 = \frac{x \frac{\text{d"}{"d"x}( cos(y-x) ) - cos(y-x)frac{"d"}{"d} x}{x}}{{x}^{2}}$

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

$0 = x \sin \left(y - x\right) \left(\frac{\text{d"y}{"d} x}{-} 1\right) + \cos \left(y - x\right)$

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}

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

$= 1 - \cot \frac{y - x}{x}$