# What is the implicit derivative of 1=xysinxy?

##### 1 Answer
Jun 23, 2017

$\frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{y}{x}$

#### Explanation:

We have a product of three functions. The product rule for three functions is essentially same for that with only two functions:

$\frac{d}{\mathrm{dx}} \left(u v w\right) = \frac{\mathrm{du}}{\mathrm{dx}} v w + u \frac{\mathrm{dv}}{\mathrm{dx}} w + u v \frac{\mathrm{dw}}{\mathrm{dx}}$

So here, we get:

$\frac{d}{\mathrm{dx}} \left(1\right) = \frac{d}{\mathrm{dx}} \left(x y \sin x y\right)$

$0 = \left(\frac{d}{\mathrm{dx}} x\right) y \sin x y + x \left(\frac{d}{\mathrm{dx}} y\right) \sin x y + x y \left(\frac{d}{\mathrm{dx}} \sin x y\right)$

The derivative of $\sin x y$ will require the chain rule:

$0 = y \sin x y + x \frac{\mathrm{dy}}{\mathrm{dx}} \sin x y + x y \cos x y \left(\frac{d}{\mathrm{dx}} x y\right)$

The derivative of $x y$ requires the product rule:

$0 = y \sin x y + x \sin x y \frac{\mathrm{dy}}{\mathrm{dx}} + x y \cos x y \left[\left(\frac{d}{\mathrm{dx}} x\right) y + x \left(\frac{d}{\mathrm{dx}} y\right)\right]$

$0 = y \sin x y + x \sin x y \frac{\mathrm{dy}}{\mathrm{dx}} + x y \cos x y \left(y + x \frac{\mathrm{dy}}{\mathrm{dx}}\right)$

Expanding:

$0 = y \sin x y + x \sin x y \frac{\mathrm{dy}}{\mathrm{dx}} + x {y}^{2} \cos x y + {x}^{2} y \cos x y \frac{\mathrm{dy}}{\mathrm{dx}}$

Isolating the terms with the derivative, $\frac{\mathrm{dy}}{\mathrm{dx}}$:

$- y \sin x y - x {y}^{2} \cos x y = \frac{\mathrm{dy}}{\mathrm{dx}} \left(x \sin x y + {x}^{2} y \cos x y\right)$

Thus:

$\frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{y \sin x y + x {y}^{2} \cos x y}{x \sin x y + {x}^{2} y \cos x y}$

Or:

$\frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{y \left(\sin x y + x y \cos x y\right)}{x \left(\sin x y + x y \cos x y\right)}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{y}{x}$