# How do you implicitly differentiate -1=xysin(x/y) ?

Mar 19, 2016

$x \cos \left(\frac{x}{y}\right) - {x}^{2} / y \cos \left(\frac{x}{y}\right) \frac{\mathrm{dy}}{\mathrm{dx}} + x \frac{\mathrm{dy}}{\mathrm{dx}} \sin \left(\frac{x}{y}\right) + y \sin \left(\frac{x}{y}\right) = 0$

#### Explanation:

Implicit differentiation is used when a function, here $y$, is not explicitly written in terms $x$. One will need the formula for differentiation of product & ratios of functions as well as chain formula to solve the given function.

For this, we take differential of both sides and using product formula gives us

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

Now differentiating $\frac{d}{\mathrm{dx}} \sin \left(\frac{x}{y}\right)$ in first term gives

$\cos \left(\frac{x}{y}\right) \frac{d}{\mathrm{dx}} \left(\frac{x}{y}\right) = \cos \left(\frac{x}{y}\right) \left[x \cdot \frac{d}{\mathrm{dx}} \left(\frac{1}{y}\right) + \frac{1}{y}\right]$

= $\cos \left(\frac{x}{y}\right) \left[x \cdot \left(- \frac{1}{y} ^ 2\right) \frac{\mathrm{dy}}{\mathrm{dx}} + \frac{1}{y}\right]$

= $\cos \left(\frac{x}{y}\right) \left\{\frac{1}{y} - \frac{x}{y} ^ 2 \frac{\mathrm{dy}}{\mathrm{dx}}\right\}$

Hence derivative of given function is

$x y \left[\cos \left(\frac{x}{y}\right) \left\{\frac{1}{y} - \frac{x}{y} ^ 2 \frac{\mathrm{dy}}{\mathrm{dx}}\right\}\right] + x \frac{\mathrm{dy}}{\mathrm{dx}} \sin \left(\frac{x}{y}\right) + y \sin \left(\frac{x}{y}\right) = 0$ or

$x \cos \left(\frac{x}{y}\right) - {x}^{2} / y \cos \left(\frac{x}{y}\right) \frac{\mathrm{dy}}{\mathrm{dx}} + x \frac{\mathrm{dy}}{\mathrm{dx}} \sin \left(\frac{x}{y}\right) + y \sin \left(\frac{x}{y}\right) = 0$