# How do you find dy/dx by implicit differentiation of y=sin(xy)?

Apr 16, 2017

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

,OR,

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{{y}^{2} \sqrt{1 - {y}^{2}}}{y - \sqrt{1 - {y}^{2}} a r c \sin y} .$

#### Explanation:

$y = \sin \left(x y\right) .$

$\therefore \frac{\mathrm{dy}}{\mathrm{dx}} , \text{ using the Chain Rule,}$

$= \frac{d}{\mathrm{dx}} \left(\sin \left(x y\right)\right) = \left\{\cos \left(x y\right)\right\} \left\{\frac{d}{\mathrm{dx}} \left(x y\right)\right\} , \text{ &, using the Product Rule,}$

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

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

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

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

Otherwise, $y = \sin \left(x y\right) \Rightarrow a r c \sin y = x y , \mathmr{and} , x = \frac{a r c \sin y}{y} .$

Hence, diff.ing both sides w.r.t. $y ,$ we have, by the Quotient Rule,

$\frac{\mathrm{dx}}{\mathrm{dy}} = \frac{y \cdot \frac{d}{\mathrm{dy}} \left(a r c \sin y\right) - \left(a r c \sin y\right) \cdot \frac{d}{\mathrm{dy}} \left(y\right)}{y} ^ 2 ,$

$= \frac{y \cdot \left(\frac{1}{\sqrt{1 - {y}^{2}}}\right) - \left(a r c \sin y\right) \cdot 1}{y} ^ 2 ,$

$= \frac{y - \sqrt{1 - {y}^{2}} a r c \sin y}{{y}^{2} \cdot \sqrt{1 - {y}^{2}}} ,$

Therefore, $\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{{y}^{2} \sqrt{1 - {y}^{2}}}{y - \sqrt{1 - {y}^{2}} a r c \sin y} .$

I leave it to the Questioner to show that both Answers tally with each other.

Enjoy Maths.!