# What is the implicit derivative of 25=sin(xy)/(3xy)?

May 7, 2017

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

#### Explanation:

Given: $25 = \sin \frac{x y}{3 x y}$

Differentiate all of the terms:

$\frac{d \left(25\right)}{\mathrm{dx}} = \frac{d \left(\sin \frac{x y}{3 x y}\right)}{\mathrm{dx}}$

The derivative of a constant is 0

$0 = \frac{d \left(\sin \frac{x y}{3 x y}\right)}{\mathrm{dx}}$

Use the quotient rule:

$0 = \frac{d \left(\sin \frac{x y}{3 x y}\right)}{\mathrm{dx}} = \frac{\frac{d \left(\sin \left(x y\right)\right)}{\mathrm{dx}} 3 x y - \sin \left(x y\right) \frac{d \left(3 x y\right)}{\mathrm{dx}}}{9 {x}^{2} {y}^{2}}$

Using the chain rule, $\frac{d \left(\sin \left(x y\right)\right)}{\mathrm{dx}} = \cos \left(x y\right) \frac{d \left(x y\right)}{\mathrm{dx}}$

$0 = \frac{\cos \left(x y\right) \frac{d \left(x y\right)}{\mathrm{dx}} 3 x y - \sin \left(x y\right) \frac{d \left(3 x y\right)}{\mathrm{dx}}}{9 {x}^{2} {y}^{2}}$

Use the linear property of the derivative:

$0 = \frac{\cos \left(x y\right) \frac{d \left(x y\right)}{\mathrm{dx}} 3 x y - 3 \sin \left(x y\right) \frac{d \left(x y\right)}{\mathrm{dx}}}{9 {x}^{2} {y}^{2}}$

Factor the $\frac{d \left(x y\right)}{\mathrm{dx}}$ out of the numerator:

$0 = \frac{\cos \left(x y\right) 3 x y - 3 \sin \left(x y\right)}{9 {x}^{2} {y}^{2}} \frac{d \left(x y\right)}{\mathrm{dx}}$

$0 = \frac{\cos \left(x y\right) 3 x y - 3 \sin \left(x y\right)}{9 {x}^{2} {y}^{2}} \left(y + x \frac{\mathrm{dy}}{\mathrm{dx}}\right)$

Eliminate the leading coefficient by dividing it into 0:

$0 = y + x \frac{\mathrm{dy}}{\mathrm{dx}}$

Subtract y from both sides

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

Divide both sides by x:

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