# How do you implicitly differentiate 7xy- 3 lny= 4/x?

Dec 29, 2015

$\frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{7 {x}^{2} {y}^{2} + 4 y}{7 {x}^{3} y - 3 {x}^{2}}$

#### Explanation:

Find the derivative of each part.

Use product rule:

$\frac{d}{\mathrm{dx}} \left(7 x y\right) = 7 y + 7 x y '$

Recall that $\frac{d}{\mathrm{dx}} \left(\ln u\right) = \frac{u '}{u}$ through the chain rule:

$\frac{d}{\mathrm{dx}} \left(- 3 \ln y\right) = - \frac{3 y '}{y}$

Rewrite $\frac{4}{x}$ as $4 {x}^{-} 1$.

$\frac{d}{\mathrm{dx}} \left(4 {x}^{-} 1\right) = - 4 {x}^{-} 2 = - \frac{4}{x} ^ 2$

Add all these derivatives to find the complete differentiated expression, then solve for $y '$.

$7 y + 7 x y ' - \frac{3 y '}{y} = - \frac{4}{x} ^ 2$

Multiply everything by ${x}^{2} y$ to clear fractions.

$7 {x}^{2} {y}^{2} + 7 {x}^{3} y \left(y '\right) - 3 {x}^{2} y ' = - 4 y$

$y ' \left(7 {x}^{3} y - 3 {x}^{2}\right) = - 7 {x}^{2} {y}^{2} - 4 y$

$y ' = - \frac{7 {x}^{2} {y}^{2} + 4 y}{7 {x}^{3} y - 3 {x}^{2}} = \frac{\mathrm{dy}}{\mathrm{dx}}$