# How do you implicitly differentiate y / [ x - 7y ] = x^4 + 9?

Jun 16, 2016

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{5 {x}^{4} - 28 {x}^{3} y + 9}{7 {x}^{4} + 64}$

#### Explanation:

$\frac{y}{x - 7 y} = {x}^{4} + 9$ $\Leftrightarrow y = \left({x}^{4} + 9\right) \left(x - 7 y\right)$

Hence $\frac{\mathrm{dy}}{\mathrm{dx}} = \left({x}^{4} + 9\right) \left(1 - 7 \frac{\mathrm{dy}}{\mathrm{dx}}\right) + 4 {x}^{3} \left(x - 7 y\right)$

or $\frac{\mathrm{dy}}{\mathrm{dx}} = {x}^{4} + 9 - 7 \frac{\mathrm{dy}}{\mathrm{dx}} \left({x}^{4} + 9\right) + 4 {x}^{4} - 28 {x}^{3} y$

or $\frac{\mathrm{dy}}{\mathrm{dx}} + 7 \frac{\mathrm{dy}}{\mathrm{dx}} \left({x}^{4} + 9\right) = 5 {x}^{4} - 28 {x}^{3} y + 9$

or $\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{5 {x}^{4} - 28 {x}^{3} y + 9}{1 + 7 \left({x}^{4} + 9\right)}$

or $\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{5 {x}^{4} - 28 {x}^{3} y + 9}{7 {x}^{4} + 64}$