# How do you differentiate sin^2x/y^2-sin^2y/x^2=16?

Jul 31, 2016

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{x {\sin}^{2} x + 2 {x}^{2} \sin x \cos x - 16 x {y}^{2}}{16 {x}^{2} y + y {\sin}^{2} y + {y}^{2} \sin y \cos y}$

#### Explanation:

${\sin}^{2} \frac{x}{y} ^ 2 - {\sin}^{2} \frac{y}{x} ^ 2 = 16$ and multiplying both sides by ${x}^{2} {y}^{2}$

$\Leftrightarrow {x}^{2} {\sin}^{2} x - {y}^{2} {\sin}^{2} y = 16 {x}^{2} {y}^{2}$

Now differentiating both sides, with respect to $x$, note whenever we have a function of $y$, we differentiate w.r.t. $y$ and multiply by $\frac{\mathrm{dy}}{\mathrm{dx}}$

$2 x \times {\sin}^{2} x + {x}^{2} \times 2 \sin x \times \cos x - 2 y {\sin}^{2} y \times \frac{\mathrm{dy}}{\mathrm{dx}} - {y}^{2} \times 2 \sin y \times \cos y \times \frac{\mathrm{dy}}{\mathrm{dx}} = 16 \times 2 x \times {y}^{2} + 16 {x}^{2} \times 2 y \times \frac{\mathrm{dy}}{\mathrm{dx}}$ or

$2 x {\sin}^{2} x + 4 {x}^{2} \sin x \cos x - 2 y {\sin}^{2} y \frac{\mathrm{dy}}{\mathrm{dx}} - 2 {y}^{2} \sin y \cos y \frac{\mathrm{dy}}{\mathrm{dx}} = 32 x {y}^{2} + 32 {x}^{2} y \frac{\mathrm{dy}}{\mathrm{dx}}$ or

$32 {x}^{2} y \frac{\mathrm{dy}}{\mathrm{dx}} + 2 y {\sin}^{2} y \frac{\mathrm{dy}}{\mathrm{dx}} + 2 {y}^{2} \sin y \cos y \frac{\mathrm{dy}}{\mathrm{dx}} = 2 x {\sin}^{2} x + 4 {x}^{2} \sin x \cos x - 32 x {y}^{2}$ or

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{2 x {\sin}^{2} x + 4 {x}^{2} \sin x \cos x - 32 x {y}^{2}}{32 {x}^{2} y + 2 y {\sin}^{2} y + 2 {y}^{2} \sin y \cos y}$

= $\frac{x {\sin}^{2} x + 2 {x}^{2} \sin x \cos x - 16 x {y}^{2}}{16 {x}^{2} y + y {\sin}^{2} y + {y}^{2} \sin y \cos y}$