Question #bf859

Nov 7, 2016

$\frac{{d}^{2} y}{\mathrm{dx}} ^ 2 = \frac{4 {y}^{2} {\sec}^{2} \left(2 x\right) \tan \left(2 x\right) + {y}^{2} - {\sec}^{4} \left(2 x\right) - 2 x {\sec}^{2} \left(2 x\right) - {x}^{2}}{y} ^ 3$

Explanation:

${y}^{2} - {x}^{2} = \tan 2 x$

$2 y \frac{\mathrm{dy}}{\mathrm{dx}} - 2 x = 2 {\sec}^{2} \left(2 x\right)$

$2 y \frac{\mathrm{dy}}{\mathrm{dx}} = 2 {\sec}^{2} \left(2 x\right) + 2 x$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{2 {\sec}^{2} \left(2 x\right) + 2 x}{2 y}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{{\sec}^{2} \left(2 x\right) + x}{y}$

You can use the quotient rule or just do implicit differentiation again and then substitute in $\frac{{\sec}^{2} \left(2 x\right) + x}{y}$ for $\frac{\mathrm{dy}}{\mathrm{dx}}$

$\frac{{d}^{2} y}{\mathrm{dx}} ^ 2 = \frac{4 \sec \left(2 x\right) \sec \left(2 x\right) \tan \left(2 x\right) + 1}{y} - \frac{{\sec}^{2} \left(2 x\right) + x}{y} ^ 2 \frac{\mathrm{dy}}{\mathrm{dx}}$

$\frac{{d}^{2} y}{\mathrm{dx}} ^ 2 = \frac{4 \sec \left(2 x\right) \sec \left(2 x\right) \tan \left(2 x\right) + 1}{y} - \frac{{\sec}^{2} \left(2 x\right) + x}{y} ^ 2 \cdot \frac{{\sec}^{2} \left(2 x\right) + x}{y}$

$\frac{{d}^{2} y}{\mathrm{dx}} ^ 2 = \frac{4 {\sec}^{2} \left(2 x\right) \tan \left(2 x\right) + 1}{y} - \frac{{\sec}^{4} \left(2 x\right) + 2 x {\sec}^{2} \left(2 x\right) + {x}^{2}}{y} ^ 3$

$\frac{{d}^{2} y}{\mathrm{dx}} ^ 2 = \frac{4 {y}^{2} {\sec}^{2} \left(2 x\right) \tan \left(2 x\right) + {y}^{2} - {\sec}^{4} \left(2 x\right) - 2 x {\sec}^{2} \left(2 x\right) - {x}^{2}}{y} ^ 3$