Find y'' for siny=cos2x?

differentiated first stage giving me (2sin2x)/cosy

Mar 21, 2018

$\frac{{d}^{2} y}{{\mathrm{dx}}^{2}} = 4 \left\{\frac{{\sin}^{2} 2 x \setminus \sin y - {\cos}^{2} y \setminus \cos 2 x}{{\cos}^{3} y}\right\}$

Explanation:

Given that:

$\sin y = \cos 2 x$

We can differentiate Implicitly to get:

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

$\therefore \frac{\mathrm{dy}}{\mathrm{dx}} = - 2 \frac{\sin 2 x}{\cos} y$

And differentiating again, we get:

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

$\setminus \setminus \setminus \setminus \setminus \setminus \setminus = - 2 \left\{\frac{\left({\cos}^{2} y\right) \frac{2 \cos 2 x}{\cos} y + \left(\sin 2 x\right) \left(\sin y \left(- 2 \frac{\sin 2 x}{\cos} y\right)\right)}{\cos y} ^ 2\right\}$

$\setminus \setminus \setminus \setminus \setminus \setminus \setminus = - 4 \left\{\frac{{\cos}^{2} y \setminus \cos 2 x - {\sin}^{2} 2 x \setminus \sin y}{{\cos}^{3} y}\right\}$

$\setminus \setminus \setminus \setminus \setminus \setminus \setminus = 4 \left\{\frac{{\sin}^{2} 2 x \setminus \sin y - {\cos}^{2} y \setminus \cos 2 x}{{\cos}^{3} y}\right\}$