# Let siny + cosx = 1 then find (d^2y)/(dx^2)?

May 30, 2018

$\frac{{d}^{2} y}{{\mathrm{dx}}^{2}} = \sin \frac{y}{\cos} y \setminus {\sin}^{2} \frac{x}{\cos} ^ 2 y + \cos \frac{x}{\cos} y$

#### Explanation:

We have:

$\sin y + \cos x = 1$

If we differentiate (implicitly) wrt $x$ we get:

$\cos y \setminus \frac{\mathrm{dy}}{\mathrm{dx}} - \sin x = 0$ .... [A]

If we differentiate (implicitly) wrt $x$ again whilst applying the product rule we get:

$\left(\cos y\right) \setminus \frac{{d}^{2} y}{{\mathrm{dx}}^{2}} - \left(\sin y \setminus \frac{\mathrm{dy}}{\mathrm{dx}}\right) \setminus \frac{\mathrm{dy}}{\mathrm{dx}} - \cos x = 0$

$\therefore \frac{{d}^{2} y}{{\mathrm{dx}}^{2}} - \sin \frac{y}{\cos} y \setminus {\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}^{2} - \cos \frac{x}{\cos} y = 0$

And, from [A] we get $\frac{\mathrm{dy}}{\mathrm{dx}} = \sin \frac{x}{\cos} y$, so substituting this result in the above we find that:

$\frac{{d}^{2} y}{{\mathrm{dx}}^{2}} - \sin \frac{y}{\cos} y \setminus {\left(\sin \frac{x}{\cos} y\right)}^{2} - \cos \frac{x}{\cos} y = 0$

$\therefore \frac{{d}^{2} y}{{\mathrm{dx}}^{2}} = \sin \frac{y}{\cos} y \setminus {\sin}^{2} \frac{x}{\cos} ^ 2 y + \cos \frac{x}{\cos} y$