# How do you find (d^2y)/(dx^2) given xsin2y=ycos2x?

May 7, 2018

We find using implicit differentiation twice. On way we find $\frac{\mathrm{dy}}{\mathrm{dx}}$ and substitute to get $\frac{{d}^{2} y}{{\mathrm{dx}}^{2}}$ as shown below.

#### Explanation:

We use implicit differentiation. Implicit differentiation is a special case of the chain rule for derivatives. Generally differentiation problems involve functions i.e. $y = f \left(x\right)$ - written explicitly as functions of $x$.

However, some functions y are written implicitly as functions of $x$. So what we do is to treat $y$ as $y = y \left(x\right)$ and use chain rule. This means differentiating $y$ w.r.t. $y$, but as we have to derive w.r.t. $x$, as per chain rule, we multiply it by $\frac{\mathrm{dy}}{\mathrm{dx}}$.

Here we are given $x \sin 2 y = y \cos 2 x$. Differentiating we get

$1 \cdot \sin 2 y + x \cdot 2 \cos 2 y \cdot \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\mathrm{dy}}{\mathrm{dx}} \cos 2 x - y \cdot 2 \sin 2 x$

or $\sin 2 y + 2 x \cos 2 y \frac{\mathrm{dy}}{\mathrm{dx}} = \cos 2 x \frac{\mathrm{dy}}{\mathrm{dx}} - 2 y \sin 2 x$ ...(A)

or $\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\sin 2 y + 2 y \sin 2 x}{\cos 2 x - 2 x \cos 2 y}$ ...(B)

Differentiating (A) again

$2 \cos 2 y \cdot \frac{\mathrm{dy}}{\mathrm{dx}} + 2 \cos 2 y \frac{\mathrm{dy}}{\mathrm{dx}} - 4 x \sin 2 y {\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}^{2} + 2 x \cos 2 y \frac{{d}^{2} y}{{\mathrm{dx}}^{2}} = - 2 \sin 2 x \frac{\mathrm{dy}}{\mathrm{dx}} + \cos 2 x \frac{{d}^{2} y}{{\mathrm{dx}}^{2}} - 4 y \cos 2 x - 2 \sin 2 x \frac{\mathrm{dy}}{\mathrm{dx}}$

or $\left(\cos 2 x - 2 x \cos 2 y\right) \frac{{d}^{2} y}{{\mathrm{dx}}^{2}} = \left(4 \cos 2 y + 4 \sin 2 x\right) \frac{\mathrm{dy}}{\mathrm{dx}} - 4 x \sin 2 y {\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}^{2} + 4 y \cos 2 x$

= $\left(4 \cos 2 y + 4 \sin 2 x\right) \frac{\sin 2 y + 2 y \sin 2 x}{\cos 2 x - 2 x \cos 2 y} - 4 x \sin 2 y {\left(\frac{\sin 2 y + 2 y \sin 2 x}{\cos 2 x - 2 x \cos 2 y}\right)}^{2} + 4 y \cos 2 x$

Here above we have put the value of $\frac{\mathrm{dy}}{\mathrm{dx}}$ from (B) and this can give $\frac{{d}^{2} y}{{\mathrm{dx}}^{2}}$. Remember to divide RHS by $\left(\cos 2 x - 2 x \cos 2 y\right)$.