If $\mathit{y}\mathbf{=}\mathit{y}\mathit{z}$and $\mathit{y}\mathbf{=}\mathbf{2}\mathit{s}\mathit{i}\mathit{n}\left(y+z\right)$, find$\frac{\mathbf{d}\mathbf{x}}{\mathbf{d}\mathbf{y}}\mathit{a}\mathit{n}\mathit{d}\frac{{\mathbf{d}}^{\mathbf{2}}\mathbf{x}}{\mathbf{d}{\mathbf{y}}^{\mathbf{2}}}$ .

If any function $v=v\left(x,y\right)$has independent variables which are also functions of some independent variables such that $x=x(s,t)andy=y(s,t)$, then chain rule for the function is given by:

$$\begin{gathered} \frac{\partial v}{\partial s}=\frac{\partial v}{\partial x}\frac{\partial x}{\partial s}+\frac{\partial v}{\partial y}\frac{\partial y}{\partial s} \\ \frac{\partial v}{\partial t}=\frac{\partial v}{\partial x}\frac{\partial x}{\partial t}+\frac{\partial v}{\partial y}\frac{\partial y}{\partial t} \end{gathered}$$

The given functions are: $x=yz$and $y=2\sin (y+z)$.

Differentiate $x=yz$as follows:

$$\begin{gathered} x=yz \\ dx=ydz+zdy.........\left(1\right) \end{gathered}$$

Similarly, differentiate $y=2\sin (y+z)$as follows:

localid="1664263161407" $$\begin{gathered} y=2\sin (y+z) \\ dy=2\cos (y+z)dy+2\cos (y+z)dz \\ dz=\frac{1-2\cos (y+z)}{2\cos (y+z)}dy \\ dz=\left[\frac{1}{2\cos (y+z)}-1\right]dy...........\left(2\right) \end{gathered}$$

From equations (1) and (2):

Now, $\frac{d^{2}x}{d{y}^2}=\frac{d}{dy}\left[\frac{dx}{dy}\right]=\frac{d}{dy}\left[\tan \left(y+Z\right)-y+z\right]$

Evaluating these, we get:

$$\begin{gathered} \frac{d^{2}x}{d{y}^2}=se{c}^2(y+z)\left[1+\frac{dz}{dy}\right]-1+\frac{dz}{dy} \\ =se{c}^2(y+z)\left[1+\frac{1}{2\cos (y+z)}-1\right]-1+\frac{1}{2\cos (y+z)}-1 \\ =\frac{1}{2}se{c}^3(y+z)+\frac{1}{2}sec(y+z)-2 \end{gathered}$$

Hence, the required answers are:

$$\begin{gathered} \frac{dx}{dy}=\tan \left(y+z\right)-y+z \\ \frac{d^{2}x}{d{y}^2}=\frac{1}{2}se{c}^3(y+z)+\frac{1}{2}sec(y+z)-2 \end{gathered}$$