91Ó°ÊÓ

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:

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

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:

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