91Ó°ÊÓ

$f\left(x\right)$be a differentiable function of a single variable x.

The graph of$f\left(x\right)$ is given by the equation $y=f\left(x\right)$.

This graph is plotted between the xy-axis.

The graph of $g(x,y)=f\left(x\right)$ is given by the equation

$$\begin{gathered} z=g(x,y) \\ z=f\left(x\right) \end{gathered}$$

This appears to be the same graph of $f\left(x\right)$, but now on xz-plane.

(b) The partial derivative $\frac{∂g}{∂y}$of $g(x,y)$is determined by differentiating the function with respect to ' y ', keeping ' x ' as constant.

Hence, the derivative $\frac{∂g}{∂y}$exists for all points in the domain of $g(x,y)$, where ' x ' exists.

Similarly, the partial derivative $\frac{∂g}{∂x}$of $$ is determined by differentiating the function with respect to ' x ', keeping ' y ' as constant.

Since the function $g(x,y)=f\left(x\right)$ does not involve ' y ', the derivative will exist only if the function ' f ' can be differentiated for ' x '.

Hence, the derivative $\frac{∂g}{∂x}$exists for all points in the domain of $g(x,y)$, where $f^{\text{'}}\left(x\right)$ exists.

(c) The function $g(x,y)=f\left(x\right)$is a function in terms of only one variable ' x '.

Differentiating it partially with respect to ' x ' is actually differentiating it completely with respect to ' x '.

$$\begin{gathered} \frac{∂g}{∂x}=\frac{∂}{∂x}\left(g\right(x,y\left)\right) \\ =\frac{∂}{∂x}\left(f\right(x\left)\right) \\ =\frac{d}{dx}\left(f\right(x\left)\right) \\ =f^{\text{'}}\left(x\right) \end{gathered}$$

Differentiating it partially with respect to ' y ', means to treat ' x ' as constant.

The function $g(x,y)=f\left(x\right)$, is a function in terms of only one variable ' x '.

$\frac{∂g}{∂y}=\frac{∂}{∂y}\left(g\right(x,y\left)\right)$

role="math" localid="1650746484348" $$\begin{gathered} =\frac{∂}{∂y}\left(f\right(x\left)\right) \\ =0 \end{gathered}$$