Q. 6 - Calculus

Question

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

(a) What is the relationship between the graph of $f\left(x\right)$ and the graph of the function of two variables, $g(x, y)=f\left(x\right)$ ?

(b) For what values of x and y do the first-order partial derivatives of g exist?

(c) What are $\frac{\partial g}{\partial x}$ and $\frac{\partial g}{\partial y}$ ? Why do these partial derivatives make sense?

Step-by-Step Solution

Answer

Ans: 

part (a). The graph of  $g(x, y)=f\left(x\right)$ can be said to be the solid surface formed by translating the graph of $f\left(x\right)$along y-axis.

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

part (c). $$\begin{gathered} \frac{\partial g}{\partial x}=f^{\text{'}}\left(y\right) \\ \frac{\partial g}{\partial y}= 0 \end{gathered}$$

Differentiating it partially with respect to ' y ' is actually differentiating a constant function.

Step 1. Given information:

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

Step 2. Solving part (a):

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.

Step 3. Continue:

Thus, it is a similar graph as $y=f\left(x\right)$. The difference lies in the fact that this graph is to be plotted between $xyz$-axis.

Since the variable ' y ' is not involved in the equation of the graph, it is clear that the graph is the same for each value of ' y '. Hence, the graph of $g(x, y)=f\left(x\right)$ can be said to be formed by repeating indefinitely the graph of $z=f\left(x\right)$ along y-axis.

Hence, the graph of $g(x, y)=f\left(x\right)$ can be said to be the solid surface formed by translating the graph of $f\left(x\right)$ along the y-axis.

Step 4. Solving part (b):

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

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

Step 5. Continue:

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

Step 6. Solving part (c):

(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 '.

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

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 '.

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

$$\begin{gathered} =\frac{\partial }{\partial y}\left(f\right(x\left)\right) \\ =0 \end{gathered}$$

Differentiating it partially with respect to ' y ' is actually differentiating a constant function.