Q. 7 - Calculus

Question

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

• (a) What is the relationship between the graph of $f\left(y\right)$ and the graph of the function of two variables, $g(x, y)=f\left(y\right)$ ?
• (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 $g(x, y)=f\left(y\right)$ can be said to be the solid surface formed by translating the graph of $f\left(y\right)$ along x-axis.

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

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

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

Step 1. Given information:

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

Step 2. Solving part (a):

• .The graph of $f\left(y\right)$ is given by the equation $x=f\left(y\right)$.
• This graph is plotted between the xy-axis.
• The graph of $g(x, y)=f\left(y\right)$ is given by the equation

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

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

Step 3. Continue:

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

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

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

Step 4. Solving part (b):

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

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

Step 5. Continue:

Similarly, 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.

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

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

Step 6. Solving part (c):

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

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

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

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

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

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

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

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