Q. 10 - Calculus

Question

Two operations commute if they can be done in either order. Does multiplying a function by a constant commute with differentiation? Does adding two functions commute with differentiation? What about products and quotients and the operation of differentiation?

Step-by-Step Solution

Answer

$$\begin{gathered} \mathrm{The} \mathrm{multiplication} \mathrm{of} \mathrm{a} \mathrm{function} \mathrm{is} \mathrm{commute} \mathrm{with} \mathrm{differentiation}. \\ \mathrm{The} \mathrm{addition} \mathrm{of} \mathrm{two} \mathrm{functions} \mathrm{is} \mathrm{commute} \mathrm{with} \mathrm{differentiation}. \\ \mathrm{The} \mathrm{multiplication} \mathrm{of} \mathrm{two} \mathrm{functions} \mathrm{is} \mathrm{commute} \mathrm{with} \mathrm{differentiation}. \\ \mathrm{The} \mathrm{division} \mathrm{of} \mathrm{two} \mathrm{functions} \mathrm{is} \mathrm{not} \mathrm{commute} \mathrm{with} \mathrm{differentiation}. \end{gathered}$$

Step 1. Given information is:

Two operations commute if they can be done in either order. 

Step 2. Checking when constant is multiplied to a function

$$\begin{gathered} \mathrm{Consider}: \mathrm{f}\left(\mathrm{x}\right) = \mathrm{k} \mathrm{g}\left(\mathrm{x}\right) \\ \mathrm{To} \mathrm{perform} \mathrm{the} \mathrm{differentiation} \mathrm{follows} \mathrm{the} \mathrm{steps}: \\ \frac{d\left[\mathrm{f}\left(\mathrm{x}\right)\right]}{d\mathrm{x}}=\frac{d\left[\mathrm{k} \mathrm{g}\left(\mathrm{x}\right)\right]}{d\mathrm{x}} \\ \frac{d\left[\mathrm{f}\left(\mathrm{x}\right)\right]}{d\mathrm{x}}=\mathrm{k}\frac{d\left[\mathrm{g}\left(\mathrm{x}\right)\right]}{d\mathrm{x}} \\ \frac{d\left[\mathrm{f}\left(\mathrm{x}\right)\right]}{d\mathrm{x}}=\mathrm{k} \mathrm{g}\text{'}\left(\mathrm{x}\right) \\ \mathrm{Again} \\ \frac{d\left[\mathrm{f}\left(\mathrm{x}\right)\right]}{d\mathrm{x}}=\frac{d\left[\mathrm{g}\left(\mathrm{x}\right) \mathrm{k}\right]}{d\mathrm{x}} \\ \frac{d\left[\mathrm{f}\left(\mathrm{x}\right)\right]}{d\mathrm{x}}=\frac{d\left[\mathrm{k} \mathrm{g}\left(\mathrm{x}\right)\right]}{d\mathrm{x}} \\ \frac{d\left[\mathrm{f}\left(\mathrm{x}\right)\right]}{d\mathrm{x}}=\mathrm{k} \mathrm{g}\text{'}\left(\mathrm{x}\right) \\ \mathrm{Thus} \mathrm{the} \mathrm{multiplication} \mathrm{of} \mathrm{a} \mathrm{function} \mathrm{is} \mathrm{commute} \mathrm{with} \mathrm{differentiation}. \end{gathered}$$

Step 3. Checking when two functions are added

$$\begin{gathered} \mathrm{Consider}: \mathrm{f}\left(\mathrm{x}\right) = \mathrm{g}\left(\mathrm{x}\right) + \mathrm{h}\left(\mathrm{x}\right) \\ \mathrm{To} \mathrm{perform} \mathrm{the} \mathrm{differentiation} \mathrm{follows} \mathrm{the} \mathrm{steps}: \\ \frac{d\left[\mathrm{f}\left(\mathrm{x}\right)\right]}{d\mathrm{x}}=\frac{d\left[\mathrm{g}\left(\mathrm{x}\right) + \mathrm{h}\left(\mathrm{x}\right)\right]}{d\mathrm{x}} \\ \frac{d\left[\mathrm{f}\left(\mathrm{x}\right)\right]}{d\mathrm{x}}=\frac{d\left[\mathrm{g}\left(\mathrm{x}\right)\right]}{d\mathrm{x}} +\frac{d\left[\mathrm{h}\left(\mathrm{x}\right)\right]}{d\mathrm{x}} \\ \frac{d\left[\mathrm{f}\left(\mathrm{x}\right)\right]}{d\mathrm{x}}=\mathrm{g}\text{'}\left(\mathrm{x}\right) + \mathrm{h}\text{'}\left(\mathrm{x}\right) \\ \mathrm{Again} \\ \frac{d\left[\mathrm{f}\left(\mathrm{x}\right)\right]}{d\mathrm{x}}=\frac{d\left[\mathrm{h}\left(\mathrm{x}\right) + \mathrm{g}\left(\mathrm{x}\right)\right]}{d\mathrm{x}} \\ \frac{d\left[\mathrm{f}\left(\mathrm{x}\right)\right]}{d\mathrm{x}}=\frac{d\left[\mathrm{h}\left(\mathrm{x}\right)\right]}{d\mathrm{x}} +\frac{d\left[\mathrm{g}\left(\mathrm{x}\right)\right]}{d\mathrm{x}} \\ \frac{d\left[\mathrm{f}\left(\mathrm{x}\right)\right]}{d\mathrm{x}}=\mathrm{h}\text{'}\left(\mathrm{x}\right) + \mathrm{g}\text{'}\left(\mathrm{x}\right) \\ \frac{d\left[\mathrm{f}\left(\mathrm{x}\right)\right]}{d\mathrm{x}}=\mathrm{g}\text{'}\left(\mathrm{x}\right) + \mathrm{h}\text{'}\left(\mathrm{x}\right) \\ \mathrm{Thus} \mathrm{the} \mathrm{addition} \mathrm{of} \mathrm{two} \mathrm{functions} \mathrm{is} \mathrm{commute} \mathrm{with} \mathrm{differentiation}. \end{gathered}$$

Step 4. Checking when two functions are multiplied

$$\begin{gathered} \mathrm{Consider}: \mathrm{f}\left(\mathrm{x}\right) = \mathrm{g}\left(\mathrm{x}\right) . \mathrm{h}\left(\mathrm{x}\right) \\ \mathrm{To} \mathrm{perform} \mathrm{the} \mathrm{differentiation} \mathrm{follows} \mathrm{the} \mathrm{steps}: \\ \frac{d\left[\mathrm{g}\left(\mathrm{x}\right) . \mathrm{h}\left(\mathrm{x}\right)\right]}{d\mathrm{x}}=\frac{d\left[\mathrm{g}\left(\mathrm{x}\right)\right]}{d\mathrm{x}} \mathrm{h}\left(\mathrm{x}\right) + \frac{d\left[\mathrm{h}\left(\mathrm{x}\right)\right]}{d\mathrm{x}} \mathrm{g}\left(\mathrm{x}\right) \\ \frac{d\left[\mathrm{g}\left(\mathrm{x}\right) . \mathrm{h}\left(\mathrm{x}\right)\right]}{d\mathrm{x}}=\frac{d\left[\mathrm{h}\left(\mathrm{x}\right)\right]}{d\mathrm{x}} \mathrm{g}\left(\mathrm{x}\right) + \frac{d\left[\mathrm{g}\left(\mathrm{x}\right)\right]}{d\mathrm{x}} \mathrm{h}\left(\mathrm{x}\right) \\ \frac{d\left[\mathrm{g}\left(\mathrm{x}\right) . \mathrm{h}\left(\mathrm{x}\right)\right]}{d\mathrm{x}} = \frac{d\left[\mathrm{h}\left(\mathrm{x}\right) . \mathrm{g}\left(\mathrm{x}\right)\right]}{d\mathrm{x}} \\ \mathrm{Thus} \mathrm{the} \mathrm{multiplication} \mathrm{of} \mathrm{two} \mathrm{functions} \mathrm{is} \mathrm{commute} \mathrm{with} \mathrm{differentiation}. \end{gathered}$$

Step 5. Checking when two functions are divided

$$\begin{gathered} \mathrm{Consider}: \mathrm{f}\left(\mathrm{x}\right) = \frac{\mathrm{g}\left(\mathrm{x}\right)}{ \mathrm{h}\left(\mathrm{x}\right)} \\ \mathrm{To} \mathrm{perform} \mathrm{the} \mathrm{differentiation} \mathrm{follows} \mathrm{the} \mathrm{steps}: \\ \frac{d\left[\frac{\mathrm{g}\left(\mathrm{x}\right)}{ \mathrm{h}\left(\mathrm{x}\right)}\right]}{d\mathrm{x}}=\frac{\frac{d\left[\mathrm{g}\left(\mathrm{x}\right)\right]}{d\mathrm{x}} \mathrm{h}\left(\mathrm{x}\right) - \frac{d\left[\mathrm{h}\left(\mathrm{x}\right)\right]}{d\mathrm{x}} \mathrm{g}\left(\mathrm{x}\right)}{{\left[\mathrm{h}\left(\mathrm{x}\right)\right]}^2} \\ \frac{d\left[\frac{\mathrm{g}\left(\mathrm{x}\right)}{ \mathrm{h}\left(\mathrm{x}\right)}\right]}{d\mathrm{x}}=\frac{\mathrm{g}\text{'}\left(\mathrm{x}\right) \mathrm{h}\left(\mathrm{x}\right) - \mathrm{h}\text{'}\left(\mathrm{x}\right) \mathrm{g}\left(\mathrm{x}\right)}{{\left[\mathrm{h}\left(\mathrm{x}\right)\right]}^2} \\ \mathrm{Again} \\ \frac{d\left[\frac{\mathrm{h}\left(\mathrm{x}\right)}{ \mathrm{g}\left(\mathrm{x}\right)}\right]}{d\mathrm{x}}=\frac{\frac{d\left[\mathrm{h}\left(\mathrm{x}\right)\right]}{d\mathrm{x}} \mathrm{g}\left(\mathrm{x}\right) - \frac{d\left[\mathrm{g}\left(\mathrm{x}\right)\right]}{d\mathrm{x}} \mathrm{h}\left(\mathrm{x}\right)}{{\left[\mathrm{g}\left(\mathrm{x}\right)\right]}^2} \\ \frac{d\left[\frac{\mathrm{h}\left(\mathrm{x}\right)}{ \mathrm{g}\left(\mathrm{x}\right)}\right]}{d\mathrm{x}}=\frac{\mathrm{h}\text{'}\left(\mathrm{x}\right) \mathrm{g}\left(\mathrm{x}\right) - \mathrm{g}\text{'}\left(\mathrm{x}\right) \mathrm{h}\left(\mathrm{x}\right)}{{\left[\mathrm{g}\left(\mathrm{x}\right)\right]}^2} \\ \mathrm{Thus} \mathrm{the} \mathrm{division} \mathrm{of} \mathrm{two} \mathrm{functions} \mathrm{is} \mathrm{not} \mathrm{commute} \mathrm{with} \mathrm{differentiation}. \end{gathered}$$