Question

Prove the product rule for derivatives of complex valued function.

Short answer

The solution is $\frac{d}{dt}\left(z_1{z}_2\right)=z_2\frac{d}{dt}\left(z_1\right)+z_1\frac{d}{dt}\left(z_2\right)$.

Step-by-step solution

Simplify the complex function z1(t)z2(t). 

Consider the complex valued functions$z_1\left(t\right)$and$z_2\left(t\right)$where$z_1\left(t\right)=x_1\left(t\right)+i{y}_1\left(t\right)$and$z_2\left(t\right)=x_2\left(t\right)+i{y}_2\left(t\right)$.

Multiply the equations $z_1\left(t\right)=x_1\left(t\right)+i{y}_1\left(t\right)$and$z_2\left(t\right)=x_2\left(t\right)+i{y}_2\left(t\right)$ as follows.

$\begin{array}{c}{z}_1\left(t\right)z_2\left(t\right)=\{x_1\left(t\right)+i{y}_1\left(t\right)\}\{x_2\left(t\right)+i{y}_2\left(t\right)\}\\ z_1\left(t\right)z_2\left(t\right)=x_1\left(t\right)x_2\left(t\right)-y_1\left(t\right)y_2\left(t\right)+i\{x_1\left(t\right)y_2\left(t\right)+x_2\left(t\right)y_1\left(t\right)\}\end{array}$

Determine the value of z2(t)dz1(t)dt+z1(t)dz2(t)dt . 

Differentiate the equation$z_1\left(t\right)=x_1\left(t\right)+i{y}_1\left(t\right)$ both side with respect to $t$as follows.

$\begin{array}{c}{z}_1\left(t\right)=x_1\left(t\right)+i{y}_1\left(t\right)\\ \frac{d{z}_1\left(t\right)}{dt}=\frac{d{x}_1\left(t\right)}{dt}+i\frac{d{y}_1\left(t\right)}{dt}\end{array}$

Differentiate the equation $z_2\left(t\right)=x_2\left(t\right)+i{y}_2\left(t\right)$both side with respect t to as follows.

$\begin{array}{c}{z}_2\left(t\right)=x_2\left(t\right)+i{y}_2\left(t\right)\\ \frac{d{z}_2\left(t\right)}{dt}=\frac{d{x}_2\left(t\right)}{dt}+i\frac{d{y}_2\left(t\right)}{dt}\end{array}$

Substitute the values $x_1\left(t\right)+i{y}_1\left(t\right)$for$z_1\left(t\right)$ ,$\frac{d{x}_1\left(t\right)}{dt}+i\frac{d{y}_1\left(t\right)}{dt}$ for $\frac{d{z}_1\left(t\right)}{dt}$, $x_2\left(t\right)+i{y}_2\left(t\right)$for$z_2\left(t\right)$ and$\frac{d{x}_2\left(t\right)}{dt}+i\frac{d{y}_2\left(t\right)}{dt}$ for$\frac{d{z}_2\left(t\right)}{dt}$ in $z_2\left(t\right)\frac{d}{dt}\left(z_1\left(t\right)\right)+z_1\left(t\right)\frac{d}{dt}\left(z_2\left(t\right)\right)$as follows.

$\begin{array}{l}{z}_2\left(t\right)\frac{d}{dt}\left(z_1\left(t\right)\right)+z_1\left(t\right)\frac{d}{dt}\left(z_2\left(t\right)\right)=\left[\begin{array}{l}\{x_2\left(t\right)+i{y}_2\left(t\right)\}\{\frac{d{x}_1\left(t\right)}{dt}+i\frac{d{y}_1\left(t\right)}{dt}\}\\ +\{x_1\left(t\right)+i{y}_1\left(t\right)\}\{\frac{d{x}_2\left(t\right)}{dt}+i\frac{d{y}_2\left(t\right)}{dt}\}\end{array}\right]\\ z_2\left(t\right)\frac{d}{dt}\left(z_1\left(t\right)\right)+z_1\left(t\right)\frac{d}{dt}\left(z_2\left(t\right)\right)=\left[\begin{array}{l}\{x_2\left(t\right)\frac{d{x}_1\left(t\right)}{dt}+i{x}_2\left(t\right)\frac{d{y}_1\left(t\right)}{dt}+i{y}_2\left(t\right)\frac{d{x}_1\left(t\right)}{dt}-y_2\left(t\right)\frac{d{y}_1\left(t\right)}{dt}\}\\ +\{x_1\left(t\right)\frac{d{x}_2\left(t\right)}{dt}+i{x}_1\left(t\right)\frac{d{y}_2\left(t\right)}{dt}+i{y}_1\left(t\right)\frac{d{x}_2\left(t\right)}{dt}-y_1\left(t\right)\frac{d{y}_2\left(t\right)}{dt}\}\end{array}\right]\\ z_2\left(t\right)\frac{d}{dt}\left(z_1\left(t\right)\right)+z_1\left(t\right)\frac{d}{dt}\left(z_2\left(t\right)\right)=\left[\begin{array}{l}\{x_2\left(t\right)\frac{d{x}_1\left(t\right)}{dt}-y_2\left(t\right)\frac{d{y}_1\left(t\right)}{dt}+i\{x_2\left(t\right)\frac{d{y}_1\left(t\right)}{dt}+y_2\left(t\right)\frac{d{x}_1\left(t\right)}{dt}\}\}\\ +\{x_1\left(t\right)\frac{d{x}_2\left(t\right)}{dt}-y_1\left(t\right)\frac{d{y}_2\left(t\right)}{dt}+i\{x_1\left(t\right)\frac{d{y}_2\left(t\right)}{dt}+y_1\left(t\right)\frac{d{x}_2\left(t\right)}{dt}\}\}\end{array}\right]\\ z_2\left(t\right)\frac{d}{dt}\left(z_1\left(t\right)\right)+z_1\left(t\right)\frac{d}{dt}\left(z_2\left(t\right)\right)=\left[\begin{array}{l}\{x_2\left(t\right)\frac{d{x}_1\left(t\right)}{dt}-y_2\left(t\right)\frac{d{y}_1\left(t\right)}{dt}{x}_1\left(t\right)\frac{d{x}_2\left(t\right)}{dt}-y_1\left(t\right)\frac{d{y}_2\left(t\right)}{dt}\}\\ +i\{x_2\left(t\right)\frac{d{y}_1\left(t\right)}{dt}+y_2\left(t\right)\frac{d{x}_1\left(t\right)}{dt}+x_1\left(t\right)\frac{d{y}_2\left(t\right)}{dt}+y_1\left(t\right)\frac{d{x}_2\left(t\right)}{dt}\}\end{array}\right]\end{array}$

Therefore, the value of $z_2\left(t\right)\frac{d}{dt}\left(z_1\left(t\right)\right)+z_1\left(t\right)\frac{d}{dt}\left(z_2\left(t\right)\right)$is $\left[\begin{array}{l}\{x_2\left(t\right)\frac{d{x}_1\left(t\right)}{dt}-y_2\left(t\right)\frac{d{y}_1\left(t\right)}{dt}{x}_1\left(t\right)\frac{d{x}_2\left(t\right)}{dt}-y_1\left(t\right)\frac{d{y}_2\left(t\right)}{dt}\}\\ +i\{x_2\left(t\right)\frac{d{y}_1\left(t\right)}{dt}+y_2\left(t\right)\frac{d{x}_1\left(t\right)}{dt}+x_1\left(t\right)\frac{d{y}_2\left(t\right)}{dt}+y_1\left(t\right)\frac{d{x}_2\left(t\right)}{dt}\}\end{array}\right]$.

Determine the value of  ddt{z1(t)z2(t)}

Differentiate the equation$z_1\left(t\right)z_2\left(t\right)=x_1\left(t\right)x_2\left(t\right)-y_1\left(t\right)y_2\left(t\right)+i\{x_1\left(t\right)y_2\left(t\right)+x_2\left(t\right)y_1\left(t\right)\}$ both side with respect to $t$as follows.

$\begin{array}{c}{z}_1\left(t\right)z_2\left(t\right)=x_1\left(t\right)x_2\left(t\right)-y_1\left(t\right)y_2\left(t\right)+i\{x_1\left(t\right)y_2\left(t\right)+x_2\left(t\right)y_1\left(t\right)\}\\ \frac{d}{dt}\left\{z_1\left(t\right)z_2\left(t\right)\right\}=\frac{d}{dt}\{x_1\left(t\right)x_2\left(t\right)-y_1\left(t\right)y_2\left(t\right)+i\{x_1\left(t\right)y_2\left(t\right)+x_2\left(t\right)y_1\left(t\right)\}\}\\ \frac{d}{dt}\left\{z_1\left(t\right)z_2\left(t\right)\right\}=\left[\begin{array}{c}{x}_1\left(t\right)\frac{d}{dt}{x}_2\left(t\right)+x_2\left(t\right)\frac{d}{dt}{x}_1\left(t\right)-y_1\left(t\right)\frac{d}{dt}{y}_2\left(t\right)-y_2\left(t\right)\frac{d}{dt}{y}_1\left(t\right)\\ +i\{x_1\left(t\right)\frac{d}{dt}{y}_2\left(t\right)+y_2\left(t\right)\frac{d}{dt}{x}_1\left(t\right)+y_1\left(t\right)\frac{d}{dt}{x}_2\left(t\right)+x_2\left(t\right)\frac{d}{dt}{y}_1\left(t\right)\}\end{array}\right]\end{array}$

Show that ddt{z1(t)z2(t)}=z2(t)ddt{z1(t)}+z1(t)ddt{z2(t)}.

As $\frac{d}{dt}\left\{z_1\left(t\right)z_2\left(t\right)\right\}=\left[\begin{array}{c}{x}_1\left(t\right)\frac{d}{dt}{x}_2\left(t\right)+x_2\left(t\right)\frac{d}{dt}{x}_1\left(t\right)-y_1\left(t\right)\frac{d}{dt}{y}_2\left(t\right)-y_2\left(t\right)\frac{d}{dt}{y}_1\left(t\right)\\ +i\{x_1\left(t\right)\frac{d}{dt}{y}_2\left(t\right)+y_2\left(t\right)\frac{d}{dt}{x}_1\left(t\right)+y_1\left(t\right)\frac{d}{dt}{x}_2\left(t\right)+x_2\left(t\right)\frac{d}{dt}{y}_1\left(t\right)\}\end{array}\right]$, substitute the values $z_2\left(t\right)\frac{d}{dt}\left(z_1\left(t\right)\right)+z_1\left(t\right)\frac{d}{dt}\left(z_2\left(t\right)\right)$for $\left[\begin{array}{c}{x}_1\left(t\right)\frac{d}{dt}{x}_2\left(t\right)+x_2\left(t\right)\frac{d}{dt}{x}_1\left(t\right)-y_1\left(t\right)\frac{d}{dt}{y}_2\left(t\right)-y_2\left(t\right)\frac{d}{dt}{y}_1\left(t\right)\\ +i\{x_1\left(t\right)\frac{d}{dt}{y}_2\left(t\right)+y_2\left(t\right)\frac{d}{dt}{x}_1\left(t\right)+y_1\left(t\right)\frac{d}{dt}{x}_2\left(t\right)+x_2\left(t\right)\frac{d}{dt}{y}_1\left(t\right)\}\end{array}\right]$in the equation $\frac{d}{dt}\left\{z_1\left(t\right)z_2\left(t\right)\right\}=\left[\begin{array}{c}{x}_1\left(t\right)\frac{d}{dt}{x}_2\left(t\right)+x_2\left(t\right)\frac{d}{dt}{x}_1\left(t\right)-y_1\left(t\right)\frac{d}{dt}{y}_2\left(t\right)-y_2\left(t\right)\frac{d}{dt}{y}_1\left(t\right)\\ +i\{x_1\left(t\right)\frac{d}{dt}{y}_2\left(t\right)+y_2\left(t\right)\frac{d}{dt}{x}_1\left(t\right)+y_1\left(t\right)\frac{d}{dt}{x}_2\left(t\right)+x_2\left(t\right)\frac{d}{dt}{y}_1\left(t\right)\}\end{array}\right]$as follows.

$\begin{array}{c}\frac{d}{dt}\left\{z_1\left(t\right)z_2\left(t\right)\right\}=\left[\begin{array}{c}{x}_1\left(t\right)\frac{d}{dt}{x}_2\left(t\right)+x_2\left(t\right)\frac{d}{dt}{x}_1\left(t\right)-y_1\left(t\right)\frac{d}{dt}{y}_2\left(t\right)-y_2\left(t\right)\frac{d}{dt}{y}_1\left(t\right)\\ +i\{x_1\left(t\right)\frac{d}{dt}{y}_2\left(t\right)+y_2\left(t\right)\frac{d}{dt}{x}_1\left(t\right)+y_1\left(t\right)\frac{d}{dt}{x}_2\left(t\right)+x_2\left(t\right)\frac{d}{dt}{y}_1\left(t\right)\}\end{array}\right]\\ =z_2\left(t\right)\frac{d}{dt}\left(z_1\left(t\right)\right)+z_1\left(t\right)\frac{d}{dt}\left(z_2\left(t\right)\right)\\ \frac{d}{dt}\left\{z_1\left(t\right)z_2\left(t\right)\right\}=z_2\left(t\right)\frac{d}{dt}\left(z_1\left(t\right)\right)+z_1\left(t\right)\frac{d}{dt}\left(z_2\left(t\right)\right)\end{array}$

Hence, the product rule for derivative formula for the complex valued function is derived