Question

(a): For a differentiable complex valued function$\mathit{z}\mathbf{\left(}\mathbf{t}\mathbf{\right)}$, find the derivative of$\frac{\mathbf{1}}{\mathbf{z}\mathbf{\left(}\mathbf{t}\mathbf{\right)}}$.

(b): Prove the quotient rule for the derivatives of the complex valued functions.

Short answer

(a) The solution is $\frac{d}{dt}\left(\frac{1}{z\left(t\right)}\right)=-\frac{1}{z{\left(t\right)}^2}\frac{dz}{dt}$.

(b) The solution is $\frac{d}{dt}\left(\frac{g\left(t\right)}{f\left(t\right)}\right)=\frac{f\left(t\right)\frac{dg}{dt}-g\left(t\right)\frac{df}{dt}}{{\left[f\left(t\right)\right]}^2}$if$f\left(t\right)\ne 0$ .

Step-by-step solution

(a)Step 1: Simplify the function 1z(t).

Consider the complex valued functions $\frac{1}{z\left(t\right)}$.

Simplify $\frac{1}{z\left(t\right)}$as follows.

$\frac{1}{z\left(t\right)}=z{\left(t\right)}^{-1}$

Determine the derivative of the function z(t)−1 .

The derivative of the function$z^n\left(t\right)$ with respect to is $n{z}^{n-1}\left(t\right)\frac{dz}{dt}$.

Using the definition, differentiate the equation $\frac{1}{z\left(t\right)}=z{\left(t\right)}^{-1}$both side with respect to $t$as follows.

$\begin{array}{c}\frac{d}{dt}\left\{\frac{1}{z\left(t\right)}\right\}=\frac{d}{dt}\left\{z{\left(t\right)}^{-1}\right\}\\ =(-1)z{\left(t\right)}^{(-1-1)}\frac{dz}{dt}\\ \frac{d}{dt}\left\{\frac{1}{z\left(t\right)}\right\}=-z{\left(t\right)}^{-2}\frac{dz}{dt}\end{array}$

Hence, the derivative of$\frac{1}{z\left(t\right)}$ is $-z{\left(t\right)}^{-2}\frac{dz}{dt}$.

(b) Step 3: Simplify the complex function g(t)f(t).

Consider the complex valued functions$\frac{g\left(t\right)}{f\left(t\right)}$ such that $f\left(t\right)\ne 0$.

Simplify$\frac{g\left(t\right)}{f\left(t\right)}$ as follows.

$\frac{g\left(t\right)}{f\left(t\right)}=g\left(t\right)\cdot f{\left(t\right)}^{-1}$

Determine the derivative of the function g(t)×f(t)-1.

The derivative of the function$f\left(t\right)\cdot g\left(t\right)$with respect to $t$is $f\left(t\right)\frac{dg}{dt}+g\left(t\right)\frac{df}{dt}$.

Differentiate the equation$\frac{g\left(t\right)}{f\left(t\right)}=g\left(t\right)\cdot f{\left(t\right)}^{-1}$both side with respect to$t$as follows.

$\begin{array}{c}\frac{g\left(t\right)}{f\left(t\right)}=g\left(t\right)\cdot f{\left(t\right)}^{-1}\\ \frac{d}{dt}\left\{\frac{g\left(t\right)}{f\left(t\right)}\right\}=\frac{d}{dt}\{g\left(t\right)\cdot f{\left(t\right)}^{-1}\}\\ =g\left(t\right)\frac{d}{dt}\left\{f{\left(t\right)}^{-1}\right\}+f{\left(t\right)}^{-1}\frac{d}{dt}\left\{g\left(t\right)\right\}\\ \frac{d}{dt}\left\{\frac{g\left(t\right)}{f\left(t\right)}\right\}=g\left(t\right)\cdot \{-f{\left(t\right)}^{-1-1}\frac{df}{dt}\}+f{\left(t\right)}^{-1}\frac{dg}{dt}\end{array}$

Further, simplify the equation as follows.

$\begin{array}{c}\frac{d}{dt}\left\{\frac{g\left(t\right)}{f\left(t\right)}\right\}=g\left(t\right)\cdot \{-f{\left(t\right)}^{-1-1}\frac{df}{dt}\}+f{\left(t\right)}^{-1}\frac{dg}{dt}\\ =-g\left(t\right)\cdot f{\left(t\right)}^{-2}\frac{df}{dt}+f{\left(t\right)}^{-1}\frac{dg}{dt}\\ =-g\left(t\right)\cdot \frac{1}{f{\left(t\right)}^2}\frac{df}{dt}+\frac{1}{f\left(t\right)}\frac{dg}{dt}\\ \frac{d}{dt}\left\{\frac{g\left(t\right)}{f\left(t\right)}\right\}=-g\left(t\right)\cdot \frac{1}{f{\left(t\right)}^2}\frac{df}{dt}+\frac{f\left(t\right)}{f{\left(t\right)}^2}\frac{dg}{dt}\end{array}$

Further, simplify the equation as follows.

$\begin{array}{c}\frac{d}{dt}\left\{\frac{g\left(t\right)}{f\left(t\right)}\right\}=-g\left(t\right)\cdot \frac{1}{f{\left(t\right)}^2}\frac{df}{dt}+\frac{f\left(t\right)}{f{\left(t\right)}^2}\frac{dg}{dt}\\ \frac{d}{dt}\left\{\frac{g\left(t\right)}{f\left(t\right)}\right\}=\frac{f\left(t\right)\frac{dg}{dt}-g\left(t\right)\frac{df}{dt}}{f{\left(t\right)}^2}\end{array}$

Hence, the quotient rule for the derivatives of the complex valued functions is verified