91影视

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

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

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

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

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

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

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

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

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

$\frac{g\left(t\right)}{f\left(t\right)}=g\left(t\right)鈰�f{\left(t\right)}^{鈭�1}$

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

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

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

Further, simplify the equation as follows.

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

Further, simplify the equation as follows.

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

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