Chapter 10.3, Problem 49E

To determine

To show: In case the question requires to prove a particular result.

Explanation of Solution

Given information :

The given function:

  $\frac{d}{dx}{e}^{(a+b)x}=\frac{d}{dx}\left[e^{ax}(\cos bx+i\sin bx)\right]$

Calculation :

The given function is, $\frac{d}{dx}{e}^{(a+b)x}=\frac{d}{dx}\left[e^{ax}(\cos bx+i\sin bx)\right]$ differentiating both the sides $\frac{d}{dx}{e}^{(a+b)x}=\frac{d}{dx}\left[e^{ax}(\cos bx+i\sin bx)\right]$

Taking right hand side, $\frac{d}{dx}\left[e^{ax}(\cos bx+i\sin bx)\right]=\left(e^{ax}\right)(-b\sin bx+bi\cos bx)+\left(a{e}^{ax}\right)(\cos bx+i\sin bx)$

  $=\left(e^{ax}\right)\left[\left(b{i}^2\sin bx+bi\cos bx\right)+a(\cos bx+i\sin bx)\right]$

  $=\left(e^{ax}\right)[bi(\cos bx+i\sin bx)+a(\cos bx+i\sin bx)]=(a+bi)\left(e^{ax}\right)(\cos bx+i\sin bx)$

  $=(a+bi)e^{(a+ib)x}$

  $\begin{array}{cc}\text{Therefore,}\frac{d}{dx}{e}^{(a+ib)x}& =\frac{d}{dx}\left[e^{ax}(\cos bx+i\sin bx)\right]\sqrt{\frac{d}{dx}{e}^{(a+ib)x}=(a+ib)e^{(a+b)x}}\text{thus, the rule}\\ & =(a+ib)e^{(a+bb)x}\end{array}$

  $\frac{d}{dx}{e}^{kx}=k{e}^{kx}$ also holds for complex values of $k$