# To prove: If F ( x ) = e r x then, F ′ ( x ) = r e r x where r = a + b i .

### Single Variable Calculus: Concepts...

4th Edition
James Stewart
Publisher: Cengage Learning
ISBN: 9781337687805

### Single Variable Calculus: Concepts...

4th Edition
James Stewart
Publisher: Cengage Learning
ISBN: 9781337687805

#### Solutions

Chapter I, Problem 49E
To determine

## To prove: If F(x)=erx then, F′(x)=rerx where r=a+bi.

Expert Solution

### Explanation of Solution

Formula used:

Euler’s formula:

eiy=cosy+isiny

Calculation:

It is given that, if u(x)=f(x)+ig(x) is a complex valued function then, the derivative of the function u(x) is u(x)=f(x)+ig(x).

Consider the function F(x)=erx and rewrite the function by the use of Euler’s formula.

F(x)=erx=e(a+bi)x=eax+bxi=eax(cosbx+isinbx)

That is, the function is F(x)=eaxcosbx+i(eaxsinbx).

Obtain the derivative of the function F(x)=eaxcosbx+i(eaxsinbx).

F(x)=(eaxcosbx)+i(eaxsinbx)=(eaxddx(cosbx)+cosbxddx(eax))+i(eaxddx(sinbx)+sinbxddx(eax))=(aebxcosbxbeaxsinbx)+i(aeaxsinbx+beaxcosbx)=aeax(cosbx+isinbx)+beax(sinbx+icosbx)

Further simplified as,

F(x)=a[eax(cosbx+isinbx)]+beax((i2)sinbx+icosbx)[i2=1]=aerx+ibeax(cosbx+isinbx)=aerx+iberx[eax(cosbx+isinbx)=erx]=erx(a+ib)=rerx

That is, the derivative F(x)=rerx.

Hence the proof.

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