# Thread: Derivative of complex exponential with high powers

1. ## Derivative of complex exponential with high powers

evaluate:

d ^15/dx ^15 exp [(1 + sqrt(3)i) x]

i dont know how to approach this problem with the drivative to the 15th power.

firstly 1 + sqrt(3)i must be converted to polar form.. then im stuck

pls help me.

2. ## Uhh...

Originally Posted by tasukete
evaluate:

d ^15/dx ^15 exp [(1 + sqrt(3)i) x]

i dont know how to approach this problem with the drivative to the 15th power.

firstly 1 + sqrt(3)i must be converted to polar form.. then im stuck

pls help me.
Maybe I am being stupid but wouldnt it be $\displaystyle f^{(15)}(x)=(\sqrt{3}i)^{15}e^{1+\sqrt{3}ix}=-2187\sqrt{3}ie^{1+\sqrt{3}ix}$?

3. Originally Posted by Mathstud28
Maybe I am being stupid but wouldnt it be $\displaystyle f^{(15)}(x)=(\sqrt{3}i)^{15}e^{1+\sqrt{3}ix}=-2187\sqrt{3}ie^{1+\sqrt{3}ix}$?

eh? im having trouble understanding as to how u got that. i thought the answer would be in the form of trig and expo functions??? am i wrong?

4. ## Ok

Originally Posted by tasukete
eh? im having trouble understanding as to how u got that. i thought the answer would be in the form of trig and expo functions??? am i wrong?
if you think about it $\displaystyle \frac{D[e^{u(x)}]}{dx}=u'(x)\cdot{e^u(x)}$ right? Now if $\displaystyle u'(x)$ is a constant then we dont need to use chain rule we jsut have keep multiplying it so for example...remember $\displaystyle u(x)$ is a constant then we would have $\displaystyle f(x)=e^{u(x)}$...then $\displaystyle f'(x)=u'(x)\cdot{e^{u(x)}}$...now since it is a constant we just yank it out so $\displaystyle f''(x)=u'(x)\cdot{\frac{D[e^{u(x)}]}{dx}}$...so using this same concept of yanking out the constant we get $\displaystyle f^{(n)}(x)=u'(x)^{n}\cdot{e^{u(x)}}$...and in this case $\displaystyle u'(x)$ is a consant...so yeah

5. Originally Posted by Mathstud28
Maybe I am being stupid but wouldnt it be $\displaystyle f^{(15)}(x)=(\sqrt{3}i)^{15}e^{1+\sqrt{3}ix}=-2187\sqrt{3}ie^{1+\sqrt{3}ix}$?

after obtaining an expression for the derivative...i got a different answer to ures... since the constant would be of high power.. (1 + sqrt(3) i ) ^ 15... expanding it would be silly.

so wouldnt u need to convert it into polar form and use De Moivre's theorem to simplify it?

if that is so i got something different to ures... i had -2^15 ????

hrmm... i just noticed u only included the imaginary part in the constant....y is that?

6. Hello,

Originally Posted by tasukete
after obtaining an expression for the derivative...i got a different answer to ures... since the constant would be of high power.. (1 + sqrt(3) i ) ^ 15... expanding it would be silly.

so wouldnt u need to convert it into polar form and use De Moivre's theorem to simplify it?

if that is so i got something different to ures... i had -2^15 ????

hrmm... i just noticed u only included the imaginary part in the constant....y is that?
Indeed, mathstud got wrong because he forgot 1.

(1 + sqrt(3) i ) ^ 15
This is correct.
But don't bother with De Moivre's theorem

$\displaystyle (1+\sqrt{3} i)^2=(1-3+2\sqrt{3} i)=2(-1+\sqrt{3} i)$

So $\displaystyle (1+\sqrt{3} i)^3=(1+\sqrt{3} i)^2(1+\sqrt{3} i)=2\underbrace{(-1+\sqrt{3} i)(1+\sqrt{3} i)}_{(a-b)(a+b)=a^2-b^2}=2(-3-1)=-2^3$

So $\displaystyle (1+\sqrt{3} i)^{15}=((1+\sqrt{3} i)^3)^5=(-2^3)^5=-2^{15}$

7. ## Wait

Originally Posted by Moo
Hello,

Indeed, mathstud got wrong because he forgot 1.

This is correct.
But don't bother with De Moivre's theorem

$\displaystyle (1+\sqrt{3} i)^2=(1-3+2\sqrt{3} i)=2(-1+\sqrt{3} i)$

So $\displaystyle (1+\sqrt{3} i)^3=(1+\sqrt{3} i)^2(1+\sqrt{3} i)=2\underbrace{(-1+\sqrt{3} i)(1+\sqrt{3} i)}_{(a-b)(a+b)=a^2-b^2}=2(-3-1)=-2^3$

So $\displaystyle (1+\sqrt{3} i)^{15}=((1+\sqrt{3} i)^3)^5=(-2^3)^5=-2^{15}$
I dont think you are right Moo...I just put the answer in my calculator and it says I am right

8. Oh yeah ?

You forgot the brackets...

It's (1+sqrt(3) i)x, not 1+sqrt(3) ix

Calculators are evil especially when the one who uses it got wrong himself

9. Originally Posted by Moo
Oh yeah ?

You forgot the brackets...

It's (1+sqrt(3) i)x, not 1+sqrt(3) ix

Calculators are evil especially when the one who uses it got wrong himself

hrmmm...was wondering why i couldnt get stud's answer..

11. ## OHhhh

Originally Posted by Moo
Oh yeah ?

You forgot the brackets...

It's (1+sqrt(3) i)x, not 1+sqrt(3) ix

Calculators are evil especially when the one who uses it got wrong himself
I see why I got it wrong....there needs to be a mandatory LaTeX requirement for clarity in questions

12. Originally Posted by Mathstud28
I see why I got it wrong.......there needs to be a mandatory LaTeX requirement for clarity in questions
His question was clear...all brackets were here...