What is the derivative of y(x) = e^{5ix} when 5i is a complex number?

Trenthan
Messages
50
Reaction score
0
Hey guys/ girls and thanks in advance

Okay this may seem like the dumbest question on the planet. But i figure i better ask before i waste the lecture's time of booking a consult.

Whats the derivative of

y(x) = e^{5ix}

where 5i is a complex number

both the partial and standard derivative would be

\frac{dy}{dx} =\frac{\partial y}{\partial x} = 5i*e^{5ix}

In the notes the 'i' is missed at various stages but he arrives at the correct solution which I've verified with other sources. Just the derivation i question... since it uses

\frac{dy}{dx} =\frac{\partial y}{\partial x} = 5*e^{5ix}

Any particular reason this could even be true... ? (Im just thinking a big typo but its over many pages so yea...)

Cheers Trent
 
Physics news on Phys.org
It is most likely a typo. On the other hand, if he is only looking for the imaginary part of the derivative, then his answer is correct.
 
Last edited:
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
Back
Top