Derivative of Taylor Series

[Ragib]

hey guys i was just wondering if anyone knew any functions that are equal to their own derivative other than e^x?

[Severian]

No. You can solve [math]\frac{df}{dx}=f[/math] and you get the solution [math]f=Ce^x[/math] where C is a constant, so there are no other possibilities.

 

ie. [math]\frac{df}{dx}=f

\Rightarrow \int \frac{df}{f}= \int dx = x + \log C

\Rightarrow \log f = x + \log C

\Rightarrow f= C e^x[/math]

[the tree]

The derivative of [math]f(x)=0[/math] is [math]f'(x)=0[/math], but I don't think that counts.

[timo]

The derivative of [math]f(x)=0[/math] is [math]f'(x)=0[/math], but I don't think that counts.

It does. And giving Severians post a 2nd thought you´ll see that this case was included there.