What is the derivative of e^(x/2)?

Factual Questions
1

Not homework; just trying to remember how.

∫e[sup]x/2[/sup]*dx

IIRC, you take the derivative of the ‘inner’ function (x/2) and multiply it by the derivative of the ‘outer’ function (e[sup]x/2[/sup]). So the derivative of the ‘inner’ function is (1/2) * x[sup]0[/sup] = (1/2) * 1 = 1/2 and the derivative of the ‘outer’ function is e[sup]x/2[/sup], and the derivative of e[sup]x/2[/sup] is (1/2)*e[sup]x/2[/sup], right?

2

FYI: Your words say derivitive, but I get an integral symbol on my computer.

But otherwise your right. d/dx e^(x/2) = .5*e^(x/2)

3

Sorry, I meant integral.

Thanks.

4

d exp(f(x))/ dx = (d(f(x))/dx) exp(f(x))

5

Then your answer is wrong. The derivative is half the original function, the integral is twice the original function.

6

You’re right that the derivative of e[sup]x/2[/sup] is (1/2)e[sup]x/2[/sup]. But that means that the indefinite integral, or antiderivative, of (1/2)e[sup]x/2[/sup] is e[sup]x/2[/sup] + C, and then the integral of e[sup]x/2[/sup] is 1/(1/2)e[sup]x/2[/sup] + C, or 2e[sup]x/2[/sup] + C.

7

Let me WolframAlpha that for you.

8

OK, how do you get from ∫e[sup]x/2[/sup] to 2e[sup]x/2[/sup] + C? (I know where the C comes from, since I didn’t include limits.)

Wait, let me think… If the coefficient of e is one, then to get the derivative 1/2 the coefficient of e needs to be 2. But how do I get there logically?

Also: Why do I get these things in my head and spend time on them?

9

u-substitution works:

∫e[sup]x/2[/sup] dx = ∫e[sup]x/2[/sup] 2 d(x/2) = 2 ∫e[sup]u[/sup] du = 2 e[sup]u[/sup] + c = 2e[sup]x/2[/sup] + c

10

U-substitution! :smack: I remember that!

Thanks.

11

Did I mention I suck at math? Oh, yes. In several threads. :smiley:

And yet… I can’t help needlessly torturing myself. :rolleyes:

12

Only on this board would someone working calculus problems think that they suck at math. :slight_smile:

13

The rules for derivatives are all well-established and algorithmic, such that any standard function or combination of standard functions has a derivative that is a combination of standard functions, and that combination can be found straightforwardly. The same is not, however, true of integrals. So when one is presented with a pair of functions and the claim that the one is the integral of the other, the simple way to check this claim is to perform the derivative, not to perform the integral.