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[Ecrofirt]

My calc 3 semester just started, and I've gotten a bit rusty over the summer. As such, I've got a calc question.

What's the derivative of te^t?

I think the derivative of e^t is te^t, so I'd imagine the derivative of te^t is t²e^t + e^t, but this doesn't seem correct.

The reason I'm asking is because I've got to find dy/dx when t=0 for these parametric equations:

x=te^t
y=t + e^t

and doing it my way has me coming up with someting / 0, when the answer is supposed to be 2.

 

[ghostface]

I think the derivative of e^t is te^t

No, the derivative of e^t is e^t. The other part of your product rule is correct. The answer is then e^t + te^t, or e^t (t +1 ).

Then again, my calc sucks so I could be all wrong.

 

[Ecrofirt]

Well, it makes much more sense with what you said. That must be it! Thanks!

 

[Macam]

The derivative of e^t is e^t, that's correct. Thus, the derivative with respect to t (d/dt) of te^t is (1)e^t + t*e^t via the Product Rule.

For the parametric equations:

x= te^t
y= t + e^t

The answers, assuming you're looking for d/d(t), would be:

x = (1)e^t + t*e^t
y = 1 + e^t

So when t=0, x = 1 + 0 = 1 and y = 1 + 1 = 2. I don't think dy/dx makes sense for what you're doing, if you're doing parametric equations. I believe that should match up with what your book is saying if I understood everything correctly.

Makes ya wish -jinx- was still around, doesn't it? Fuckin' hate thread.

 

[Ecrofirt]

Yes, jinx always helped so well when I had a math question.

I'm now trying to figure out why the derivative of e^SQR(t) is e^SQR(t)/(2SQR(t))

with sqr being square root.

edit:

Nevermind. That was an easy one. I hope I start remembering my simple calc stuff, or I'll be dead in the water

 

[ghostface]

Well, like we agreed earlier, the derivative of e^something is equal to the derivative of the something, times e^something.

So in this case the derivative of e^SQR(t) is equal to the derivative of SQR(t) times e^SQR(t). In case you forgot, to get the derivative of SQR(t):

Change SQR(t) to t^(1/2). Then the derivative is simply 1/2 * t^(-1/2), or 1/2t^(1/2), or 1/2SQR(t).

Edit: Glad you got it. Even the easiest calc can be tricky if you havent practiced in a while. I suggest you dig out a basic calculus book (your first one, if you can find it) and take a full day and just practice simple shit. And use a site with step by step solutions like this one to help you out. Good luck.

 

[Macam]

Good luck, is all I have to say. I took Cal III after about a year or two's hiatus, and it was rather killer, notably when trying to recall trigonometric derivatives, polar/spherical coordinates, multivariable derivatives, etc. Crackin' on Diffrential Equations at the moment with more to follow. Joy.