1. ## Find the Derivative

Ok, I have another problem that's bugging me pretty badly.

Find the derivative:
y=13t^2+40sqrt(t)+15/t

I know that the derivative of 13t^2 = 26t
but I'm unsure about what to do with the 40sqrt(t).
Would it be something like:
(.5)40x^-.5=1/20x?

2. Originally Posted by bpdude
Ok, I have another problem that's bugging me pretty badly.

Find the derivative:
y=13t^2+40sqrt(t)+15/t

I know that the derivative of 13t^2 = 26t
but I'm unsure about what to do with the 40sqrt(t).
Would it be something like:
(.5)40x^-.5=1/20x?

Does it help if written in the form of exponents?
$y=13t^2+40\sqrt{t}+\frac{15}{t} = 13t^2 + 40t^{0.5}+15t^{-1}$

Spoiler:
$26t + \frac{20}{\sqrt{t}} - \frac{15}{t^2} = 26t^3+20t\,\sqrt{t}-15$

3. It does help a bit, but I'm still stuck
I have:
13t^2=26t
40sqrt(t)=20t^-.5(?)
15/t=15t^-1=-15
so:
26t-(1/20t)-15?

Am I doing this right?

4. Originally Posted by bpdude
It does help a bit, but I'm still stuck
I have:
13t^2=26t
40sqrt(t)=20t^-.5(?) e^(i*pi): Nay, because the square root is in the numerator it should be +0.5

15/t=15t^-1=-15 e^(i*pi): No, the exponent is only on the t, the 15 is separate from t.
so:
26t-(1/20t)-15?

Am I doing this right?
...

5. e^(i*pi), could you possibly show me from the start how to go about solving this, so that maybe I can catch what I'm doing wrong? I'm not quite understanding your reply (I understand the sqrt(t) being +.5, but not the 15/t part).

6. Originally Posted by bpdude
e^(i*pi), could you possibly show me from the start how to go about solving this, so that maybe I can catch what I'm doing wrong? I'm not quite understanding your reply (I understand the sqrt(t) being +.5, but not the 15/t part).
I did derive it in Post 2.

Do you know how to derive

$
y=13t^2 + 40t^{0.5}+15t^{-1}
$

This will be using the fact that $\frac{d}{dx}(ax^n) = anx^{n-1}$ as you did for the first term

7. Originally Posted by bpdude
Ok, I have another problem that's bugging me pretty badly.

Find the derivative:
y=13t^2+40sqrt(t)+15/t

I know that the derivative of 13t^2 = 26t
but I'm unsure about what to do with the 40sqrt(t).
Would it be something like:
(.5)40x^-.5=1/20x?

$y=13t^2+40t^{0.5}+15t^{-1}$
$y'=13*2t^{2-1}+40*0.5t^{0.5-1}+15*(-1)t^{-1-1}$
$y'=26t^{1}+20t^{-0.5}-15t^{-2}$

8. Wouldn't it be:
26t+20t^-.5-15?
Or maybe I'm missing something?

I'm sorry if I am only being a 'headache' here.

9. Originally Posted by bpdude
Wouldn't it be:
26t+20t^-.5-15?
Or maybe I'm missing something?

I'm sorry if I am only being a 'headache' here.
Sry man, it wouldn't

$15/t=15*t^{-1}$ and $(t^{-1})'=-1t^{-2}$

10. ialbrekht, YOU ROCK!

So the overall solution was:
y=26t+20t^-.5-15t^-2

Thanks a lot guys! I really appreciate the responses!