Find the Derivative

**bpdude** · October 3rd 2009, 07:02 AM

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?

but then what about 15/t?

**e^(i*pi)** · October 3rd 2009, 07:15 AM

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?

but then what about 15/t?

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:

**bpdude** · October 3rd 2009, 07:35 AM

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?

**e^(i*pi)** · October 3rd 2009, 07:38 AM

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?

...

**bpdude** · October 3rd 2009, 07:48 AM

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).

**e^(i*pi)** · October 3rd 2009, 08:54 AM

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

$<br /> y=13t^2 + 40t^{0.5}+15t^{-1}<br />$

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

**ialbrekht** · October 3rd 2009, 09:00 AM

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?

but then what about 15/t?