Differentiate a fraction

• Sep 15th 2012, 09:01 AM
Mukilab
Differentiate a fraction
How would I differentiate this?

$\displaystyle x^2+\frac{2000}{x}$

I thought it would go like this (but it doesn't work). Where am I going wrong?::
$\displaystyle x^2+2000*x^{-1}$>>>>>$\displaystyle x^2+2000^{-1}*x^{-2}=x^2+\frac{1}{2000}*\frac{1}{x^2}$

• Sep 15th 2012, 09:05 AM
MaxJasper
Re: Differentiate a fraction
$\displaystyle x^2+2000*x^{-1}>>>>>2* x - 2000*x^{-2}$
• Sep 15th 2012, 09:10 AM
MarkFL
Re: Differentiate a fraction
I would write the function as:

$\displaystyle f(x)=x^2+2000x^{-1}$

To differentiate, use the power rule term by term: $\displaystyle \frac{d}{dx}\left(kx^r \right)=krx^{r-1}$
• Sep 15th 2012, 09:14 AM
Mukilab
Re: Differentiate a fraction
Thanks guys :)
• Sep 15th 2012, 12:01 PM
Mukilab
Re: Differentiate a fraction
What something such as $\displaystyle \frac{24300x-225x^3}{16}$

Surely since the 16^(-1) differentiates to 0, the whole equation differentiates to 0 (since everything is multiplied by 16^(-1))
• Sep 15th 2012, 12:16 PM
skeeter
Re: Differentiate a fraction
no ...

$\displaystyle \frac{d}{dx} \left[ c \cdot f(x)\right] = c \cdot f'(x)$

$\displaystyle y = \frac{1}{16}(24300x - 225x^3)$

$\displaystyle y' = \frac{1}{16}(24300 - 675x^2)$

fyi, problems involving derivatives belong in the calculus forum
• Sep 15th 2012, 12:18 PM
MaxJasper
Re: Differentiate a fraction
$\displaystyle f'(x)=\frac{1}{16} \left(24300-675 x^2\right)$
• Sep 15th 2012, 12:27 PM
Mukilab
Re: Differentiate a fraction
Quote:

Originally Posted by skeeter
no ...

$\displaystyle \frac{d}{dx} \left[ c \cdot f(x)\right] = c \cdot f'(x)$

$\displaystyle y = \frac{1}{16}(24300x - 225x^3)$

$\displaystyle y' = \frac{1}{16}(24300 - 675x^2)$

fyi, problems involving derivatives belong in the calculus forum

Quote:

Originally Posted by MaxJasper
$\displaystyle f'(x)=\frac{1}{16} \left(24300-675 x^2\right)$

Wow I was always told to multiply out before doing any differentiating. Thanks for that!
• Sep 15th 2012, 12:32 PM
Mukilab
Re: Differentiate a fraction
Can I do the same for integration? I.e. ignoring 15/16 until after I've integrated the thing in the brackets.
• Sep 15th 2012, 12:47 PM
skeeter
Re: Differentiate a fraction
$\displaystyle \int k \cdot f(x) \, dx = k \int f(x) \, dx$

• Sep 15th 2012, 12:50 PM
Mukilab
Re: Differentiate a fraction
Quote:

Originally Posted by skeeter
$\displaystyle \int k \cdot f(x) \, dx = k \int f(x) \, dx$