Differentiate a fraction

**Mukilab** · September 15th 2012, 09:01 AM

How would I differentiate this?

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

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

Please help me out. Thank you in advance.

**MaxJasper** · September 15th 2012, 09:05 AM

$x^2+2000*x^{-1}>>>>>2* x - 2000*x^{-2}$

**MarkFL** · September 15th 2012, 09:10 AM

I would write the function as:

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

To differentiate, use the power rule term by term: $\frac{d}{dx}\left(kx^r \right)=krx^{r-1}$

**Mukilab** · September 15th 2012, 09:14 AM

Thanks guys

**Mukilab** · September 15th 2012, 12:01 PM

What something such as $\frac{24300x-225x^3}{16}$

Surely since the 16^(-1) differentiates to 0, the whole equation differentiates to 0 (since everything is multiplied by 16^(-1))

**skeeter** · September 15th 2012, 12:16 PM

no ...

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

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

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

fyi, problems involving derivatives belong in the calculus forum

**MaxJasper** · September 15th 2012, 12:18 PM

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

**Mukilab** · September 15th 2012, 12:27 PM

Originally Posted by skeeter

no ...

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

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

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

fyi, problems involving derivatives belong in the calculus forum

Originally Posted by MaxJasper

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

Wow I was always told to multiply out before doing any differentiating. Thanks for that!

**Mukilab** · September 15th 2012, 12:32 PM

Can I do the same for integration? I.e. ignoring 15/16 until after I've integrated the thing in the brackets.

**skeeter** · September 15th 2012, 12:47 PM

$\int k \cdot f(x) \, dx = k \int f(x) \, dx$

look over your basic rules ... they're in your text.

**Mukilab** · September 15th 2012, 12:50 PM

Originally Posted by skeeter

$\int k \cdot f(x) \, dx = k \int f(x) \, dx$

look over your basic rules ... they're in your text.

You would be surprised.

How does one determine f(x)? Is it just anything involving an x? In other words, can I make anything k as long as it multiplies into the f(x) and does not involve x?

**skeeter** · September 15th 2012, 01:39 PM

k is a constant factor of the function

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