1. ## 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?::
$\displaystyle x^2+2000*x^{-1}$>>>>>$\displaystyle x^2+2000^{-1}*x^{-2}=x^2+\frac{1}{2000}*\frac{1}{x^2}$

2. ## Re: Differentiate a fraction

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

3. ## 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: $\displaystyle \frac{d}{dx}\left(kx^r \right)=krx^{r-1}$

Thanks guys

5. ## 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))

6. ## 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

7. ## Re: Differentiate a fraction

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

8. ## Re: Differentiate a fraction

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
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!

9. ## 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.

10. ## Re: Differentiate a fraction

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

11. ## Re: Differentiate a fraction

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