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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?::
$\displaystyle x^2+2000*x^{-1}$>>>>>$\displaystyle x^2+2000^{-1}*x^{-2}=x^2+\frac{1}{2000}*\frac{1}{x^2}$
Please help me out. Thank you in advance.
$\displaystyle x^2+2000*x^{-1}>>>>>2* x - 2000*x^{-2}$
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 :)
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))
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
$\displaystyle f'(x)=\frac{1}{16} \left(24300-675 x^2\right)$
Quote:
Originally Posted by skeeter
Wow I was always told to multiply out before doing any differentiating. Thanks for that!Quote:
Originally Posted by MaxJasper
Can I do the same for integration? I.e. ignoring 15/16 until after I've integrated the thing in the brackets.
$\displaystyle \int k \cdot f(x) \, dx = k \int f(x) \, dx$
look over your basic rules ... they're in your text.
You would be surprised.Quote:
Originally Posted by skeeter
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?
k is a constant factor of the function
