Where did I go wrong?
well you can if you raise the denominators to the power of -1
i.e. $\displaystyle A = x^{2}.4\pi^{-1} + (x^{2} - 80x + 1600).16^{-1} $ but thats not reall that useful for what you want to do.
Just use the quotient rule twice and sum the two answes to get $\displaystyle A' $
Cant see how this is university level though...
I think I should stick to differentiating the left and than the right, because that's what we do in class. Heres where I got my answer from doing the quotient rule on x^2 / 4pi:
f'(x) = g'(x)h(x) - g(x)h'(x) / (4pi)^2
= (2x)(4pi) - (x^2)(0) / 4pi^2
= 8pix - 0 / 4pi^2
= 8pix / 4pi^2
Where did I mess up? I used a derivative calculator online and it said the derivative would be x / 2pi
Sorry again harish thanks for that technique but I don't believe I should be doing it that way in the class im in
Ok here goes...
let $\displaystyle u = x^{2} $ and $\displaystyle v = 4\pi $
differentiating u w.r.t x we get $\displaystyle \frac{du}{dx} = 2x $ and
differentiating v w.r.t x we get $\displaystyle \frac{dv}{dx} = 0 $ This is always the case when differentiating constants.
Applying the quotient rule we get;
$\displaystyle \frac{dy}{dx} = \frac{4\pi.2x - x^{2}.0}{4\pi^{2}} $
=> $\displaystyle \frac{dy}{dx} = \frac{8\pi.x}{(4\pi)^{2}} $
=> $\displaystyle \frac{dy}{dx} = \frac{8\pi.x}{16\pi^{2}} $
Now we can cancel out $\displaystyle 8\pi $ as it is common to both the top and bottom line.
This leaves us with $\displaystyle \frac{dy}{dx} = \frac{x}{2\pi} $
$\displaystyle \frac{1}{4\pi} \frac{d}{dx}x^2 = \frac{1}{4 \pi} 2x = \frac{2x}{4 \pi} = \frac{x}{2 \pi}$
that is the derivative of the first expression of your term(note that this is the same as what you got from the derivative calculator).. you have to do the same to find the derivative of the second term, and add those two to get your answer..