# Did I derive this right?

• May 1st 2010, 08:27 AM
kmjt
Did I derive this right?
• May 1st 2010, 08:44 AM
CaptainBlack
Quote:

Originally Posted by kmjt

For a start:

How do you get from the first line to the second?

CB
• May 1st 2010, 08:57 AM
kmjt
For somee reason I thought you can bring the denominator up and attach it, I guess you can't (Headbang)
• May 1st 2010, 09:17 AM
Tekken
Quote:

Originally Posted by kmjt
For somee reason I thought you can bring the denominator up and attach it, I guess you can't (Headbang)

well you can if you raise the denominators to the power of -1

i.e. $A = x^{2}.4\pi^{-1} + (x^{2} - 80x + 1600).16^{-1}$ 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 $A'$

Cant see how this is university level though...
• May 1st 2010, 09:24 AM
harish21
Quote:

Originally Posted by kmjt

Notice that $4\pi \text{and} 16$ in the denominators of the first and second expressions are constants. So your derivative would be:

$A' = \frac{1}{4 \pi} \frac{d}{dx} x^2 + \frac{1}{16} \frac{d}{dx} (x^2-80x+1600)$
• May 1st 2010, 09:34 AM
kmjt
So far I used the quotient rule on x^2/4pi and got 8pi x / 4pi^2, is that correct?
• May 1st 2010, 09:42 AM
Tekken
Quote:

Originally Posted by kmjt
So far I used the quotient rule on x^2/4pi and got 8pi x / 4pi^2, is that correct?

Nope, try again. i'll tell you the answer after another effort. Using Harish's method is easier perhaps
• May 1st 2010, 09:42 AM
harish21
Quote:

Originally Posted by kmjt
So far I used the quotient rule on x^2/4pi and got 8pi x / 4pi^2, is that correct?

No. Look at the post above. Because 4pi and 16 are constants, you can isolate those terms while differentiating

[(1/4pi) times the derivative of $x^2$ ] +

[(1/16) times the derivative of $x^2-80x+1600$]
• May 1st 2010, 09:53 AM
kmjt
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 (Happy)
• May 1st 2010, 10:06 AM
Tekken
Quote:

Originally Posted by kmjt
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

You have to remember that $(4\pi)^{2} = 4\pi.4\pi = 16\pi^{2}$ and NOT equal to $4\pi^{2}$

Sorry again harish thanks for that technique but I don't believe I should be doing it that way in the class im in (Happy)

Ok here goes...

let $u = x^{2}$ and $v = 4\pi$

differentiating u w.r.t x we get $\frac{du}{dx} = 2x$ and
differentiating v w.r.t x we get $\frac{dv}{dx} = 0$ This is always the case when differentiating constants.

Applying the quotient rule we get;

$\frac{dy}{dx} = \frac{4\pi.2x - x^{2}.0}{4\pi^{2}}$

=> $\frac{dy}{dx} = \frac{8\pi.x}{(4\pi)^{2}}$

=> $\frac{dy}{dx} = \frac{8\pi.x}{16\pi^{2}}$

Now we can cancel out $8\pi$ as it is common to both the top and bottom line.

This leaves us with $\frac{dy}{dx} = \frac{x}{2\pi}$
• May 1st 2010, 10:06 AM
skeeter
Quote:

Originally Posted by kmjt
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 ... denominator should be $\textcolor{red}{16 \pi^2}
$

...
• May 1st 2010, 10:06 AM
harish21
Quote:

Originally Posted by kmjt
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 (Happy)

$\frac{1}{4\pi} \frac{d}{dx}x^2 = \frac{1}{4 \pi} 2x = \frac{2x}{4 \pi} = \frac{x}{2 \pi}$

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..
• May 1st 2010, 10:10 AM
kmjt
I see thanks (Rofl)