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Gernald · Apr 9, 2009, 08:08 PM

Hi all I'm having an issue with finding the derivative of (x+2)(sqrt. -x)
could somebody help me out and show me how to do it.
It was on my last test, and I'm trying to correct it with no luck.

Thanks!

while I'm here I might as well also ask about another question: I have a cylinder with a volume of 16pi inches squared and I need to find the height and radius if the least amount of material is used.
I have no idea where to even start :-(

Thanks again!

ROLCAM · Apr 9, 2009, 09:29 PM

volume of cylinder = pi *r^2 *L = 16pi

= r^2 * L = 16

Radius 2 inches
Length 4 inches.

This answer would satisfy the equation .

There are others of course.
Radius 1 inch
Length 16 inches is another.

Perito · Apr 10, 2009, 06:37 AM

Originally Posted by Gernald

finding the derivative of (x+2)(sqrt. -x)

Boy, it's too long since I did any complex derivatives. I'm not sure if this will help or not:

$(x+2)\sqrt{-x}\,=\,(x+2)\,i\,\sqrt{x}$

$=\,i\,(x^{(\frac{3}{2})}\,+\,2x^{(\frac{1}{2})})$

I don't know where to go from there. Sorry. Maybe someone can teach both you and me ;-)

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As for the second question,

$volume = \pi r^2 L = 16 \pi$

$area = 2 \pi r^2 + 2 \pi r L$

You want to maximize the volume and minimize the area. This is done by differentiating an equation (must be of one independent variable) and setting that equal to zero. This gives a maximum or a minimum, so you have to check to make sure you get what you want.

From the first equation (volume)

$L=\frac{16}{r^2}$

So, substituting, the area equation becomes

$A = 2 \pi r^2 + 2 \pi r \frac{16}{r^2}$

$A = 2 \pi r^2 + 32 \pi \frac{1}{r}$

$\frac{dA}{dr} = 0 = 4 \pi r - 32 \pi \frac{1}{r^2}$

simplifying

$32 \pi = 4 \pi r^3$

$r = 8^{(\frac 13)} = 2$ which is Rolcam's first answer that satisfies the equation. The difference is that we know this is either a maximum or a minimum.

galactus · Apr 10, 2009, 08:32 AM

You can find the derivative as usual. If x were negative, we would have a real number and it is differentiable

$\frac{d}{dx}[(x+2)\sqrt{-x}]=\frac{-3x-2}{2\sqrt{-x}}$

Let's use first principals on $\sqrt{-x}$

$\lim_{h\to \infty}\frac{\sqrt{-(x+h)}-\sqrt{-x}}{h}\cdot\frac{\sqrt{-(x+h)}+\sqrt{-x}}{\sqrt{-(x+h)}+\sqrt{-x}}$

$=\lim_{h\to 0}\frac{-1}{\sqrt{-(x+h)}+\sqrt{-x}}=\frac{-1}{2\sqrt{-x}}$

Now, if x were positive, then we have the complex domain.

If x>0, then we have $\frac{i}{2\sqrt{x}}$

Therefore, for the whole problem, when x<0, we have a derivative of $\fbox{\frac{3x+2}{2\sqrt{x}}\cdot i}$

If x>0, $\fbox{\frac{-3x-2}{2\sqrt{-x}}}$

Gernald · Apr 12, 2009, 05:30 PM

Okay... thanks all! You've been a huge help!!