1. ## Derivative Help

I am having a lot of trouble with the following problem:

Recall that and that .
The ratio is a constant number. Its value is....

I keep getting: 8(cosx)/sin^3x

Doesn't seem to be the right answer. Any help would be great! Thanks!

2. Originally Posted by jwebb19
I am having a lot of trouble with the following problem:

Recall that and that .
The ratio is a constant number. Its value is....

I keep getting: 8(cosx)/sin^3x

Doesn't seem to be the right answer. Any help would be great! Thanks!
$\frac{d}{dx}cot(x) = -(cot^2(x) + 1) = -csx^2(x)$

To understand how this is happening, see if you can derive the previous formulas I used.

-Dan

3. Thanks!

Now a new problem.......

I cannot get the derivative of

I get $(3.5+(3/(2sqrt(3x))))/sqrt(7x+sqrt(3x))$

4. Originally Posted by jwebb19
Thanks!

Now a new problem.......

I cannot get the derivative of

I get $(3.5+(3/(2sqrt(3x))))/sqrt(7x+sqrt(3x))$

$y = \sqrt{7x + \sqrt{3x}}$

Use the chain rule:
$\frac{dy}{dx} = \frac{1}{2\sqrt{7x + \sqrt{3x}}} \cdot \left ( 7 + \frac{1}{2\sqrt{3x}} \cdot 3 \right )$

You simplify it.

-Dan

5. Originally Posted by topsquark

$y = \sqrt{7x + \sqrt{3x}}$

Use the chain rule:
$\frac{dy}{dx} = \frac{1}{2\sqrt{7x + \sqrt{3x}}} \cdot \left ( 7 + \frac{1}{2\sqrt{3x}} \right ) \cdot 3$

You simplify it.

-Dan

$\frac{dy}{dx} = \frac{1}{2\sqrt{7x + \sqrt{3x}}} \cdot \left ( 7 + \frac{\sqrt{3}}{2\sqrt{x}} \right )$

6. Originally Posted by CaptainBlack
$\frac{dy}{dx} = \frac{1}{2\sqrt{7x + \sqrt{3x}}} \cdot \left ( 7 + \frac{\sqrt{3}}{2\sqrt{x}} \right )$
Thanks for the spot. I fixed it in the post.

-Dan