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!
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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! $\displaystyle \frac{d}{dx}cot(x) = -(cot^2(x) + 1) = -csx^2(x)$ To understand how this is happening, see if you can derive the previous formulas I used. -Dan
Thanks! Now a new problem....... I cannot get the derivative of I get $\displaystyle (3.5+(3/(2sqrt(3x))))/sqrt(7x+sqrt(3x))$
Originally Posted by jwebb19 Thanks! Now a new problem....... I cannot get the derivative of I get $\displaystyle (3.5+(3/(2sqrt(3x))))/sqrt(7x+sqrt(3x))$ Please post new questions in new threads. $\displaystyle y = \sqrt{7x + \sqrt{3x}}$ Use the chain rule: $\displaystyle \frac{dy}{dx} = \frac{1}{2\sqrt{7x + \sqrt{3x}}} \cdot \left ( 7 + \frac{1}{2\sqrt{3x}} \cdot 3 \right )$ You simplify it. -Dan
Last edited by topsquark; Oct 4th 2007 at 04:34 AM. Reason: Doh!
Originally Posted by topsquark Please post new questions in new threads. $\displaystyle y = \sqrt{7x + \sqrt{3x}}$ Use the chain rule: $\displaystyle \frac{dy}{dx} = \frac{1}{2\sqrt{7x + \sqrt{3x}}} \cdot \left ( 7 + \frac{1}{2\sqrt{3x}} \right ) \cdot 3$ You simplify it. -Dan $\displaystyle \frac{dy}{dx} = \frac{1}{2\sqrt{7x + \sqrt{3x}}} \cdot \left ( 7 + \frac{\sqrt{3}}{2\sqrt{x}} \right )$
Originally Posted by CaptainBlack $\displaystyle \frac{dy}{dx} = \frac{1}{2\sqrt{7x + \sqrt{3x}}} \cdot \left ( 7 + \frac{\sqrt{3}}{2\sqrt{x}} \right )$ Thanks for the spot. I fixed it in the post. -Dan
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