Pls Chk Work (dx/dt)

**Yogi_Bear_79** · August 7th 2006, 04:52 PM

Find dx/dt given x = sqrt(1 + cot 3t)?

dx/dt = 1/2 ( 1+cos3t) ^(-½) [0 - cosec² 3t(3)]

dx/dt = 1/2{ 1/(1+cost3t)½ } {-3cosec² 3t}

= [ (-3/2 ) (cosec²3t) {1/(1+cos3t)½} ]

**galactus** · August 7th 2006, 05:08 PM

$\frac{-3csc^{2}(3t)}{2\sqrt{1+cot(3t)}}$

Looks good.

**Yogi_Bear_79** · August 7th 2006, 05:17 PM

Could you convert the entire solution using the code you used? I am new to posting online and the example would be useful, and easier to read!

**galactus** · August 7th 2006, 05:46 PM

Click quote and see what the code is. With a little practice you, too, can learn LaTex. I converted what you had.

Find $\frac{dx}{dt}$ given $x = \sqrt{1 + cot 3t}$ ?

$\frac{dx}{dt} = \frac{1}{2}( 1+cot(3t))^{\frac{-1}{2}}[0 - csc²(3t)]$

$\frac{dx}{dt} = \frac{1}{2}\frac{1}{\sqrt{1+cot(3t)}}{(-3csc² (3t))}$
= $(\frac{-3}{2}) (csc²(3t)) {\frac{1}{\sqrt{1+cot(3t)}}$

= $\frac{-3csc^{2}(3t)}{2\sqrt{1+cot(3t)}}$

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