CALCULUS





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Discus: SAT/ACT Tests and Test Preparation: October 2003 Archive: CALCULUS
By Student9 (Student9) on Monday, October 06, 2003 - 09:50 pm: Edit

Find the derivative implicitly:

xsiny = ycosx


Find the derivative:

cot^2(3x)

Thanks.

By Omnislasher3333 (Omnislasher3333) on Tuesday, October 07, 2003 - 09:58 pm: Edit

I am a little rusty, but these are what i got . .

x sin y = y cos x
x (cos y)dy/dx + (1)(sin y) = y (-sin x) + (1)(cos x)dy/dx

x cos y (dy/dx) - cos x (dy/dx) = -y sin x - sin y

dy/dx = (-y sin x - sin y)/(x cos y - cos x)


[d/dx]cot^2(3x) = 2(cot 3x)(-csc^2(3x))(3)

= -6 (cot (3x))(-csc^2(3x))

By Quarky (Quarky) on Wednesday, October 08, 2003 - 12:39 pm: Edit

Omni got em right except for #2 -- there is only one negative, so remove the negative sign in front of cosecant.


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