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d/dx [|u|] = (u/|u|)(u'), u ≠ 0 |
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d/dx [ln(5x3)] = 15x2/5x3 = 3/x |
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d/dx [e5x^3] = (e5x^3)(15x2) |
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d/dx[log520x3] = 60x2/(ln 5)20x3 = 3/(x(ln 5)) |
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d/dx [sec u] = (sec u tan u)u' |
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d/dx [sec 2x8] = (sec 2x8 tan 2x8)16x7 |
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d/dx[csc u] = -(csc u cot u)u' |
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d/dx[csc 3x2] = -(csc 3x2 cot 3x2)6x |
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d/dx[53x^3] = (ln 5)(53x^3)9x2 |
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∫55xdx = (1/5) (55x) (1/ln(5)) + C |
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∫tan(u)du = -ln|cos(u)|+C |
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∫tan(5x)dx = -(1/5)ln|cos(5x)| + C |
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∫cot(8x)dx = (1/8)ln|sin(8x)| + C |
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∫sec(u)du = ln|sec(u) + tan(u)| + C |
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∫sec(25x)dx = (1/25) ln|sec(25x) + tan(25x)| + C |
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∫csc(u)du = -ln|csc(u) + cot(u)| + C |
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∫csc(2x)dx = (1/2) -ln|csc(2x) + cot(2x)| + C |
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∫sec(u) tan(u) du = sec(u) + C |
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∫sec(51x) tan(51x) dx = ? |
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∫sec(51x) tan(51x) = (1/51) sec(51x) + C |
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∫csc(u) cot(u) du = -csc(u) + C |
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∫csc(16x) cot(16x) dx = ? |
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∫csc(16x) cot(16x) dx = (1/16) - csc(16x) + C |
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What is the disk method for volumes of revolution around the horizontal axis? |
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Think about how to come up the area of a circle by cutting the circle into a pie shape and turning it into a rectangle |
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https://www.google.com/search?q=understanding+area+of+a+circle&rlz=1C1CHBF_enUS849US849&oq=understanding+area+of+a+&aqs=chrome.0.0j69i57j0l4.3823j1j7&sourceid=chrome&ie=UTF-8#kpvalbx=_iIpaXaWyHo-zggfvkbHQAQ21 |
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What is the shell method to finding the volume through integration around the vertical axis? |
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V=2π∫xy dy. The equation must be substituted into the x value so that everything can be integrated with respect to y. |
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cos2A = 1/2 + (cos(2A))/2 |
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sin2A = 1/2 - (cos(2A))/2 |
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