what is the dy/dx of sinhx?
350899
coshx\nand the derivative of coshx is sinhx
148647
x+sinx\nLim ------        \n    x+2sinx\nx->0\n \n    1+cosx\nlim-------   [according to Hospital's theory]\n    1+2cosx\nx->0\n\n   1+cos0\n=----------\n   1+2cos0\n\n  1+1\n=-----\n  1+2\n\n  2\n=----\n  3\n ans 2/3.\nThanks.\n  3
1182723
sinx+cosecx=2\nsin+(1/sinx)=2\nfrom here try to cross multiply!
274505
dy/du=6u^2\ndy/du=(2x^2)/3
803550
With respect to what?\n\nd/dx xy = y + x dy/dx\n\nd/dy xy = x + y dx/dy\n\nd/dt xy = x dy/dt + y dx/dt\n\nIt's just the Product Rule.
738333
(1+sin^2x)dx/\n(1+sin^2x)(1-sin^2x)\n=dx/(1-sin^2x)\ndx/cos^2x\nsec^2xdx\ntanx+C
901088
4sinxcosx =1 (assumed for 0 ≤ x ≤ 2π)\nSo 2 sin2x = 1 (since sin2x = 2sinxcosx)\nSo sin2x = ½ (0 ≤ 2x ≤ 4π)\n2x = π/6, 5π/6, 13π/6, 17π/6\nThus x = π/12, 5π/12, 13π/12, 17π/12\n\nsin4x= -2sin2x (assumed for 0 ≤ x ≤ 2π)\nSo 2sin2xcos2x + 2sin2x = 0\nie 2sin2x(cos2x + 1) = 0\nso sin 2x = 0 or cos 2x = -1/2 (0 ≤ 2x ≤ 4π)\n2x = 0, π, 2π, 3π, 4π or 2x = 2π/3, 4π/3, 8π/3, 10π/3\nwhence x = 0, π/3, π/2, 2π/3, π, 4π/3, 3π/2, 5π/3, 2π
1059709
i think ur question meant sinx = cosecx\nsinx = 1/sinx\nsin^2x = 1\nsinx = 1\nsinx is 1 at 90degree\nthus  x = 90 degree
702823
want the proof? Here it is ( i am assuming you know the dervatives of sinx and cosx)\n\ny = tanx\ny = sinx/cosx\ny' = [cosx(cosx) - sinx(-sinx)]/(cosx)^2 --> quotient rule\ny'= [(cosx)^2 + (sinx)^2]/(cosx)^2\ny' = 1/(cosx)^2\ny' = (secx)^2\n\nvoila
940656
dy/dx = -(e^x + e^(-x)) / -(y (e^x - e^(-x) + 1)³)\n\nSo ∫ydy = ∫(e^x + e^(-x)) dx/(e^x - e^(-x) + 1)³ \n\nLet u = e^x - e^(-x) + 1 thus du = (e^x + e^(-x) ) dx\n\nSo ∫ydy = ∫du/u³\n\nSo ½y² = -½*1/u² + ½A\n\nie y² = A - 1/u²\n= A - 1/(e^x - e^(-x) + 1)²
197333
i think \nx = 2664.173
894390
2sin3x*[-15sin3x]\n=-30*sin^23x
433960
ok. well if y = sin^-1 (x/a), then that means that \nsin y = (x/a). \nso if you do implicit differentiation, you get \ncos y * dy/dx = 1/a.\n\nso thatz not hard. now if you draw a right angle triangle, and label 1 of the angles y. so sin of y is x/a, but it is also equal to opposite/hypotenuse. so then that meanz the opposite is equal to x and the hypotenuse is equal to a. now if you use pythagoras theorem, the other side (the adjacent side) is sqrt(hyp^2 - opp^2). because \nadj^2 + opp^2 = hyp^2\nso adj = sqrt(hyp^2 - opp^2)\nso adj = sqrt(a^2 - x^2).\n\nso now you need cos of y, which is adj/hyp\nso cos y = sqrt(a^2-x^2) / a.\n\nnow back to the first equation\ncos y * dy/dx = 1/a\nsqrt (a^2 - x^2)/a  * dy/dx = 1/a\nsqrt (a^2-x^2) * dy/dx = 1\nso dy/dx = 1/sqrt(a^2-x^2)\n\nyou can do this with any inverse trig derivative stuff. i hope you understood that. thatz the way i remember it but there might be other easier ways (like just memorize the relations), but there you go. hope that helped. let me know if you need more help
888072
Recall that ∫ cosx dx = sinx + C\n\nLet u = 5x, so du/dx = 5, dx = 1/5 du\n\nNow let's use substitution:\n\n∫ cos5x dx = ∫ cosu 1/5 du = \n1/5 ∫ cosu du  = 1/5 sinu + C = 1/5 sin5x + C\n\nHope that  helped!\n\n~ ♥ ~
746686
well, here goes nothing.  i have an answer, but i'm not 100% sure if it's right.  as you said, (V0/g) is a constant.  the derivitive of sin(2x) is 2cos(2x).  set equal to 0 and i got:\n(V0/g)2cos(2x)=0\ndivide both sides of equation by g/2V0:\ncos(2x)=0\nthen i graphed the function of cos(x) and saw that it equalled 0 at (pi/2) and (3pi/2).  we're looking for minimum so we keep (pi/2) and discard (3pi/2).  since we're dealing with 2x and not x we have:\n2x=(pi/2)\ndivide both sides by 2\nx=(pi/4) or 45 degrees.  hope this matches up to the back of the book
816208
RHS  (simplify it into sin and cos terms)\ncsc^2(x)/cotx\nsinx/cosx*1/sin^2x\nsinx/sinx*1/cosxsinx\n2/2sinxcosx\n2/sin2x\n2csc2x\nhence LHS=RHS
676902
Incomplete question. Are you looking for the anti-derivative of x, a, or b???\n\nput dx, da, or db at the end of the equation so we'll know.