what is the asymptote of y=sinhx?
953056
No asymptote.  It just keeps growing. That is, as x─►±∞, y ─►±∞.
831435
The graph of sin x goes up to 1 when x = pi/2. When you graph y = sin (2x), it shrinks the x-values by half - so it will be 1 when x = pi/4.
768619
the horizontal asymptote is the limit as x->∞, the vertical as x->0.  Since this equation can be written -9 + 1/x^2, when x gets large, 1/x^2 gets small with respect to -9, and can be ignored, leaving  -9 as the symptote.   As x->0, the 1/x^2 dominates, so the vertical asymptote is 1/x^2.  There is a discontinuity at x=0 where the function is undefined.
292407
Let sinh(K*arcsinh(x))=y then\n\ny= ((x+Sqrt(1+x^2))^(2K)-1) /(2(x+Sqrt(1+x^2))^K)\n\n\n(x+Sqrt(1+x^2))^(2K)-1\n-------------------------------- = y\n     2(x+Sqrt(1+x^2))^K\n\nExample: for K=1/2 and x=2\ny= ((1+Sqrt(5))/2)/Sqrt(2+Sqrt(5))\nY= 0.78615 = sinh(arcsinh(2)/2)\n\nCuriously in this example\ny= 1/Sqrt(Phi)\n\nwhere Phi= 0.5*(1+Sqrt(5))= 1.618...
829160
y'=2 cos(7x)*(-sin 7x)*7 By the Chain Rule
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
715464
y' = (sinx-cosx) / (sinx+cosx)^2\n\nTo find the tangent at (0,1) you need its coefficient, which is y'(0) = -1.\n\nYour tangent should be y - 1 = -1.(x - 0)\ny = -x + 1
934522
y' = [(x-1) - x] / (x-1)^2 = -1/(x-1)^2, so no critical points.  y' < 0 for all x where y is defined, so it's always decreasing.\n\ny'' = 2/(x-1)^3, which is positive when x>1, and negative for x<1.  Therefore the curve is concave up for x>1, and concave down when x<1.\n\nThere is a vertical asymptote at x=1, and lim x->inf x/(x-1) = \nlim x->inf 1/(1-1/x) = 1 (same for x -> -inf) so there is a horizontal asymptote at y=1.
776830
X= (2k + 1)/n * (pi/2)\n\nThis tells you the vertical asymptotes for y = tan (nx)\n\nk is any integer\neg the graph of y = tan 3x has vertials asymptotes at:\n\nx = π/2, ±3/3 π/2, ±5/3 π/2, ±7/3 π/2, .....\n\n2k + 1 = ±1, ±3, ±5, ±7, ....... and n = 3
832532
dy/dx=cos x\n\nIntegrating both sides, we will get:\ny = sinx + c\n\ny(0) = sin0 + c = 0 => c=0\n\nSo y=sinx.
963822
The vertical asymptotes occur where you are evaluating y = tan(pi/2 + k pi):\n\n2x - pi/4 = pi/2 + k pi\n2x = 3pi / 4 + k pi\nx = 3 pi / 8 + k pi / 2\n\nWithin the domain 0 <= x <= 2pi, this gives:\n\nx = 3 pi / 8, 7 pi / 8, 11 pi / 8, or 15 pi / 8
929562
∫ax(2 -x)dx\n= ∫(2ax - ax²)dx\n= 2a * ½x² - a * ⅓x³ + c\n= ax² - ⅓ax³ + c
218516
If that is supposed to be a function of x, then\nf(x) = (-2x + 7)/(4x - 3)\nSince the denominator cannot be zero, equate it with zero and you get the vertical asymptote:\n4x - 3 = 0\n4x = 3\nx = 3/4 --- Vertical asymptote.\n\nTo get the horizontal asymptote, get the inverse function.\ny = (-2x + 7)/(4x - 3)\n4xy - 3y = -2x + 7\n4xy + 2x = 3y + 7\nx = (3y + 7)/(4y + 2)\n\nThen the horizontal asymptote is the denominator equated to zero.\n4y + 2 = 0\n4y = -2\ny = -1/2 --- horizontal asymptote\n\n^_^
806885
dy/dx of xsinx = x(d/dx sinx) + sinx(d/dx x) = xcosx + sinx(1) = xcosx + sinx\n\nHope that helps - it's the first thing multiplied by the differential of the second, plus the second multiplied by the differential of the first. :)
782645
1/x=y+3\nit resembles the equation of the asymptotes of hyperbola
796769
The point P moves in such a way that its Cartesian coordinates with respect to an origin O at time t is given by x= e^(-t) , y= 2te^(-t). the distance OP is denoted by r and the angle between OP and the x-axis by θ.\n(a)Using an analytical method, find the value of t when r^2 is stationary. Determine the nature of the stationary point.\n\nr² = Δx² + Δy²\n= (e^(-t) - 0)² + (2te^(-t) - 0)²\n= e^(-2t) + 4t²e^(-2t)\n= e^(-2t)(1 + 4t²)\nd(r²)/dt = -2e^(-2t)(1 + 4t²) +8te^(-2t)\n= 2e^(-2t)(1 - 4t + 4t²)\n= 0 for stationary points\nSo 1 - 4t + 4t² = 0 as e^(-2t) > 0 for all t\nie (1 - 2t)² = 0, ie t = ½, ½\nWhen t <½, d(r²)/dt > 0 and when t >½, d(r²)/dt > 0\nThus when t = ½ there is a stationary inflexion\n\n(b) Find, in terms of t, the rate of change of θ with respect to\n\ntanθ = Δy/Δx\n= 2te^(-t)/e^(-t)\n=2t\n\nsec²θ * dθ/dt = 2\nie\n\ndθ/dt = 2/sec²θ\n= 2/(1 + tan²θ)\n= 2/(1 + 4t²)\n\nCan you show the non analytical method for part (a) too!\n\nx = e^(-t) → -t = lnx ie t = -lnx\n\ny = 2te^(-t) \n= -2xlnx\n\nx: .. 0 ...... 0.25 ... 0.5 .... 0.75 . 1.0 .. 1.25 .. 1.5\ny: undef . -0.15 . -0.36 . -0.22 ... 0 .... 0.28 . 0.61\n... (→0)\n\nLim x → 0 (x lnx) = 0 and OP increases with x \n(limiting min value of OP = 0)\n\nMin value of xlnx occurs when lnx + 1 = 0 ie x = -1/e and at this point OP² will show its inflexion as it cannot be a minimum and nor can it be a maximum so all it can do is slow its rate of growth.
1149867
We have x=rcos(θ), and y=rsin(θ).  \n\n\nTherefore x=4sin(θ)cos(θ) and y=4sin(θ)sin(θ).  \nAnd x^2+y^2= 16sin(θ)^2(cos(θ)^2+sin(θ)^2) =16sin(θ)=4y.  \nThus x^2+y^2-4y+4-4 =x^2+(y-2)^2-4=0 and x^2+(y-2)^2=4.\n\nAre you sure the second one is given by:\nρsin(θ)=2cos(θ) and not ρsin(φ)=cos(θ)?\n\nif it is ρsin(φ)=2cos(θ), then first notice that φ≠kπ for any integer k.\nThen we have ρ=2cos(θ)/sin(φ)\nNext we have x=ρcos(θ)sin(φ)=2cos(θ)^2, and y=ρsin(θ)sin(φ)=2cos(θ)sin(θ).\n\nThus x^2+y^2= 4cos(θ)^2(cos(θ)^2+sin(θ)^2) =2x,\nand x^2-2x+1-1+y^2= (x-1)^2-1+y^2=0,\nso (x-1)^2+y^2=1