find dy/dx: y=[sin(3/x)]^5/2?
349210
dy/dx = -(15cos(3/x)/2x^2) * (sin(3/x))^3/2
709530
let x=-pi, then y=3{sin(-pi+(pi/3))]=3{sin(-2pi/3)}= (- 3) (sqrt(3))/2)\nx=pi y=3sin(4pi/3)= (- 3) (sqrt(3))/2)
970213
Use the Chain Rule repeatedly:\n\nOriginal function: y = (tan(sin x))^2\n\ndy/dx = 2 tan(sin x) (d/dx) tan(sin x)\n\n= 2 tan(sin x) (sec(sin x))^2 (d/dx) sin x\n\n= 2 tan(sin x) (sec(sin x))^2 cos x, QED
374181
a) Equally inclined to both axes is parallel to y = x; hence you want dy/dx = 1.\n\n(y - 3)^2 = 4 - x^2\ny = (4 - x^2)^(1/2) + 3\n\ndy/dx = (1/2) * (4 - x^2)^(-1/2) * (-2x)\ndy/dx = x / (4 - x^2)^(1/2)\n\nSo:\n\nx^2 = (4 - x^2)\nx = +/- SQRT(2)\ny = 3 + SQRT(2)\n\n------------------------------------------------------\n\nb)  I might edit and do later if I've got time.
1010282
df(x)/dx = (x)^(2/3)*(2) -2/3(x)^(-1/3)(2x-5) [using product rule]\ndf(x)/dx = 2(x)^(2/3) - 4/3 (x)^(2/3) - 10/3(x)^(-1/3) [simplifying]\ndf(x)/dx = (2/3)(x)^(2/3) - (10/3)(x)^(-1/3)   [ after more simplification]
360927
y'=dy/dx=\n\n(6√x)' + (3/x^2)' + (10)'=\n\n[6*x^(1/2)]' + [3x^(-2)]' + 0= \nnow use the power rule\n\n6*(1/2)* x^[(1/2)-1] + 3*(-2)*x^(-2-1)=\n\n3*x^(-1/2)-6x^(-3)=\n\n3/x^(1/2) -6/x^3=\n\n3/√x - 6/x^3
729738
rate of change=dy/dx=slope
806655
dx=2xlny-2z\ndy=x^2*(1/y)-2z\ndz= x^2lny-2
897617
This is an equation with separated variables. Rewrite as \n\ndu/u^3 = (t^3-t)dt \n\nand integrate (find antideritive) of each side separately.
218622
differentiate wrt x.. dy/dx = 2x + 3 = 0\nx = -3/2 is the vertex (-3/2. -49/4)\n\nif you don't know differentiation,\nexpress in terms of (x-a)^2 + b = y, \ny = (x - -3/2)^2 + 31/4\n(-3/2, -49/4) is the vertex. (Value of y is found by substituting value of x in equation)\n\nx intercept is found by placing y=0, (-5,0) and (2,0)\ny intercept is found by placing x=0, (0,-10)
846505
Well...\nIf Y/Y'=k/y then Y'/Y=y/k\nso (lnY)'=y/k\nIntegrating\nlnY=ln A+ (y^2)/2k      A- constant\nExponentiating\nY=A*exp( (y^2)/2k )\nAm I right?
667880
a) dy/dx = x - 1/x^2\ndy = (x-x^(-2))dx\ny = x^2/2 + 1/x + C\ny(1) = 1^2/2 + 1/1 + C = 2\n1/2 + 1 + C = 2 implies C = 1/2\nAnswer: y = x^2/2 + 1/x + 1/2, x>0\n\nb) The integration is already finished.  Just plug in -1 for x.\ny(-1) = (-1)^2/2 + 1/-1 + C = 1\n-1/2 - 1 + C = 1 implies C = 3/2\nAnswer: x^2/2 + 1/x + 3/2, x<0\n\nc) Just take the derivative of each one.  C1 and C2 are constants, so their derivative is 0.  \n\nd) C1 = 1/2 and C2 = 3/2\n\ne) Follow parts (a) and (b) except plug in 2 for x and set it equal to -1.  Similarly for y(-2) = 2.\nCheck that C1 = -7/2 and C2 = 1/2
757602
[(x^3)/(2 - x)]^4/3 \ndy/dx = (4/3)[(x^3)/(2 - x)]^1/3[3x^2(2 - x) + (x^3)]/(2-x)^2\ndy/dx = (4)[(x^3)/(2 - x)]^1/3[6x^2 - 3x^3 + (x^3)]/(3(2-x)^2)\ndy/dx = 8x^3(3 - x)/(3(2-x)^7/3)
1051597
start with   d = m / v        multiply both side by v; \nyou get     v * d = m        divide both sides by d; \nyou get     v = m / d
847056
dy/dx=y² sin x\ndy/y^2=sinxdx\n\nintegrate\n-1/y = -cosx +K\n1/y  = cosx -K\nuse initial conditions\n1=1-K\nK=0\ny= secx\n\nremember, int{sin}= -cos\n\ni hope that this helps
888343
Use the chain rule:  dy/dx = d/dx[(ln x)^x] = x(ln x)^(x-1)d/dx[ln x]
831486
Differentiate both sides, we have \n\ndy =  (1/x)dx + (1 + tan^2(y))/tany dy\n\ntherefore\n(1 - (1 + tan^2(y))/tany)dy=(1/x)dx,\n\ni.e.\n\ndy/dx=tany/(x(tany - tan^2- 1)).