find the derrivative of y with respect to θ: y=tan^2 (sinθ)?
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
786513
put x^2=t\n2xdx=dt\nxdx=1/2dt\nthe integral now is\n(1/2dt/1+t^2\n(1/2)tan^-1t+C\n=(1/2)tan^-1(x^2)+C
610143
I think ur right cos^2(theta) should be cos(theta). then the answer would be perfect.
716883
Trying for wow #2---\n\nR = .15 m\nω = 2π/T = 13.963 rad/sec\nA little vector diagram shows that tanθ = rω²m/mg = rω²/g\nbut r = Rsinθ, so tanθ = sinθ/cosθ = Rsinθω²/g  from which\n\n1/cosθ = secθ = Rω²/g = .15*13.963²/9.8 = 2.984 → θ = 70.42°\n\nYou get to do b)\nGood luck..
346140
Well there are the obvious values that are known of arctan. But if your constant K doesn't fall into these categories I'm not sure how you would do this without taylor expansion.\n\nTry this link, it gives the well known values on tan(x) so reading it backwards gives the arctan values
575459
Sorry but I don't quite understand your qn.\nCan you make it clearer?\n\n(to ozert:\ncos A and sin A could be negative values if tan A is in Quadrant 3)
567245
simple x=r*cos(theta), y=r*sin(theta), r^2=x^2+y^2\nsubstitude, not hard
921171
If tan( t ) = -2 then t = atan( -2 ) = -1.107 + multiples of π.\n\nIn the range from 0 to 2π the required angles are\n\n-1.107 + π and -1.107 + 2π, or\n\n2.034 and 5.176
291468
Oh yes i have cracked it at last....\n\nsolution...\n\nLet x=tan-1 (y) (i mean tan inverse)\nthen tan x = y,and cot x = 1/y (coz tan-1 (x) = cot-1 (1/x))\n\nthen ur given equation's Left hand side part (LHS) will be\n\n(1/y)/(1-y) + (y)/(1-(1/y))\n\n= 1/(y-y2)+ y/(1-(1/y))\n\n= 1/(y-y2) + y2/(y-1)\n\n= y2(y-1) - 1/y(y-1)\n\n= y3/y(y-1)\n\nI hope u understood till this step.\n\nNow we have (a3-b3) = (a-b) (a2+ab+b2)\n\nthen the LHS becomes...\n\n(y-1)(y2+y+1)/y(y-1)\n\n= (y2+y+1)/y\n\n= y+1+(1/y)\n\nsubstitute back y=tanx, 1/y = cot x and leave 1 as it is...\n\nU have LHS = 1 + tan x + cot x...Thats ur RHS !!!
278456
So you know the ones for sin(2a) and cos(2a). Cool.\n\nNow, tan(2a) is the same as sin(2a)/cos(2a), since tan = sin/cos (definition of tangent).\n\nSub in the formulae for sin(2a) and cos(2a) :\n\n2sin(a)cos(a)\n----------------\ncos(a)^2-sin(a)^2\n\nDivide both numerator and denominator by (cos(a))^2\n\n2tan(a)\n----------\n1-tan(a)^2
761613
Using implicit differentiation, you get:\n\n-csc^2(y)(dy/dx) = 1 - dy/dx\n-csc^2(y)(dy/dx) + dy/dx = 1\n(dy/dx)[1 - csc^2(y)] = 1\n\nTherefore, dy/dx = 1/[1 - csc^2(y)]\n\nNow plug in your ordered pair to get dy/dx.  Remember csc (y) = 1/sin(y):\n\ndy/dx = 1/[1-1], which is undefined.  Therefore, you cannot write a tangent line for this ordered pair (there is a vertical asymptote at this point)\n\n------------\n\nHope this helps
954869
1. h(t) = 4 sin t - 5 cos t\n\nThe derivative of sin t is  cos t, and the derivative of cos t is -sin t:\n\nh ' (t) = 4 cos t - ( -5 sin t) = 4 cos t + 5 sin t\n\nUsing the same two derivative rules as before and taking the derivative again:\n\nh " (t) = -4 sin t + 5 cos t\n\n2. g(t) = 3 tan t\n\nThe derivative formula for y = tan x is dy/dx = sec^2 (x). I will derive it for you now using the quotient rule just so you can see how to derive the other three trigonometric formulas if you don't happen to have them memorized (you should have them memorized!):\n\ny = tan x = sin x / cos x\ndy/dx = [ (cos x)(cos x) - (sin x)(- sinx) ] / (cos x) ^2\ndy/dx = [cos ^2 (x) + sin^2 (x) ] / cos ^2 (x)\ndy/dx = 1 / cos ^2 (x)\ndy/dx = sec^2 (x)\n\nGetting back to the problem then:\n\ng ' (t) = 3 sec ^2 (t)\n\nFor the second derivative you need to use chain rule, and you need to know that the derivative of sec t is (sec t)(tan t):\n\ng " (t) = 3 (2 sec t) (sec t * tan t)\n\ng " (t) = 6 sec ^2 (t) tan t\n\n(When you try to simplify it, the trig gives you sine / cosine ^3, which can't really be made any simpler than the way it is now, so I leave the answer in its form).\n\nIt's just a matter of memorizing the appropriate formulas, being able to derive them from scratch if you don't happen to remember them, and being able to use these formulas in conjunction with the power rule, product rule, quotient rule, and chain rule- hopefully you know these already.
853051
We have to find derivative of, g(x) = the integral (with lower limit tanx and upper limit x^2) of 1 / [sqrt(2+t^4)] dt.\n\nNow, the derivative of the integrand f(t)=1 / [sqrt(2+t^4)] is \nf'(t)= -2*t^3 / (2+t^4)^(3/2) [By Quotient Rule]\n\nd(g(x)) /dt= d {int(tan x)(x^2) {1 / [sqrt(2+t^4)] }dt} /dt\n=>dy /dt= d {f(t)}/dt (at tan x) - d {f(t)}/dt (at x^2)\n=>dy /dt= f'(t) (at tan x) - f(t) (at x^2)\n=>dy /dt= [2*t^3 / (2+t^4)^(3/2)] (at tan x) - [2*t^3 / (2+t^4)^(3/2)] (at x^2)\n=>dy /dt= [2*(tan x) ^3 / (2+(tan x)^4)^(3/2)]  - [2*(x^2)^3 / (2+(x^2)^4)^(3/2)] \n=>dy /dt= [2*(tan x) ^3 / (2+(tan x)^4)^(3/2)]  - [2*(x^8) / (2+(x^16))^(3/2)]\n\nI hope that answers your question.
1160901
you have to use partial deriavative to solve the problem\n\nlet's z=16x^3-24xy+6y^2 = f(x,y)\n\nfirst deriavative respects to x , f'x =32x^2 - 24y = 0\nfirst deriavative respects to y , f'y = -24x +12y = 0\n\nSolve simultaneously, you will get:\ny = 0, x =0, z=0 and y = 3, x = 3/2, z= 0\n\nSo, you have two stationary points.\n\nDo the second deriavation repects to x and y \nfxx = 64x\nfyy = 12\nfxy = fyx = -24\n\nformula, (fxx) × (fyy) - (fxy)^2 = S\nif S>0, it is minimum if fxx>0 or fyy >0 otherwise it is maximum\nif S<0, it is saddle point\n\nX=0, y =0, z=0 ( 0,0,0)<--coordinate\nS = -576 ----> saddle point\n\n\nx= 3/2, y=3, z=0 / (3/2, 3, 0) <--coordinate\nS = 576, fxx >0 ----->> local minimum
710047
Since theta is between 90 and 180 degrees, it is in the II quadrant. Tan = opp/adj. The y value (opp) is +3, and the x value (adj) is -4.  By the Pythagorean theorem, hyp^2 = opp^2 + adj^2. So the hypotenuse  is the sqrt of 3^2 +4^2 and is equal to 5. Cos is add/hyp, and is -4/5. So the answer is -4/5.
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
702411
Rewrite equation as:\ny = x - x^-1\n\nDifferentiate:\ny' = 1 + x^-2\nAt (1,0) the gradient is:\ny' = 1 + 1 = 2\n\nSo equation of tangent is:\ny-0=2(x-1)\ny=2x-2