Calculus question-finding second derivative involving trigonometry functions (should be easy)?
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.
879956
2 - csc²z =\n2 - (1/sin²z) = \n(2sin²z - 1) / sin²z = \n(sin²z + sin²z - 1 ) / sin²z =\n(sin²z - cos²z) / sin²z =  {using trig id sin²+cos²=1}\n1 - (cos²z / sin²z) =\n1 - cot²z\n\n~ ~ ~ ~\n\ncot³t / csct =\n(cos³t/sin³t) / (1/sint) =\n(cos³t/sin³t) * sint =\ncos³t/sin²t =\ncost (cos²t/sin²t) =\ncost (cot²t) = \ncost (csc² -1)  {using trig id 1+cot²=csc²}\n\nHope that helped!\n\n~ ♥ ~
781707
Find the first derivative and set it equal to zero.\nf'(x) = 6x^2 - 30x + 24\n6x^2 - 30x + 24 = 0\nDivide everything by 6\nx^2 - 5x + 4 = 0\nFactor\n(x-1)(x-4) = 0\nx = 1 or x = 4\nTo test if these are max or min's, we can use the second derivative.\nf''(x) = 12x - 30\nFor x = 1, f''(1) = 12(1) - 30 = -18 < 0\nThis means f(x) is concave down at x = 1; hence, x = 1 is a local max.\nFor x = 4, f''(4) = 12(4) - 30 = 18 > 0\nThis means f(x) is concave up at x = 4; thus, x = 4 is a local min.\n\nIf you are looking for the points, then you will need to substitute x = 1 and x = 4 into f(x) to find the corresponding y value for each.
761290
well first you have to get the first derivative\n\nso: the 1st derivative= t*cost+1*sint\n                                  = cost^2+sint\nthen you take the derivative of that\nso: the 2nd derivative= 2cost*-sint+cost\n                                   = -2sintcost+cost
858889
QUESTION 1:\n\nAsk yourself: WHICH TRIG FUNCTION IS OPPOSITE OVER ADJACENT?  IT IS THE TANGENT FUNCTION, RIGHT?\n\nThe bottom of the tree is 25 feet away from a point that measures 30 degrees from ground level to the top of the tree.\n\nAnswer: tan30degrees = x/25feet...choice 1.\n\nNotice that they are NOT asking you to solve for x.  Get it?  \n\nQUESTION 2:\n\nYou need to write the given equation 3x - 2y = 12 in the y =mx + b\nform.\n\nLet's do that right now.\n\nWe have:\n\n3x - 2y = 12 \n\nI will subtract -3x from both sides of the equation.\n\nWe get:\n\n-2y = -3x + 12...See what I did?  \n\nNext:\n\nDivide BOTH sides of the equation by -2 to find the value of y.\n\nWe get this:\n\ny = (-3x + 12)/-2\n\nWe can reduce the equation to this:\n\ny = 3/2 (x) - 6\n\nSlope is 3/2 and the y-intercept is -6.\n\nFinal answer: choice 1\n\nGuido
748572
sin(x)tan(x)=(cos(x)csc^2(x))/(cot^2(x)+cot^4(x))\nStep 1. Starting with the right side (the obviously more complicated one) factor out a cotangent squared out of the denominator.\n\n(cos x csc^2 x)/ cot^2 x (1+cot^2 x)\n\nStep 2. Use the identity csc^2 x= cot^2 x+1 in the numerator  and separate the cosine x in the top and the cotangent squared x in the bottom\n\n(cosx/cot^2 x) * (cot^2 x+1/ cot^2 x +1)\n\nStep 3. This leaves you with a cosine x divided by cotangent squared x times 1.  Use the identity that says cot x = cos x/ sin x\n\n(cos x) / (cos^2 x/ sin^2 x)\n\nStep 4. Divide the double fraction and get\n\n(sin^2 x cosx)/ (cos^2 x)\n\nStep 5. Separate your multiple factors from the squares and separate the fraction into three parts, one will be the sine x, the other will be the tangent x (because tan x = sin x/ cos x), and the cosine x divided by cosine x will go to 1.\n\nsin x * (sin x/ cos x) *(cos x/ cos x)\n\nStep 6.  Write final answer\n\nsin x tan x = sin x tan x
764433
Finding max's and mins is done thru the first derivitive and setting it to 0, so\n\nf(x)= sinx - cosx\nf1(x)= cosx + sinx\n0=cosx + sinx\n-cosx = sin x\nthis occurs when x= (3pi)/4\nmax is root 2, as was stated.\n\nThe answer that used to be directly above is wrong.
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.
1101719
Step 1: Find the slope of the line x-2y = 2.  Call it m. \n\nStep 2: Find the derivative f'(x) of the function f(x) = (x-1)/(x+1).\n\nStep 3: Find all values of x where f'(x) = m.\n\nStep 4: For each such value, find its corresponding y-coordinate by setting y=f(x), and then find the equation of the line that goes through that point with slope m.\n\nGood luck!
282780
1.sin(2x)=2sin(x)cos(x)\n2.tan(x)=sin(x)/cos(x)\n3.cos(x)=(1-sin^{2}(x))^(1/2)\n\nwe can use (3) in (2) and (1) to eliminate cos(x). Then substitute sin(x)=y:\n\n1.sin(2x)=2y.(1-y^{2})^(1/2)\n2.tan(x)= y/(1-y^{2})^(1/2)\n\ntherefore,\n\nf(y)=2y(1-y^{2})^{1/2}\n     +y/(1-y^{2})^{1/2}\n\nI hope that's what you wanted.
184556
Let x(t) = A*cos( w*t + theta )\n\nthen x(0) = A*cos( theta )\nand dx/dt (0) = -A*w*sin(theta)\n\nso if you know x and its derivative in t=0 u have two equations and two unknowns ( A and theta ) . the rest is simple algebra .
890169
x = r( t - sint); y = r(1 - cost)\ndx/dt = r(1 - cost); dy/dt = rsint\n\ndy/dx = (dy/dt)/(dx/dt)\n= (rsint)/(r(1 - cost))\n= sint/(1 - cost)\n\nSo at the point where t = a\n\nSlope of tangent (m) = sina/(1 - cosa)\nWhen t = a. x = r(a - sina) and y = r(1 - cosa)\nSo tangent is of the form\ny = mx + b\nand it passes through (r(a - sina) , r(1 - cosa))\n\nThus requred tangent is of form \n\ny = sina/(1 - cosa) x + b\nAnd r(1 - cosa) = sina/(1 - cosa) * r(a - sina) + b\nb = r(1 - cosa) (1 - cosa)\n= r[(1 - cosa)² - sina(a - sina)]/(1 - cosa)\n= r[ (1 - 2cosa + cos²a - asina + sin²a)/(1 - cosa)\n= r[(2(1 - cosa) - asina)]/(1 - cosa))\n= 2r - arsina/(1 - cosa))\nSo the equation is\ny = sina/(1 - cosa) x + 2r - arsina/(1 - cosa))\nOR\nxsina - y(1 - cosa)  + 2r(1 - cosa) - rsina = 0\n\nb) Tangent is horizontal when dy/dx = 0\nie sint/(1 - cost) = 0\nwhence sint = 0\nSo t = -π, 0, π (for -π ≤ t ≤ π)\n\nBut when t = 0, cos t = 1 \nSo dy/dx is indeterminate there (0/0)\n\nBy d'Hopitals Rule\nlim sint/(1 - cost) =\nt →0\n\nlim cost/(-sint)  → ±∞\nt →0\n\nSo horizontal tangents when t = -π, π\nie at (-πr, 2r) and (πr, 2r)\n\nThe tangent is vertical when dy/dx is undefined,\nie 1 - cost = 0\nSo cost = 1\nie t = 0 (for -π ≤ t ≤ π) and here the slope becomes undefined as t→0 (see above)\n\nSo vertical tangent at t = 0 ie at (0, 0)
1096282
cos^2(A) / ( 1 - sinA )\n=( 1 - sin^2(A) ) / ( 1 - sinA )\n=( 1 + sinA )( 1 - sinA ) / ( 1 - sinA )\n=1 + sinA
768749
A'ighty, first write sec(x) = 1/cos(x).  Now take the derivative using the Quotient Rule:\n\nd/dx (1/cos(x)) = (cos(x)*d/dx(1) - 1*d/dx(cos(x))) / cos^2(x)\n\n                        = (0 - (-sin(x))) / cos^2(x)\n\n                        = sin(x) / cos^2(x)\n\n                        = (1/cos(x)) * (sin(x) / cos(x))\n\n                        = sec(x)*tan(x).\n\nAnd there you go.  ^_^
878725
antiderivative of:\n20-12cos2pi(x-0.1)\nis:\n\n20x - 12 sin( 2pi(x-0.1)) / 2pi +C\n=20x - 6 sin( 2pi(x-0.1)) / pi +C\nto double check just take the derivative of the function and see that you get:  20-12cos2pi(x-0.1) '
892157
wait 1 sec.\nOk this is reasonably difficult, you have to use the chain rule and the quotient rule.\nUsing the chain rule:\nderivative = [ (1/2) * ( (2x + 3) / (7x+4) ) ^ (-1/2) ]  *  [d/dx (2x+3) / (7x+4) ]\n\nSo now you to use the quotient rule to find:\nd/dx (2x+3) / (7x+4) = (14x+8 - 14x + 21) / (7x+4)^2\n= 29 / (7x+4)^2\n\nSo the derivative is [1/2 * ((2x+3)/(7x+4))^(-1/2)]*[29 / (7x+4)^2]\n\n\nBy the way i got the thumbs down before I wrote all this up, so it doesn't mean it's wrong.
274298
Use implicit differentiation to find dy/dx. Standard differentiation rules apply, except when you differentiate a term with a "y" in it you must tack on a dy/dx, so:\n\n2x+8y(dy/dx)=0\ndy/dx = -x/4y\n\nThe tangent line intersects the ellipse at (x1,y1), so the slope of the tangent line will be -x1/(4*y1).\n\nWe can also compute the slope of the tangent line by plugging the two points (-5,0) and (x1,y1) into the standard slope formula m = delta(y)/delta(x) and get that the slope of the tangent line is y1/(x1+5). Since these expressions both describe the slope of the tangent line, they must be equal, so\n-x1/(4*y1)=y1/(x1+5)\nx1^2+5*x1=-4*y1^2\nx1^2+4*y1^2+5*x1=0\n\nSince (x1,y1) is a point on the ellipse, it must satisfy the equation of the ellipse, so x1^2+4*y1^2=5. Substituting this in to x1^2+4*y1^2+5*x1=0 gives us 5+5*x1=0, and x1=-1. At this point, there are actually two y1 values that will satisfy the equation of the ellipse, y1 could be either plus or minus 1 (get these by letting x=-1 in the original equation of the ellipse), and so there are actually two points on the ellipse where the tangent line passes through (-5,0), and those points are (-1,1) and (-1,-1).