How do i find the maxima and minima in differentiation?
612056
the local maxima can be found at the points where dy/dx is equal to zero and (d^2.y/(dt)^2) is less than zero. The local minima can be found at the points where dy/dx is equal to zero and (d^2.y/(dt)^2)(read as dee sqared y by dee tee squared) is greater than zero.\n\nThe explanation is that if u were to draw the graph of watever function u want to find the maxima and minima of,.... at the point where its max, the tangent is zero and the differential of the tangent is sloping downward!! And it slopes upward for the point of minima....\n\nHope this helps.\n\nYou better gimme best answer for this.
816542
Find derivatives with respect to each variable.
120944
dy/dx = -4x \n\nAt x=2, gradient is -8\n\ny - y(1) = dy/dx(x - x(1)) [x(1), y(1) means x,y coordinates]\n\ny - 1 = -8*(x - 2)\n\ny = -8x + 17
657050
1) f'(x) = 40x^4+6x^2+4\nTo find local max's and min's, set this equal to zero, but you will see that all solutions will be imaginary, so there are no local max's and min's on the real line.\n\nf''(x) = 160x^3+12x\nTo find the points of inflect, set this equal to zer, but you will see that the only place where there is an inflection is a x = 0.\n\nSo what have we learned?  f(x) is a strictly increasing function, meaning tha there could only be one real root.\n\n2. Let u = 1+t and use substituion.
806346
using implicit differentiation:\nyou get 2y(dy/dx) - [y + x(dy/dx)] -3 = 0\nthen move 3 over and factor out (dy/dx):\n(2y-x)(dy/dx)=3 + y\ndy/dx=(3+y) / (2y-x)\nnow plug in your point (0,-1)\ndy/dx = (3+ -1)/(2(-1) - 0\ndy/dx = -1\nanswer = -1\n\n(dont forget to use your chain rule when you are implicitlly differentiating)\n\n#2\nsame thing make sure you sue multiplication rule\n[y^2 + 2xy(dx/dy)] - 2(dx/dy) + 12y^2(dx/dy) = 0\nthen move over the y^2 and factor out dx/dy\n(dx/dy)[2xy - 2 + 12y^2] = y^2\ndx/dy = y^2/ (2xy -2 +12y^2)\nPlug in 1 for y\ndx/dy = 1/(2x -2 + 12)\ndx/dy = 1/ (2x +10) or (2x + 10)^(-1)\n\nAnswer (leave in terms of x) = 1/ (2x +10) or (2x + 10)^(-1)\n\nI may have made mistakes so let me know
765615
differentiate implicitly:\n\n2x + 2yy' = 0\n\nand then some algebra\n\ny' = -(x/y)\n\nthen differentiate again:\n\ny'' = (-y - (-x)*y')/y^2 = (-y + xy')/y^2\n\nthen substitute in values:\n\ny' = -0/5 = 0\n\nand y'' = (-5 + 0*0)/(5^2) = =-5/25 = -1/5
49129
To find the direction of maximum ascent, find the directional derivative (DD), and find the direction where it has its max.\nNote: This is the equation of a plane, so the directional derivative is independent of the point where it is taken.\n\nIf v is a unit vector, then the DD in the direction of v, is the vector dot product of v with the vector  formed from the two partial derivatives\nPartial Z with respect to x is 1.5\nPartial Z with respedt to y is -.8\n\nDD = (1.5, -.8) dot v\n\nSince v is  unit vector, you can represent it\n\n(sin t, cos t)\n\ngiving us DD = (1.5)sint - (.8)cost \n\nTo find direction the max, take the derivative with respect to t, and set to 0.\n\n0 = (1.5)cost + (.8)sint\n-(1.5)cost = .8sint\n-1.5/.8 = tan t\n\nThe solutions to that say that v has the same direction as either of the (non-unit) vectors\n(1.5, -.8) or (-1.5, .08). The first is maximum ascent, the second maximum descent.
320591
The first derivative of this is\n\n10x^4 - 32x^3 + 21x^2 - 20x + 10\n\nSet that equal to zero and solve for the FOUR solutions. That will give you the x values of the four turning points that are the maxima and minima of the original equation. \n\nThe second derivative will serve two purposes. It will enable to you to find the thtree inflection points, and also enable you to tell which of the four turning points of the curve are maxima and which of them are minima.\n\nThe second derivative is \n\n40x^3 - 96x^2 + 42x - 20\n\nSet that equal to zero and solve. The three solutions give you the x-values of the three inflection points. To use the second derivative to find which turning points are maxima and which are minima, plug in the value of x for each turning point into the second derivative. If the value of the second derivative is positive, the turning point in a minimum. If it is negative, the turning point is a maximum.
858220
Depends on what variable you're differentiating with respect to:\ndV/dr : 2/3*pi*r*h\ndV/dh:  1/3*pi*r^2
85418
dx/dt+ax=b :P\n\n\nmathematica had some online find-the-derivitive program, last I remember
766935
The easy way is to note that both numbers have to be equal so 56/2=28\n\n\nthe hard calculus way is to maximize xy given that y=56-x\nx+y=56 implies\nxy=x(56-x)=56x-x^2= to find maximum of 56x-x^2\ndifferentiating with respect to x gives 56-2x. The critical point is when  56-2x=0 so x=28.y=28
749284
You need to take partial derivatives of your cost function (w.r.t. x and w.r.t. y) and solve the resulting equations to get critical points. Then, generally, you need to check whether your function is maximized or minimized at the critical points..\n\nIf partial derivatives are not covered in the class, try to argue that x = y (a square achieves the maximum area of all rectangles of a given perimeter).  With this, you have only one variable in the cost equation; use ordinary differentiation...
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.
818186
.  The y intercept is simply where the curve intersects the y axis, which means x=0.  Just plug in zero for x and find what the function is equal to.\n.  The local maxima and minima occur where the first derivative (the slope of the curve) is equal to zero.\n.  Points of inflection are where the rate of change in slope (the second derivative) goes from positive to negative or vice versa, so take the second derivative.\n.  The value of the function is increasing with x where the slope, (the first derivative) is positive.  The function is decreasing where the slope and first derivative are negative.\n.  The curve is concave up when the slope is increasing with x, and concave down where the slope is decreasing; so define the intervals over which the second derivative is positive or negative, and you will have define the concave up and concave down regions.
1081882
To do optimization problems, set up an equation in terms of one variable, take its derivative, and solve for 0. In this case, write an equation describing the area of the garden. Take the derivative and set it equal to 0, then solve.\n\nI'd be more help, but I don't quite understand the question...
434625
You are trying to find d/dt (0.51) ...\n\n0.51 -> alpha [ which inturn points "CONSTANT" ]\n\nDerivative of a constant is zero\n\nso d/dt (alpha) = 0
1177010
If the you have a local max/min at an interior point, , the partial derivatives must be zero.\nSince the sum of the second partials is zero, the point must be a saddle point, not a max/min.