How do you use the first and second derivitive to determine max and mins, inflection points, concavity, etc?
819742
The first derivative is the slope of the line that is tangent to the graph.  If there is a maximum or minimum, the tangent line at that point will be horizontal, so its slope is 0.  \n\nThe second derivative is 0 when there is an inflection point. An inflection point is a place on the graph where it changes the direction of the concavity. \n\nIn the two cases above, the converse is not necessarily true.  That is, just because the first derivative is 0 doesn't mean there has to be a max/min there.\n\n If the second derivative is positive, the graph is concave upward which means like a regular "U".  If it's negative, it's concave downward like an upside-down "U".
813180
three variables are related to time ...( X,Y,Z,) related to time\n\nthe total derative can be expressed by the sum of the partials... that is dx/dt + dy/dt + dz/dt = total derative....\n\nthe partial with respect to x (dx/dt)  plus\nthe partial with respect to y (dy/dt)  plus\nthe partial with respect to z (dz/dt)  equals\n\nthe total derative\n\nthe partial gives us an idea of how the function acts on that varaible
898860
First one: d= - 1\n\nEDIT: the rest I'm too lazy to do. Do your own homework! =D
963751
find the first derivative for the curve, and then substitute the x coordinate for the slope at that point.\n\nfor example, for the graph y = x^2, the first derivative is 2x. to find the slope at where x = 1, sub x = 1 into 2x and the slope at that point is 2.\n\nalternatively, draw a tangent to the point where u want to find the point slope and the slope of the tangent is the point slope. note that this method's accuracy depends on how accurately you draw your tangent
725690
The derivative is -2x + 4\nSetting that equal to zero gives you 2 as the critical number\nSubbing 2 into the original equation gives you a corresponding y coordinate of 4. So you know you have a relative extrema at (2,4)\n\nAll you need to figure out is whether it's a maximum or a minimum. Draw a line and label the number 2 in the middle. Pick a number to the left of two (say 1) and sub it into the equation of the derivative, it gives you a positive sign (-2(1)+4=2)\n\nDo the same for a number to the right of 2 (say 3). It gives you a negative number for an answer. (-2(3)+4=-2)\n\nAccording to the first derivative test, when f prime at x changes from positive to negative at c (our critical number which is 2), then f(c) or f(2) in our case is a relative maximum of f. That means at (2,4) we get our maximum and therefore the maximum is four and to answer your question, the function has a maximum.
616629
First of all, change this:\n\n<p>Enter first numerator: <input type=text name=num1 size=20></p>\n<p>Enter first numerator: <input type=text name=num2 size=20></p>\n<p>Enter second denomenator: <input type=text name=num1 size=20></p>\n<p>Enter second denomenator: <input type=text name=num2 size=20></p>\n\n\nto this:\n\n<p>Enter first numerator: <input type=text name=num1 size=20></p>\n<p>Enter first denomenator: <input type=text name=num2 size=20></p>\n<p>Enter second numerator: <input type=text name=num3 size=20></p>\n<p>Enter second denomenator: <input type=text name=num4 size=20></p>\n\nThen based on the operation selected, implement the proper logic there (its basic math) and then your program will work.\n\nGood luck!!
920579
use synthetic  divsion method \ni think this link would help\nhttp://www.purplemath.com/modules/synthdiv.htm
695595
Each question is a function of two variables... and it has two derivatives, relative to each variable.\n\nDerivative of F relative to Z1   is   uZ1^(u-1)*Z2^v\nDerivative of F relative to Z2   is   Z1^u*vZ2^(v-1)\n\nDerivative of MRTS relative to Z1 is  -uZ2/(vZ1^2)\n\nDerivative of MRTS relative to Z2 is  u/(vZ1)
786690
u can do it by\n(n/2)(a+l)\nwhere \nn is no. of terms\na is first term\nl is last term
249704
Take the double derivative and find the inflection points. Analyze the behavior between the inflection points. If it is positive it is concave up. If it is negative it is concave down.
821550
if f(x) = ∫[t = x to 0] 1/ (1 + t + t²) dt\n\nf'(x) = -1/ (1 + x + x²)\n\nf''(x) = (1 + 2x)/(1 + x + x²)²\n\nFor concave UP f''(x) > 0\n\nSo 1 + 2x > 0 (as (1 + x + x²)² > 0 for all x)\n\nie x < -1/2
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
226178
Just doing the two partial derivatives and setting them to zero is not enough because you cam have a change of inflection (if I'm not mistaken). With one variable only, that would be equivalent to taking the derivative and equaling it to zero for f(x)= x^3. f'(x) = 3x^2 and f'(0)=0. However, x=0 is not a local extreme, it is only a candidate. This is where analisys takes place. One way to verify if this candidate is infact an extreme is to calculate the second derivative. If it is >0, it is a local minimum, if it is <0 it is a local maximum. In this case f''(0) = 0, so it is simply a point of inflection (check this terminology cause I've graduated a long time ago and am quoting from memory).\nFor a two variable function, you have to repeat this for y. So a candidate will be a point (xi, yi) in which both partial derivatives are zero and both second derivatives have the *same* sign (otherwise you have a point of saddle: I dont know if this is the correct terminology in English) and both different from zero.
736249
Curves of higher order can have more points of inflection.\n\nOrdinarily, a quadratic curve has 0 points of inflection.\nA cubic has one point of inflection.\nAn equation of the fourth order has two points of inflection, \nand so on.
876214
g(x)=-x^3+4x-2\ng'(x)=3x^2+4\ng"(x)=6x\ng"(x)=0=6x\nx=0 is the inflection point
222592
1. p'q\n\n2. p'\n\n3. pq' + p'\n\nI think i got the correct ones !
710887
SINCE THIS IS A PEROIDIC FUNC. with peroid = 2pie\n                                                         THEREFORE global maxima lies in peroid (0,2pi)\nf'(x)=2sinxcos2x + sin2xcosx\n      =2sinx(1-2cos^2x) + 2sinxcos^2(x)\n      =4sinx - 6sin^3x\n for f(x) to attain maxima f'(x)=0\n   therefore f'(x)=4sinx- 6sin^3x=0\n    it comes out sinx=0,+-(2/3)^1/2\n    at x=0 it is minima whereas at other pts it is maxima\n    hence maxima is at x=sin^-1(+-k)     k=(2/3)^(1/2)\n    sin2x=(1-cos^2(2x))\n             =(1-(1-2sin^2x)^2)= 2*k\n    therefore global maxima is =2*k*k=8/9