I neeed your help terrriblyyyyy (calculus) -- Please HELP!?
919338
Well, I see you've used the correct order of operations this time. That's good - it shows that you're improving.\n\n#1: Apply the product and chain rules to find f'(x), then evaluate at 0:\n\nf'(x) = d(√(x³+5x+121))/dx (x²+x+11) + (2x+1)√(x³+5x+121)\n\nf'(x) = 1/(2√(x³+5x+121)) * d(x³+5x+121)/dx * (x²+x+11) + (2x+1)√(x³+5x+121)\n\nf'(x) = 1/(2√(x³+5x+121)) * (3x²+5) * (x²+x+11) + (2x+1)√(x³+5x+121)\n\nNow evaluating :\n\nf'(0) = 1/(2√121)*5*11 + √121 = 5/2 + 11 = 27/2\n\n#2: First, evaluate the function at y=1 to get the value of x:\n\nxy^2 -2y + 4y^3 = 6\nx - 1 + 4 = 6\nx = 3\n\nNow, implicitly differentiate both sides:\n\nd(xy^2 -2y + 4y^3)/dx = d(6)/dx\ny² + 2xy dy/dx - 2 dy/dx + 12y² dy/dx = 0\n\nSolve for dy/dx:\n\ny² + (2xy - 2 + 12y²) dy/dx = 0\n(2xy - 2 + 12y²) dy/dx = -y²\ndy/dx = -y²/(2xy - 2 + 12y²)\n\nNow evaluate at (3, 1):\n\ndy/dx = -1/(6 - 2 + 12) = -1/16\n\n#3: First, find the derivative:\n\nf'(x) = (x²+x-2 - (2x+1)x)/(x²+x-2)² (using the quotient rule)\n\nSimplifying:\n\nf'(x) = (x²+x-2 - 2x² - x)/(x²+x-2)²\nf'(x) = (-x²-2)/(x²+x-2)²\n\nNow we must find where f'(x) < 0\n\n(-x²-2)/(x²+x-2)² < 0\n-x²-2 < 0 (we know (x²+x-2)² is nonnegative, and if the expression on the left is defined, it is also nonzero. Thus it is positive and we can multiply by it without changing the inequality)\n-x² < 2\n\nNow, since x is a real number, this expression will always be true. Thus, if f'(x) is even defined, x is decreasing. Now f'(x) is defined provied that (x²+x-2)²≠0. So:\n\n(x²+x-2)²≠0\n(x-1)²(x+2)²≠0\nx≠1 and x≠-2\n\nWhich makes sense, since those are vertical asymptotes of the original function. Thus, the function is decreasing on:\n\n(-∞, -2) ∪ (-2, 1) ∪ (1, ∞)
846527
You'll have to do this in more than one step.  Basically it will sort of have to be a program.  I would start with the restart; command and then  define the function (I would look this up in the help menu since it is dependent on 2 variables).  Then use a command to find the derivative.  Then a command to solve for the pairs (x,y) when the derivative is 0.  That will give you 1.  For 2 and 3 you will also have to do a series of commands.  Use your calc book to refresh your memory on the procedure of finding these, it should be a similar procedure in maple.
460400
You would need to know how to integrate (that is, find anti-derivatives). The proceedure forthis type of problem can usually be understood after first semester calculus. The specific integrations may need some second semester calculus.
931778
Since you have conflicting answers, I'm just chiming in to say go with ManiacalAce -- he got the correct answer.\n\nCorrect second derivative is as follows:\n\n(-9/4)x^(-3/2) - (3/4)x^(-1/2)\n\n~ ♥ ~
221164
to get an extra credit, maybe?\n\n---\n\nseriously, I cannot figure out how you can study calculus without knowing limits: even if you only deal with numerical procedures, every iterative formula, or better, EVERY FORMULA (such as those used to find the roots of a non-linear equation, approximated integrals, derivatives and so on) needs a preventive study for convergence conditions: this involves limits!\n\nOne more step back: DERIVATIVE IS A LIMIT ! How can you apply it without knowing what it is?
118732
good luck man... http://mathforum.org/library/drmath/sets/high_calculus.html
502732
Let me guess you're taking the AP Calc test tomorrow!  Me too!  :)\n\nWell first take the seperate derivatives of x and y\nto get dx/dt and dy/dt\n\nthen put you dy answer over the dx answer to have dy/dx\n\nthen take the derivative of dy/dx and put that over your dx/dt\n\n\nexample problem\nx=5t^2                 y=4t^3\n\nsolution:\n\ndx/dt= 10t       dy/dt= 12t^2\n\ndy/dx (first derivative) = (12t^2)/(10t)\n\nwhich equals (6/5)t\n\nthen take d(dy/dx)/dt (derivative of (dy/dx))\n\nso that equals (6/5)\n\nthen put that over dx/dt which is 10t\n\nso (6/5)/(10t) which equals 6/(5*10*t)\n\nso answer equals 6/(50t)\n\n\nHope it helps and if you are taking the AP Test good luck!!!
787465
Write out the exponentials and use the series for e^x. Other than that, do your own homework.
87883
First write y as a product instead of a quotient:\n\n(2+x)*(2-3x)^(-1)\n\nNow you should be able to do this. \nUse 2+x as your f and (2-3x)^(-1) as your g. \nYou will have to use the chain rule to find g'.
428665
This one is fairly clear.  Hope it helps.\n\nhttp://www.math.hmc.edu/calculus/tutorials/substitution/
175327
Try this, maybe?\n\nhttp://mathforum.org/library/drmath/drmath.elem.html
375277
Mathworld.com.  Or, if you type the subject your interested in followed by "Forum" in google, you'll find some good help that way.  Fore xample, "calculus forum".  Also, you can ask me whatever you want.
713982
use logarthmic differentiation\n\ny = x^(1/x)\n\nln(y) = (1/x)* ln(x)\n\ndiff w.r.t x\n\n(1/y) (dy/dx) = (1/x)*(1/x) +ln(x)*(-1/x^2)\n\n(dy/dx) = { y/(x^2) }  [ 1 - ln(x) ]\n\nsubstitute for y
49770
There is no closed form for the nth derivative of this function.  I symbolically calculated and factored the first 50 derivatives on MathCAD, and there was not a pattern.  Sorry, but you're out of luck....
502657
Try this link and read it all the way through.  It's short.\n\nhttp://mathcentral.uregina.ca/qq/database/QQ.09.00/tom3.html\n\nThis link is better:\n\nhttp://mathforum.org/library/drmath/view/56760.html\n\nJust scroll down.
927535
#1 \n\nIf f(x)= √(x³ + 5x +121) * (x² +x +11) \n\nthen, f(0)= √(121) * (11) \n= 11 * 11\n= 121\n\nf'(0) = -1/2 (See #3 below at ***)\n\n#2 \nIf y is a differentiable function of x, then the slope of the curve xy^2 -2y + 4y^3 = 6 at the point where y = 1 is what?\n\nWhen y = 1\n\nx(1)^2 - 2(1) + 4(1)^3 = 6 \nie x + 2 = 6\nSo x = 4\nSo y^2 + 2xydy/dx -2dy/dx + 12y^2dy/dx = 0\nSo 1 + 12 dy/dx - 2dy/dx + 12 dy/dx = 0\nThus 22dy/dx = -1\ndy/dx = -1/22\nie slope of tangent at (1, 6) is -1/22\n\n#3\nFind all open intervals on which the following function is decreasing:\nf(x) = x / (x^2 + x -2)\n= x/[(x + 2)(x - 1)]\n= 2/[3(x + 2)] + 1/[3(x - 1)] (by partial fractions)\nDiscontinuities at x = -2 and x = 1\n\nf'(x) = -2/[3(x + 2)²] - 1/[3(x - 1)²]\n\nSo f'(0) = -2/[3(2)²] - 1/[3(-1)²]\n= -2/12 - 1/3\n= -1/2 ***\n\nClearly f'(x) < 0 for all x except x = -2 and x = 1 where it is undefined \n\nie the function is decreasing fo x< -2, -2 < x < 1 and x > 1
1048691
(r*x)*(-sin(x))+(cos(x))*(r)+\n(0)*(lnrsin(x))+(r)*(cot(x)) \n\nfor the cot(x) you end up for the derivative of lnrsin(x) is (1/sin(x)) *cos(x)=cot(x)\n\nsorry it cut off on the top. hopefully this works