what's the derivative of y=(1+cos^2(7x))?
829160
y'=2 cos(7x)*(-sin 7x)*7 By the Chain Rule
1068232
Use quotient rule to get the derivative, which, simplified, is:\n\n-30*(2x+3)/((3x^2+9x+7)^2)\n\nSet it equal to zero and solve.  Since the denominator cannot be zero, set the numerator equal to zero\n\n2x+3=0\nx=(-3/2)\n\nI checked it, and it indeed is the solution. \n\nBy the way, you don't have to worry about an undefined points of this function making problems in the graph or in the max/min values in the derivative.  The denominator has imaginary roots.
830837
If ∫f(x)dx = y(x), then f(x) = d y(x) / dx;  So to get the answer, take the derivative of the expression x^2 - 2x + 1.  The value of the integral at limit 1 is  a constant and will not appear in the derivative.
934742
uh, find the derivative mathematically and then graph it.\n\nfor example to graph derivative of x^2.\n\nfind the derivative equation f(x)=2x\nand then graph y=2x.\n\nooops, didnt realize it was a picture.\n\nfind the zeroes (a high or low point i.e. tangent is a horizontal line)\nyou should be able to track poitive and negative slopes.  not too hard, you'll figure it out.
1034892
What is the function? I assume it is \n\n(dy/dx is equal to 1 only if y=x+C whre C is a constant)\n\nTo prove use the definition of the derivative which is dy/dx as a limit of delta_x appraching of (delta_y/delta_x)\nwe have \ny2-y1=x2+C-x1-C then\ndelta_y=x2-X1\n\ndelta_y/delta_x=(x2-x1)/(x2-x1)=1 QED\n(the is no need to take a limit, however in must be used. I hav eommited it for lack of better representation)\n\nBest
758976
The total differential is [∂F/∂x]dx +[ ∂F/∂y]dy\n\n∂F/∂x = 4x^3*y^-1\n∂F/∂y = x^4*(-y^-2) = -x^4y^-2
586962
(6y^2-3y)(y+7)-->(2y)(y-1)(y+7)\nif that helps
352749
By the chain rule:\n[e^(3x^(2)-7x)]*(6x-7)
477866
Let y=cos2x, then, \ndy/dx = limit (as h tends to zero) [(cos2(x+h) - cos2x)/h]\n      = lim. [(cos2x cos2h  - sin2x sin2h - cos2x)/h]\n      = lim. [(cos2x(cos2h-1) -sin2x sin2h)/h]\n      = lim. [(cos2x (-2(sin2h)^2)/h] - sin2x* lim.(sin2h/h)\n      = -2 cos2x * (lim. sin2h/h)*(lim. (sin2h))- 2 sin2x\n      = -2 cos2x * 2 * 0 - 2 sin2x\n      = -2 sin2x
921398
I just taught this section today to my calc students.  You start by finding the first and second derivative.  I recommend looking in your book for an example.  Your book probably has something like this towards the front of the section.  It would be too long winded to write everything down here.  If you don't have your book on you, the web page below has problems that you can do and then check your answer.\n\nhttp://www.math.ucdavis.edu/~kouba/CalcOneDIRECTORY/graphingdirectory/Graphing.html
180190
let f(x) = (x^2) sin(x+2)\nthe derivitive of f(x) [ f'(x) = 2x sin(x+2) + (x^2) cos(x+2)]\n\nThis is done by taking the product rule {g(x)h(x)}' = g'(x)h(x) + g(x)h'(x)( the derivitive of the first times the second plus the derivitive of the second times the first) \nwhere g(x) = x^2 and h(x) = sin(x+2)
79944
The above is the derivative.\nIntegration gives x + (1/4)x^4 + c (generic constant).
806655
dx=2xlny-2z\ndy=x^2*(1/y)-2z\ndz= x^2lny-2
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
1138866
I'll lead you in the most important part, but it is up to you to complete it.\n\nDerivative of x . y = x'.y + x.y'\n\nThen, you will have:\n\nf'(x) = (2x . ln x^2) + x^2 (ln x^2)'\n\nContinue from here, the only problem that you may have is finding the derivative of ln y , which you can see in the link below.
820064
x^2+2xy-y^2=7\n\ndy/dx\n2x+2y+2x(dy/dx)-2(dy/dx)=0\n(2x-2)(dy/dx)=-2x-2y\n\ndy/dx= (-2x-2y)/(2x-2)\ndy/dx=-3\n\nd^2y/dx^2\n[(2x-2)(-2-2(dy/dx))-\n(-2x-2y)(2)]/(2x-2)^2\n\nd^2y/dx^2=5\n\ni think thats rite sry bout all the parantheses
143223
a) (6x-7y)(6x+7y)\nb) y(4xy-7)\nc) (x-8)(x+3)