Consider the function  whose second derivative is  . If  and  , what is  ?
830127
f''(x)=8x+4sin(x)\nf'(x)=4x^2-4cos(x)+c\nf'(0)=-4+c=4\nc=8\nf'(x)=4x^2-4cos(x)+8\nf(x)=(4/3)x^3-4sin(x)+8x+c1\nf(0)=c1=2\nf(x)=(4/3)x^3-4sin(x)+8x+2\n\nputting 5 into sin(x) is weird
948927
well, to differntiate a funtion using first principle, you have to follow this procedure. \nlet f(x) be a function. then f'(x) is the derivative of f(x) with respect to x. from the first principle, it is given by\nf'(x)=  Lt        f(x+h)-f(x)\n        (h->0)    -------------\n                          h\nwhere Lt represents the limit as h tends to zero.\nnow, take f(x)=cos(x^1/2). so, f'(x) is\n         Lt        cos((x+h)^1/2) - cos(x^1/2)\n        (h->0)    -------------------------------\n                                    h\n    here we apply L'Hopital's rule, because substituting h=0 will yield an indeterminate form. so, the result is \n      Lt        -sin((x+h)^1/2)\n     (h->0)   ------------------\n                   2*((x+h)^1/2)\nsubstitute h=0, we get the result as -1/2*(sin(x^1/2)/(x^1/2))
773933
1-In general the derivative of ln|U| is U'/U. In your case U = 2x^x + 4x + 1. Using the fact that x^x = exp(xlnx) you can easily check that the derivative of \n\nU' = 2(1 + lnx)x^x + 4.\n\nThus the derivative of your function is\n\n[2(1+lnx)x^x+4]/[2x^x+4x+1].\n\n2- The derivative of U^a, where a is a real number, is aU^{a-1}U'. In your case U = x e^x and a=1/2. Since U' =(x+1)e^x, one has\n\n(sqrt(xe^x))' =1/2{(x+1)/sqrt(xe^x)}e^x = 1/2{(x+1)/sqrt(x)}exp(x/2).
706213
Wait a minute, did you want to know when the function is zero or when its derivative is zero?  \n\nFor derivative:\ny' = 2x + sin x, right?\n\nIt looks like there's one place where it's 0.  Requires and numerical solution.\n\nWell, it's 0 at x=0, right?\nI assume that x is in radians.  1 radian is about 57 deg.  As x decreases from 0 both parts become more negative.\n\nIt looks like x = 0 is the only zero of the dirivative.
1108892
Clairaut's theorem states that if the second partial derivatives of a real function exist and are continuous at a given point, then the order of differentiation in a second order mixed partial derivative does not matter at that point (i.e., d^2(z)/dxdy = d^2(z)/dydx at that point.)\n\nOf course, if the second derivatives of the function exist and are continuous over the entire domain of the function, then the order of differentiation does not matter at any point.\n\nSee source.
838016
This is basically just asking what is f(x)/g(x)\n\nso your answer would be \n\n(x+2) / (3x^2 - 5)
664811
Well,... yes. Your question is flawed though; if the derivative is continuous at a point (say, at x=a), then the derivative must exist at x=a (by definition of continuity). But then, of course, the function is differentiable at x=a since this simply means that the derivative exists at x=a
518009
d/dx: I say "d.d.x.", or the derivative with respect to x\nd^2/dx^2: The second derivative with respect to x\nd^4f/dx^4: the fourth derivative of the function with respect to x\n\nf': F prime, or the derivative of the function\nf": F double prime, or the 2nd derivative of the function\nf''': F triple prime, or the 3rd derivative of the function\nn!: N factorial
700913
a function is written in  the form f(x) or g(X)  or anything relating to that form\nan equation has and equal sign\n\ntherefore this one is a linear equation
741312
If f(t)=(t-a)*(t-b) = t^2-(a+b)*t+ab; then \nI=(1/3)*t^3 - (a/2+b/2)*t^2 + a*b*t+C
493069
First :\nSince f is continuous over its domain (R),\ndf/dx = 5 x^4 +2 > 0 for every x\nSo f is an increasing function over its domain\nHence for every value of y their exist a unique value of x\nHence f has an inverse and one-one function\n3- Use function graph to graph f-1 as you mentioned\n(df^-1/dx) (2)= the slope of the tangent to f-1 at x=2\n6- (1,2)\n7- 7\n8- 1/7
432891
If g(x) is differentiable, then it certainly is continuous, and hence integrable.  You can define H(x) as the following definite integral:\n\nH(x) = 1 + Integral from 0 to x of { g(t) dt }
832910
yes the integral of the derivative of f(x)=x^2 is x^2.\n\nand just to check:\nf(x)=x^2\nf'(x)=2x\nF(x)=2*(1/2)x^(1+1)=x^2\n\nwhether you graph it or not makes no difference.
117843
Sure.  The derivative of x^3 is 3x^2.  That should get you started.
1082751
1)   f(x) = x+1/x =(x^2+1)/x  \n      g(x) = x-2/x =(x^2-2)/x\n(f+g)(x) = (x^2+x^2+1-2)/x = (2*x^2-1)/x = 2*x -1/x\n\n2)  f(x) = -2/x\n     g(x) = (x+1) / (x-3)\n(f+g)(x) = -2/x + (x+1)/(x-3) = {-2*(x-3)+x*(x+1)}/x*(x-3)\n              = (-2x+6+x^2-x)/(x^2-3x) =(x^2-x+6)/x^2-3x
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~ ♥ ~
945648
Take the derivative of f :\nf'(x) = 5x^4 (x+7)^4 + 4 (x+7)^3 x^5\nf'(x)= x^4 (x+7)^3 [ 5 (x+7) + 4x]\nf'(x)= x^4 (x+7)^3 [ 9x + 35]\n\nSo the derivative is equal to 0 for the values :\nx= -7\nx= - 35/ 9\nx = 0\n\nthe function is increasing in [-11;-7],\nthe function is increasing in [-35/9;+12],